Solve the following system of simultaneous linear equations by the substitution method:
x + y = 14
x - y = 4
Answer
Given,
x + y = 14 ......(i)
x - y = 4 .........(ii)
From eqn. (ii) we get,
x = 4 + y.
Substituting above value of x in eqn. (i) we get,
⟹ (4 + y) + y = 14
∴ x = y + 4 = 5 + 4 = 9.
Hence, x = 9 and y = 5.
Solve the following system of simultaneous linear equations by the substitution method:
s - t = 3
s 3 + t 2 = 6 \dfrac{s}{3} + \dfrac{t}{2} = 6 3 s + 2 t = 6
Answer
Given,
s - t = 3 ........(i)
s 3 + t 2 = 6 \dfrac{s}{3} + \dfrac{t}{2} = 6 3 s + 2 t = 6
From eqn. (i) we get,
s = 3 + t.
Substituting above value of s in eqn. (ii) we get,
⇒ s 3 + t 2 = 6 ⇒ 3 + t 3 + t 2 = 6 ⇒ 2 ( 3 + t ) + 3 t 6 = 6 ⇒ 6 + 2 t + 3 t = 36 ⇒ 5 t = 36 − 6 ⇒ 5 t = 30 ⇒ t = 6. \Rightarrow \dfrac{s}{3} + \dfrac{t}{2} = 6 \\[1em] \Rightarrow \dfrac{3 + t}{3} + \dfrac{t}{2} = 6 \\[1em] \Rightarrow \dfrac{2(3 + t) + 3t}{6} = 6 \\[1em] \Rightarrow 6 + 2t + 3t = 36 \\[1em] \Rightarrow 5t = 36 - 6 \\[1em] \Rightarrow 5t = 30 \\[1em] \Rightarrow t = 6. ⇒ 3 s + 2 t = 6 ⇒ 3 3 + t + 2 t = 6 ⇒ 6 2 ( 3 + t ) + 3 t = 6 ⇒ 6 + 2 t + 3 t = 36 ⇒ 5 t = 36 − 6 ⇒ 5 t = 30 ⇒ t = 6.
∴ s = 3 + t = 3 + 6 = 9.
Hence, t = 6 and s = 9.
Solve the following system of simultaneous linear equations by the substitution method:
2x + 3y = 9
3x + 4y = 5
Answer
Given,
2x + 3y = 9 ......(i)
3x + 4y = 5 .......(ii)
Solving eqn. (i) we get,
⟹ 2x + 3y = 9
⟹ 2x = 9 - 3y
⟹ x = 9 − 3 y 2 \dfrac{9 - 3y}{2} 2 9 − 3 y
Substituting above value of x in eqn. (ii) we get,
⇒ 3 ( 9 − 3 y 2 ) + 4 y = 5 ⇒ 27 − 9 y 2 + 4 y = 5 ⇒ 27 − 9 y + 8 y 2 = 5 ⇒ 27 − y = 10 ⇒ y = 27 − 10 = 17. \Rightarrow 3\Big(\dfrac{9 - 3y}{2}\Big) + 4y = 5 \\[1em] \Rightarrow \dfrac{27 - 9y}{2} + 4y = 5 \\[1em] \Rightarrow \dfrac{27 - 9y + 8y}{2} = 5 \\[1em] \Rightarrow 27 - y = 10 \\[1em] \Rightarrow y = 27 - 10 = 17. ⇒ 3 ( 2 9 − 3 y ) + 4 y = 5 ⇒ 2 27 − 9 y + 4 y = 5 ⇒ 2 27 − 9 y + 8 y = 5 ⇒ 27 − y = 10 ⇒ y = 27 − 10 = 17.
Solving for x,
x = 9 − 3 y 2 = 9 − 3 ( 17 ) 2 = 9 − 51 2 = − 42 2 = − 21. x = \dfrac{9 - 3y}{2} = \dfrac{9 - 3(17)}{2} \\[1em] = \dfrac{9 - 51}{2} \\[1em] = \dfrac{-42}{2} \\[1em] = -21. x = 2 9 − 3 y = 2 9 − 3 ( 17 ) = 2 9 − 51 = 2 − 42 = − 21.
Hence, x = -21 and y = 17.
Solve the following system of simultaneous linear equations by the substitution method:
3x - 5y = 4
9x - 2y = 7
Answer
Given,
3x - 5y = 4 .......(i)
9x - 2y = 7 .......(ii)
Solving eqn. (i) we get,
⟹ 3x - 5y = 4
⟹ 3x = 4 + 5y
⟹ x = 4 + 5 y 3 \dfrac{4 + 5y}{3} 3 4 + 5 y
Substituting above value of x in eqn. (ii) we get,
⇒ 9 x − 2 y = 7 ⇒ 9 ( 4 + 5 y 3 ) − 2 y = 7 ⇒ 3 ( 4 + 5 y ) − 2 y = 7 ⇒ 12 + 15 y − 2 y = 7 ⇒ 12 + 13 y = 7 ⇒ 13 y = 7 − 12 ⇒ 13 y = − 5 ⇒ y = − 5 13 . \Rightarrow 9x - 2y = 7 \\[1em] \Rightarrow 9\Big(\dfrac{4 + 5y}{3}\Big) - 2y = 7 \\[1em] \Rightarrow 3(4 + 5y) - 2y = 7 \\[1em] \Rightarrow 12 + 15y - 2y = 7 \\[1em] \Rightarrow 12 + 13y = 7 \\[1em] \Rightarrow 13y = 7 - 12 \\[1em] \Rightarrow 13y = -5 \\[1em] \Rightarrow y = -\dfrac{5}{13}. ⇒ 9 x − 2 y = 7 ⇒ 9 ( 3 4 + 5 y ) − 2 y = 7 ⇒ 3 ( 4 + 5 y ) − 2 y = 7 ⇒ 12 + 15 y − 2 y = 7 ⇒ 12 + 13 y = 7 ⇒ 13 y = 7 − 12 ⇒ 13 y = − 5 ⇒ y = − 13 5 .
Solving for x by substituting value of y,
⇒ x = 4 + 5 y 3 = 4 + 5 × − 5 13 3 = 4 − 25 13 3 = 52 − 25 13 3 = 27 39 = 9 13 . \Rightarrow x = \dfrac{4 + 5y}{3} \\[1em] = \dfrac{4 + 5 \times \dfrac{-5}{13}}{3} \\[1em] = \dfrac{4 - \dfrac{25}{13}}{3} \\[1em] = \dfrac{\dfrac{52 - 25}{13}}{3} \\[1em] = \dfrac{27}{39} \\[1em] = \dfrac{9}{13}. ⇒ x = 3 4 + 5 y = 3 4 + 5 × 13 − 5 = 3 4 − 13 25 = 3 13 52 − 25 = 39 27 = 13 9 .
Hence, x = 9 13 and y = − 5 13 . \dfrac{9}{13}\text{ and y } = -\dfrac{5}{13}. 13 9 and y = − 13 5 .
Solve the following system of simultaneous linear equations by substitution method:
3x - 5y = -2
7x - 3y = -9
Answer
Given,
3x - 5y = -2 ......................(1)
7x - 3y = -9 ......................(2)
Solving equation (1), we get :
⇒ 3x - 5y = -2
⇒ 3x = -2 + 5y
⇒ x = − 2 + 5 y 3 \dfrac{-2 + 5y}{3} 3 − 2 + 5 y
Substituting above value of x in equation (2), we get :
⇒ 7 ( − 2 + 5 y 3 ) − 3 y = − 9 ⇒ 7 ( − 2 + 5 y ) − 3 y × 3 3 = − 9 ⇒ 7 ( − 2 + 5 y ) − 9 y 3 = − 9 ⇒ 7 ( − 2 + 5 y ) − 9 y = − 27 ⇒ − 14 + 35 y − 9 y = − 27 ⇒ − 14 + 26 y = − 27 ⇒ 26 y = − 27 + 14 ⇒ 26 y = − 13 ⇒ y = − 13 26 ⇒ y = − 1 2 . \Rightarrow 7\Big(\dfrac{-2 + 5y}{3}\Big) - 3y = -9\\[1em] \Rightarrow \dfrac{7(-2 + 5y) - 3y \times 3}{3} = -9\\[1em] \Rightarrow \dfrac{7(-2 + 5y) - 9y}{3} = -9\\[1em] \Rightarrow 7({-2 + 5y}) - 9y = -27\\[1em] \Rightarrow -14 + 35y - 9y = -27\\[1em] \Rightarrow -14 + 26y = -27\\[1em] \Rightarrow 26y = -27 + 14\\[1em] \Rightarrow 26y = -13\\[1em] \Rightarrow y = -\dfrac{13}{26}\\[1em] \Rightarrow y = -\dfrac{1}{2}. ⇒ 7 ( 3 − 2 + 5 y ) − 3 y = − 9 ⇒ 3 7 ( − 2 + 5 y ) − 3 y × 3 = − 9 ⇒ 3 7 ( − 2 + 5 y ) − 9 y = − 9 ⇒ 7 ( − 2 + 5 y ) − 9 y = − 27 ⇒ − 14 + 35 y − 9 y = − 27 ⇒ − 14 + 26 y = − 27 ⇒ 26 y = − 27 + 14 ⇒ 26 y = − 13 ⇒ y = − 26 13 ⇒ y = − 2 1 .
Substituting value of y in x = − 2 + 5 y 3 \dfrac{-2 + 5y}{3} 3 − 2 + 5 y
⇒ x = − 2 + 5 × − 1 2 3 ⇒ x = − 2 − 5 2 3 ⇒ x = − 4 − 5 2 3 ⇒ x = − 4 − 5 6 ⇒ x = − 9 6 ⇒ x = − 3 2 . \Rightarrow x = \dfrac{-2 + 5 \times \dfrac{-1}{2}}{3}\\[1em] \Rightarrow x = \dfrac{-2 - \dfrac{5}{2}}{3} \\[1em] \Rightarrow x = \dfrac{\dfrac{-4 - 5}{2}}{3} \\[1em] \Rightarrow x = \dfrac{-4 - 5}{6}\\[1em] \Rightarrow x = \dfrac{-9}{6}\\[1em] \Rightarrow x = -\dfrac{3}{2}. ⇒ x = 3 − 2 + 5 × 2 − 1 ⇒ x = 3 − 2 − 2 5 ⇒ x = 3 2 − 4 − 5 ⇒ x = 6 − 4 − 5 ⇒ x = 6 − 9 ⇒ x = − 2 3 .
Hence, x = − 3 2 -\dfrac{3}{2} − 2 3 − 1 2 -\dfrac{1}{2} − 2 1
Solve the following system of simultaneous linear equations by the substitution method:
5x + 4y - 4 = 0
x - 20 = 12y
Answer
Given,
5x + 4y - 4 = 0 ......(i)
x - 20 = 12y ........(ii)
From eqn. (ii) we get,
x = 12y + 20 .......(iii)
Substituting value of x from eqn. (iii) in eqn. (i) we get,
⟹ 5(12y + 20) + 4y - 4 = 0
⟹ 60y + 100 + 4y - 4 = 0
⟹ 64y + 96 = 0
⟹ 64y = -96
⟹ y = − 96 64 = − 3 2 -\dfrac{96}{64} = -\dfrac{3}{2} − 64 96 = − 2 3
Substituting value of y in eqn. (iii) we get,
x = 12 × − 3 2 + 20 = 6 × − 3 + 20 = − 18 + 20 = 2. x = 12 \times \dfrac{-3}{2} + 20 \\[1em] = 6 \times -3 + 20 \\[1em] = -18 + 20 \\[1em] = 2. x = 12 × 2 − 3 + 20 = 6 × − 3 + 20 = − 18 + 20 = 2.
Hence, x = 2 and y = − 3 2 . -\dfrac{3}{2}. − 2 3 .
Solve the following system of simultaneous linear equations by the substitution method:
2 x − 3 y 4 = 3 2x - \dfrac{3y}{4} = 3 2 x − 4 3 y = 3
5x - 2y - 7 = 0
Answer
Given,
2 x − 3 y 4 = 3 2x - \dfrac{3y}{4} = 3 2 x − 4 3 y = 3
5x - 2y - 7 = 0 .......(ii)
Multiplying eqn. (i) by 4 we get,
⇒ 4 ( 2 x − 3 y 4 ) = 3 × 4 ⇒ 8 x − 3 y = 12 ⇒ 8 x = 12 + 3 y ⇒ x = 12 + 3 y 8 . . . . . . ( i i i ) \Rightarrow 4\Big(2x - \dfrac{3y}{4}\Big) = 3 \times 4 \\[1em] \Rightarrow 8x - 3y = 12 \\[1em] \Rightarrow 8x = 12 + 3y \\[1em] \Rightarrow x = \dfrac{12 + 3y}{8} ......(iii) ⇒ 4 ( 2 x − 4 3 y ) = 3 × 4 ⇒ 8 x − 3 y = 12 ⇒ 8 x = 12 + 3 y ⇒ x = 8 12 + 3 y ...... ( iii )
Putting value of x from eqn. (iii) in eqn. (ii),
⇒ 5 ( 12 + 3 y 8 ) − 2 y − 7 = 0 ⇒ 60 + 15 y 8 − 2 y − 7 = 0 ⇒ 60 + 15 y − 16 y − 56 8 = 0 ⇒ 4 − y = 0 ⇒ y = 4. \Rightarrow 5\Big(\dfrac{12 + 3y}{8}\Big) - 2y - 7 = 0 \\[1em] \Rightarrow \dfrac{60 + 15y}{8} - 2y - 7 = 0 \\[1em] \Rightarrow \dfrac{60 + 15y - 16y - 56}{8} = 0 \\[1em] \Rightarrow 4 - y = 0 \\[1em] \Rightarrow y = 4. ⇒ 5 ( 8 12 + 3 y ) − 2 y − 7 = 0 ⇒ 8 60 + 15 y − 2 y − 7 = 0 ⇒ 8 60 + 15 y − 16 y − 56 = 0 ⇒ 4 − y = 0 ⇒ y = 4.
Substituting value of y in eqn. (iii) we get,
⇒ x = 12 + 3 y 8 = 12 + 3 ( 4 ) 8 = 12 + 12 8 = 24 8 = 3. \Rightarrow x = \dfrac{12 + 3y}{8} \\[1em] = \dfrac{12 + 3(4)}{8} \\[1em] = \dfrac{12 + 12}{8} \\[1em] = \dfrac{24}{8} \\[1em] = 3. ⇒ x = 8 12 + 3 y = 8 12 + 3 ( 4 ) = 8 12 + 12 = 8 24 = 3.
Hence, x = 3 and y = 4.
Solve the following system of simultaneous linear equations by the substitution method:
2x + 3y = 23
5x - 20 = 8y
Answer
Given,
2x + 3y = 23 ......(i)
5x - 20 = 8y ......(ii)
Solving (i) we get,
⟹ 2x + 3y = 23
⟹ 2x = 23 - 3y
⟹ x = 23 − 3 y 2 \dfrac{23 - 3y}{2} 2 23 − 3 y
Substituting value of x from eqn. (iii) in eqn. (ii) we get,
⇒ 5 ( 23 − 3 y 2 ) − 20 = 8 y ⇒ 115 − 15 y 2 − 20 = 8 y ⇒ 115 − 15 y − 40 2 = 8 y ⇒ 115 − 15 y − 40 = 16 y ⇒ 75 − 15 y = 16 y ⇒ 31 y = 75 ⇒ y = 75 31 = 2 13 31 . \Rightarrow 5\Big(\dfrac{23 - 3y}{2}\Big) - 20 = 8y \\[1em] \Rightarrow \dfrac{115 - 15y}{2} - 20 = 8y \\[1em] \Rightarrow \dfrac{115 - 15y - 40}{2} = 8y \\[1em] \Rightarrow 115 - 15y - 40 = 16y \\[1em] \Rightarrow 75 - 15y = 16y \\[1em] \Rightarrow 31y = 75 \\[1em] \Rightarrow y = \dfrac{75}{31} = 2\dfrac{13}{31}. ⇒ 5 ( 2 23 − 3 y ) − 20 = 8 y ⇒ 2 115 − 15 y − 20 = 8 y ⇒ 2 115 − 15 y − 40 = 8 y ⇒ 115 − 15 y − 40 = 16 y ⇒ 75 − 15 y = 16 y ⇒ 31 y = 75 ⇒ y = 31 75 = 2 31 13 .
Substituting value of y in eqn.(iii) we get,
⇒ x = 23 − 3 y 2 = 23 − 3 × 75 31 2 = 23 × 31 − 3 × 75 31 2 = 713 − 225 31 × 2 = 488 62 = 244 31 = 7 27 31 . \Rightarrow x = \dfrac{23 - 3y}{2} \\[1em] = \dfrac{23 - 3 \times \dfrac{75}{31}}{2} \\[1em] = \dfrac{\dfrac{23 \times 31 - 3 \times 75}{31}}{2} \\[1em] = \dfrac{713 - 225}{31 \times 2} \\[1em] = \dfrac{488}{62} \\[1em] = \dfrac{244}{31} = 7\dfrac{27}{31}. ⇒ x = 2 23 − 3 y = 2 23 − 3 × 31 75 = 2 31 23 × 31 − 3 × 75 = 31 × 2 713 − 225 = 62 488 = 31 244 = 7 31 27 .
Hence, x = 7 27 31 and y = 2 13 31 7\dfrac{27}{31}\text{ and y =} 2\dfrac{13}{31} 7 31 27 and y = 2 31 13
Solve the following system of simultaneous linear equations by the substitution method:
mx - ny = m2 + n2
x + y = 2m
Answer
Given,
mx - ny = m2 + n2 ......(i)
x + y = 2m .......(ii)
From eqn. (ii) we get,
x = 2m - y .......(iii)
Substituting value of x from eqn. (iii) in eqn. (i) we get,
⟹ m(2m - y) - ny = m2 + n2
⟹ 2m2 - my - ny = m2 + n2
⟹ 2m2 - m2 - n2 = my + ny
⟹ m2 - n2 = y(m + n)
⟹ y = m 2 − n 2 m + n = ( m − n ) ( m + n ) m + n \dfrac{\text{m}^2 - \text{n}^2}{\text{m} + \text{n}} = \dfrac{(\text{m} - \text{n})(\text{m} + \text{n})}{\text{m} + \text{n}} m + n m 2 − n 2 = m + n ( m − n ) ( m + n )
Substituting value of y in eqn. (iii) we get,
x = 2m - y = 2m - (m - n) = 2m - m + n = m + n.
Hence, x = m + n and y = m - n.
Solve the following system of simultaneous linear equations by the substitution method:
2 x a + y b = 2 \dfrac{2x}{a} + \dfrac{y}{b} = 2 a 2 x + b y = 2
x a − y b = 4 \dfrac{x}{a} - \dfrac{y}{b} = 4 a x − b y = 4
Answer
Given,
2 x a + y b = 2 \dfrac{2x}{a} + \dfrac{y}{b} = 2 a 2 x + b y = 2
x a − y b = 4 \dfrac{x}{a} - \dfrac{y}{b} = 4 a x − b y = 4
Multiplying both side of eqn (i) by ab we get,
⇒ a b ( 2 x a + y b ) = 2 a b ⇒ 2 b x + a y = 2 a b . . . . . . ( i i i ) \Rightarrow ab\Big(\dfrac{2x}{a} + \dfrac{y}{b}\Big) = 2ab \\[1em] \Rightarrow 2bx + ay = 2ab ......(iii) ⇒ ab ( a 2 x + b y ) = 2 ab ⇒ 2 b x + a y = 2 ab ...... ( iii )
Multiplying both side of eqn (ii) by ab we get,
⇒ a b ( x a − y b ) = 4 a b ⇒ b x − a y = 4 a b ⇒ b x = 4 a b + a y . . . . . . ( i v ) \Rightarrow ab\Big(\dfrac{x}{a} - \dfrac{y}{b}\Big) = 4ab \\[1em] \Rightarrow bx - ay = 4ab \\[1em] \Rightarrow bx = 4ab + ay ......(iv) ⇒ ab ( a x − b y ) = 4 ab ⇒ b x − a y = 4 ab ⇒ b x = 4 ab + a y ...... ( i v )
Substituting value of bx in eq. (iii) we get,
⟹ 2(4ab + ay) + ay = 2ab
⟹ 8ab + 2ay + ay = 2ab
⟹ 8ab + 3ay = 2ab
⟹ 3ay = 2ab - 8ab
⟹ 3ay = -6ab
⟹ y = − 6 a b 3 a \dfrac{-6ab}{3a} 3 a − 6 ab
Substituting value of y in eqn. (iv) we get,
⟹ bx = 4ab - 2ab
⟹ bx = 2ab
⟹ x = 2a.
Hence, x = 2a and y = -2b.
Solve 2x + y = 35, 3x + 4y = 65. Hence, find the value of x y . \dfrac{x}{y}. y x .
Answer
Given,
2x + y = 35 ......(i)
3x + 4y = 65 ......(ii)
From (i) we get,
y = 35 - 2x ......(iii)
Substituting value of y from eqn. (iii) in eqn. (ii) we get,
⟹ 3x + 4(35 - 2x) = 65
⟹ 3x + 140 - 8x = 65
⟹ 140 - 5x = 65
⟹ 5x = 140 - 65
⟹ 5x = 75
⟹ x = 15.
Substituting value of x in eqn. (iii) we get,
⟹ y = 35 - 2x = 35 - 2(15) = 35 - 30 = 5.
⟹ x y = 15 5 = 3. \dfrac{x}{y} = \dfrac{15}{5} = 3. y x = 5 15 = 3.
Hence, x = 15, y = 5 and x y = 3. \dfrac{x}{y} = 3. y x = 3.
Solve the simultaneous equations 3x - y = 5, 4x - 3y = -1. Hence, find p, if y = px - 3.
Answer
Given,
3x - y = 5 ......(i)
4x - 3y = -1 .....(ii)
Solving (i) we get,
⟹ y = 3x - 5 .....(iii)
Substituting value of y from eqn. (iii) in eqn. (ii) we get,
⟹ 4x - 3(3x - 5) = -1
⟹ 4x - 9x + 15 = -1
⟹ -5x = -1 - 15
⟹ -5x = -16
⟹ x = 16 5 \dfrac{16}{5} 5 16
Substituting value of x in eqn. (iii) we get,
⇒ y = 3 × 16 5 − 5 = 48 5 − 5 = 48 − 25 5 = 23 5 \Rightarrow y = 3 \times \dfrac{16}{5} - 5 \\[1em] = \dfrac{48}{5} - 5 \\[1em] = \dfrac{48 - 25}{5} \\[1em] = \dfrac{23}{5} ⇒ y = 3 × 5 16 − 5 = 5 48 − 5 = 5 48 − 25 = 5 23
Given, y = px - 3. Substituting value of x and y in equation,
⇒ 23 5 = 16 5 p − 3 ⇒ 23 5 = 16 p − 15 5 ⇒ 23 = 16 p − 15 ⇒ 16 p = 38 ⇒ p = 38 16 = 19 8 . \Rightarrow \dfrac{23}{5} = \dfrac{16}{5}p - 3 \\[1em] \Rightarrow \dfrac{23}{5} = \dfrac{16p - 15}{5} \\[1em] \Rightarrow 23 = 16p - 15 \\[1em] \Rightarrow 16p = 38 \\[1em] \Rightarrow p = \dfrac{38}{16} = \dfrac{19}{8}. ⇒ 5 23 = 5 16 p − 3 ⇒ 5 23 = 5 16 p − 15 ⇒ 23 = 16 p − 15 ⇒ 16 p = 38 ⇒ p = 16 38 = 8 19 .
Hence, x =16 5 , y = 23 5 and p = 19 8 . \dfrac{16}{5},\text{ y }= \dfrac{23}{5}\text{ and p} = \dfrac{19}{8}. 5 16 , y = 5 23 and p = 8 19 .
Solve the following systems of simultaneous linear equations by the elimination method
3x + 4y = 10
2x - 2y = 2
Answer
Given,
3x + 4y = 10 ......(i)
2x - 2y = 2 .......(ii)
Multiplying eq. (ii) by 2 we get,
4x - 4y = 4 ......(iii)
Adding eq. (i) and (iii) we get,
⇒ 3x + 4y + 4x - 4y = 10 + 4
⇒ 7x = 14
⇒ x = 2.
Substituting value of x in eq. (ii) we get,
⇒ 2(2) - 2y = 2
⇒ 4 - 2y = 2
⇒ 2y = 4 - 2
⇒ 2y = 2
⇒ y = 1.
Hence, x = 2 and y = 1.
Solve the following systems of simultaneous linear equations by the elimination method
2x = 5y + 4
3x - 2y + 16 = 0
Answer
Given,
2x = 5y + 4 or 2x - 5y - 4 = 0 ........(i)
3x - 2y + 16 = 0 ......(ii)
Multiplying eq. (i) by 3 and eq. (ii) by 2 we get,
6x - 15y - 12 = 0 .......(iii)
6x - 4y + 32 = 0 .......(iv)
Subtracting eq. (iii) from (iv) we get,
⇒ 6x - 4y + 32 - (6x - 15y - 12) = 0
⇒ 6x - 6x - 4y + 15y + 32 + 12 = 0
⇒ 11y + 44 = 0
⇒ 11y = -44
⇒ y = -4.
Substituting value of y in eq. (ii) we get,
⇒ 3x - 2(-4) + 16 = 0
⇒ 3x + 8 + 16 = 0
⇒ 3x + 24 = 0
⇒ 3x = -24
⇒ x = -8.
Hence, x = -8 and y = -4.
Solve the following systems of simultaneous linear equations by the elimination method
3 4 x − 2 3 y = 1 \dfrac{3}{4}x - \dfrac{2}{3}y = 1 4 3 x − 3 2 y = 1
3 8 x − 1 6 y = 1 \dfrac{3}{8}x - \dfrac{1}{6}y = 1 8 3 x − 6 1 y = 1
Answer
Given,
3 4 x − 2 3 y = 1 \dfrac{3}{4}x - \dfrac{2}{3}y = 1 4 3 x − 3 2 y = 1
3 8 x − 1 6 y = 1 \dfrac{3}{8}x - \dfrac{1}{6}y = 1 8 3 x − 6 1 y = 1
Multiplying eq. (ii) by 2 we get,
3 4 x − 1 3 y = 2 \dfrac{3}{4}x - \dfrac{1}{3}y = 2 4 3 x − 3 1 y = 2
Subtracting eq. (i) from (iii) we get,
⇒ 3 4 x − 1 3 y − ( 3 4 x − 2 3 y ) = 2 − 1 ⇒ 3 4 x − 3 4 x − 1 3 y + 2 3 y = 1 ⇒ 1 3 y = 1 ⇒ y = 3. \Rightarrow \dfrac{3}{4}x - \dfrac{1}{3}y - \Big(\dfrac{3}{4}x - \dfrac{2}{3}y\Big) = 2 - 1 \\[1em] \Rightarrow \dfrac{3}{4}x - \dfrac{3}{4}x - \dfrac{1}{3}y + \dfrac{2}{3}y = 1 \\[1em] \Rightarrow \dfrac{1}{3}y = 1 \\[1em] \Rightarrow y = 3. ⇒ 4 3 x − 3 1 y − ( 4 3 x − 3 2 y ) = 2 − 1 ⇒ 4 3 x − 4 3 x − 3 1 y + 3 2 y = 1 ⇒ 3 1 y = 1 ⇒ y = 3.
Substituting value of y in eq. (i) we get,
⇒ 3 4 x − 2 3 × 3 = 1 ⇒ 3 4 x − 2 = 1 ⇒ 3 4 x = 3 ⇒ x = 3 × 4 3 ⇒ x = 4. \Rightarrow \dfrac{3}{4}x - \dfrac{2}{3} \times 3 = 1 \\[1em] \Rightarrow \dfrac{3}{4}x - 2 = 1 \\[1em] \Rightarrow \dfrac{3}{4}x = 3 \\[1em] \Rightarrow x = \dfrac{3 \times 4}{3} \\[1em] \Rightarrow x = 4. ⇒ 4 3 x − 3 2 × 3 = 1 ⇒ 4 3 x − 2 = 1 ⇒ 4 3 x = 3 ⇒ x = 3 3 × 4 ⇒ x = 4.
Hence, x = 4 and y = 3.
Solve the following systems of simultaneous linear equations by the elimination method
2x - 3y - 3 = 0
2 x 3 + 4 y + 1 2 = 0 \dfrac{2x}{3} + 4y + \dfrac{1}{2} = 0 3 2 x + 4 y + 2 1 = 0
Answer
Given,
2x - 3y - 3 = 0 ...........(i)
2 x 3 + 4 y + 1 2 = 0 \dfrac{2x}{3} + 4y + \dfrac{1}{2} = 0 3 2 x + 4 y + 2 1 = 0
Multiplying eq. (ii) by 3 we get,
2 x + 12 y + 3 2 = 0 2x + 12y + \dfrac{3}{2} = 0 2 x + 12 y + 2 3 = 0
Subtracting eq. (i) from (iii) we get,
⇒ 2 x + 12 y + 3 2 − ( 2 x − 3 y − 3 ) = 0 ⇒ 2 x − 2 x + 12 y + 3 y + 3 2 + 3 = 0 ⇒ 15 y + 9 2 = 0 ⇒ 15 y = − 9 2 ⇒ y = − 9 2 × 15 ⇒ y = − 3 10 . \Rightarrow 2x + 12y + \dfrac{3}{2} - (2x - 3y - 3) = 0 \\[1em] \Rightarrow 2x - 2x + 12y + 3y + \dfrac{3}{2} + 3 = 0 \\[1em] \Rightarrow 15y + \dfrac{9}{2} = 0 \\[1em] \Rightarrow 15y = -\dfrac{9}{2} \\[1em] \Rightarrow y = \dfrac{-9}{2 \times 15} \\[1em] \Rightarrow y = -\dfrac{3}{10}. ⇒ 2 x + 12 y + 2 3 − ( 2 x − 3 y − 3 ) = 0 ⇒ 2 x − 2 x + 12 y + 3 y + 2 3 + 3 = 0 ⇒ 15 y + 2 9 = 0 ⇒ 15 y = − 2 9 ⇒ y = 2 × 15 − 9 ⇒ y = − 10 3 .
Substituting value of y in eq. (i) we get,
⇒ 2 x − 3 y − 3 = 0 ⇒ 2 x − 3 × − 3 10 − 3 = 0 ⇒ 2 x + 9 10 − 3 = 0 ⇒ 2 x + 9 − 30 10 = 0 ⇒ 2 x − 21 10 = 0 ⇒ x = 21 20 . \Rightarrow 2x - 3y - 3 = 0 \\[1em] \Rightarrow 2x - 3 \times -\dfrac{3}{10} - 3 = 0 \\[1em] \Rightarrow 2x + \dfrac{9}{10} - 3 = 0 \\[1em] \Rightarrow 2x + \dfrac{9 - 30}{10} = 0 \\[1em] \Rightarrow 2x - \dfrac{21}{10} = 0 \\[1em] \Rightarrow x = \dfrac{21}{20}. ⇒ 2 x − 3 y − 3 = 0 ⇒ 2 x − 3 × − 10 3 − 3 = 0 ⇒ 2 x + 10 9 − 3 = 0 ⇒ 2 x + 10 9 − 30 = 0 ⇒ 2 x − 10 21 = 0 ⇒ x = 20 21 .
Hence, x = 21 20 and y = − 3 10 . x = \dfrac{21}{20} \text{and y} = -\dfrac{3}{10}. x = 20 21 and y = − 10 3 .
Solve the following systems of simultaneous linear equations by the elimination method
15x - 14y = 117
14x - 15y = 115
Answer
Given,
15x - 14y = 117 .......(i)
14x - 15y = 115 .......(ii)
Multiplying eq. (i) by 14 and eq. (ii) by 15 we get,
210x - 196y = 1638 .......(iii)
210x - 225y = 1725 .......(iv)
Subtracting eq. (iii) from (iv) we get,
⇒ 210x - 225y - (210x - 196y) = 1725 - 1638
⇒ 210x - 210x - 225y + 196y = 87
⇒ -29y = 87
⇒ y = -3.
Substituting value of y in eq. (ii) we get,
⇒ 14x - 15(-3) = 115
⇒ 14x + 45 = 115
⇒ 14x = 115 - 45
⇒ 14x = 70
⇒ x = 5.
Hence, x = 5 and y = -3.
Solve the following systems of simultaneous linear equations by the elimination method
41x + 53y = 135
53x + 41y = 147
Answer
Given,
41x + 53y = 135 ......(i)
53x + 41y = 147 .......(ii)
Multiplying eq. (i) by 53 and eq. (ii) by 41 we get,
2173x + 2809y = 7155 ......(iii)
2173x + 1681y = 6027 ......(iv)
Subtracting eq. (iv) from (iii) we get,
⇒ 2173x + 2809y - (2173x + 1681y) = 7155 - 6027
⇒ 2173x - 2173x + 2809y - 1681y = 1128
⇒ 1128y = 1128
⇒ y = 1.
Substituting value of y in eq. (i) we get,
⇒ 41x + 53(1) = 135
⇒ 41x = 135 - 53
⇒ 41x = 82
⇒ x = 2.
Hence, x = 2 and y = 1.
Solve the following systems of simultaneous linear equations by the elimination method
x 6 = y − 6 \dfrac{x}{6} = y - 6 6 x = y − 6
3 x 4 = 1 + y \dfrac{3x}{4} = 1 + y 4 3 x = 1 + y
Answer
Given,
x 6 = y − 6 \dfrac{x}{6} = y - 6 6 x = y − 6
3 x 4 = 1 + y \dfrac{3x}{4} = 1 + y 4 3 x = 1 + y
Subtracting eq. (i) from (ii) we get,
⇒ 3 x 4 − x 6 = 1 + y − ( y − 6 ) ⇒ 9 x − 2 x 12 = 7 ⇒ 7 x 12 = 7 ⇒ x = 7 × 12 7 ⇒ x = 12. \Rightarrow \dfrac{3x}{4} - \dfrac{x}{6} = 1 + y - (y - 6) \\[1em] \Rightarrow \dfrac{9x - 2x}{12} = 7 \\[1em] \Rightarrow \dfrac{7x}{12} = 7 \\[1em] \Rightarrow x = \dfrac{7 \times 12}{7} \\[1em] \Rightarrow x = 12. ⇒ 4 3 x − 6 x = 1 + y − ( y − 6 ) ⇒ 12 9 x − 2 x = 7 ⇒ 12 7 x = 7 ⇒ x = 7 7 × 12 ⇒ x = 12.
Substituting value of x in eq. (i) we get,
⇒ 12 6 = y − 6 ⇒ 2 = y − 6 ⇒ y = 2 + 6 = 8. \Rightarrow \dfrac{12}{6} = y - 6 \\[1em] \Rightarrow 2 = y - 6 \\[1em] \Rightarrow y = 2 + 6 = 8. ⇒ 6 12 = y − 6 ⇒ 2 = y − 6 ⇒ y = 2 + 6 = 8.
Hence, x = 12 and y = 8.
Solve the following systems of simultaneous linear equations by the elimination method
x − 2 3 y = 8 3 x - \dfrac{2}{3}y = \dfrac{8}{3} x − 3 2 y = 3 8
2 x 5 − y = 7 5 \dfrac{2x}{5} - y = \dfrac{7}{5} 5 2 x − y = 5 7
Answer
Given,
x − 2 3 y = 8 3 x - \dfrac{2}{3}y = \dfrac{8}{3} x − 3 2 y = 3 8
2 x 5 − y = 7 5 \dfrac{2x}{5} - y = \dfrac{7}{5} 5 2 x − y = 5 7
Multiplying eq. (i) by 6 and eq. (ii) by 15 we get,
6x - 4y = 16 .......(iii)
6x - 15y = 21 ......(iv)
Subtracting eq. (iv) from (iii) we get,
⇒ 6x - 4y - (6x - 15y) = 16 - 21
⇒ 6x - 6x - 4y + 15y = -5
⇒ 11y = -5
⇒ y = − 5 11 -\dfrac{5}{11} − 11 5
Substituting value of y in eq. (iv) we get,
⇒ 6 x − 15 × − 5 11 = 21 ⇒ 6 x + 75 11 = 21 ⇒ 6 x = 21 − 75 11 ⇒ 6 x = 231 − 75 11 ⇒ 6 x = 156 11 ⇒ x = 156 66 ⇒ x = 26 11 . \Rightarrow 6x - 15 \times -\dfrac{5}{11} = 21 \\[1em] \Rightarrow 6x + \dfrac{75}{11} = 21 \\[1em] \Rightarrow 6x = 21 - \dfrac{75}{11} \\[1em] \Rightarrow 6x = \dfrac{231 - 75}{11} \\[1em] \Rightarrow 6x = \dfrac{156}{11} \\[1em] \Rightarrow x = \dfrac{156}{66} \\[1em] \Rightarrow x = \dfrac{26}{11}. ⇒ 6 x − 15 × − 11 5 = 21 ⇒ 6 x + 11 75 = 21 ⇒ 6 x = 21 − 11 75 ⇒ 6 x = 11 231 − 75 ⇒ 6 x = 11 156 ⇒ x = 66 156 ⇒ x = 11 26 .
Hence, x = 26 11 and y = − 5 11 . x = \dfrac{26}{11} \text{ and } y = -\dfrac{5}{11}. x = 11 26 and y = − 11 5 .
Solve the following systems of simultaneous linear equations by the elimination method
9 - (x - 4) = y + 7
2(x + y) = 4 - 3y
Answer
Given,
9 - (x - 4) = y + 7
⇒ 9 - x + 4 = y + 7
⇒ x + y = 13 - 7
⇒ x + y = 6 .......(i)
2(x + y) = 4 - 3y
⇒ 2x + 2y = 4 - 3y
⇒ 2x + 5y = 4 .........(ii)
Multiplying eq. (i) by 2 we get,
⇒ 2x + 2y = 12 ........(iii)
Subtracting eq. (iii) from (ii) we get,
⇒ 2x + 5y - (2x + 2y) = 4 - 12
⇒ 5y - 2y = -8
⇒ 3y = -8
⇒ y = − 8 3 -\dfrac{8}{3} − 3 8
Substituting value of y in eq. (i) we get,
⇒ x + ( − 8 3 ) = 6 ⇒ x = 6 + 8 3 ⇒ x = 26 3 . \Rightarrow x + \Big(-\dfrac{8}{3}\Big) = 6 \\[1em] \Rightarrow x = 6 + \dfrac{8}{3} \\[1em] \Rightarrow x = \dfrac{26}{3}. ⇒ x + ( − 3 8 ) = 6 ⇒ x = 6 + 3 8 ⇒ x = 3 26 .
Hence, x = 26 3 and y = − 8 3 x = \dfrac{26}{3} \text{ and } y = -\dfrac{8}{3} x = 3 26 and y = − 3 8
Solve the following systems of simultaneous linear equations by the elimination method
2 x + x − y 6 = 2 2x + \dfrac{x - y}{6} = 2 2 x + 6 x − y = 2
x − 2 x + y 3 = 1 x - \dfrac{2x + y}{3} = 1 x − 3 2 x + y = 1
Answer
Solving 1st equation,
⇒ 2 x + x − y 6 = 2 ⇒ 12 x + x − y 6 = 2 ⇒ 13 x − y = 12....... ( i ) \Rightarrow 2x + \dfrac{x - y}{6} = 2 \\[1em] \Rightarrow \dfrac{12x + x - y}{6} = 2 \\[1em] \Rightarrow 13x - y = 12 .......(i) ⇒ 2 x + 6 x − y = 2 ⇒ 6 12 x + x − y = 2 ⇒ 13 x − y = 12....... ( i )
Solving 2nd equation,
⇒ x − 2 x + y 3 = 1 ⇒ 3 x − 2 x − y 3 = 1 ⇒ x − y = 3........ ( i i ) \Rightarrow x - \dfrac{2x + y}{3} = 1 \\[1em] \Rightarrow \dfrac{3x - 2x - y}{3} = 1 \\[1em] \Rightarrow x - y = 3 ........(ii) ⇒ x − 3 2 x + y = 1 ⇒ 3 3 x − 2 x − y = 1 ⇒ x − y = 3........ ( ii )
Multiplying eq. (ii) by 13 we get,
⇒ 13x - 13y = 39 .........(iii)
Subtracting eq. (iii) from (i) we get,
⇒ 13x - y - (13x - 13y) = 12 - 39
⇒ 13x - 13x - y + 13y = -27
⇒ 12y = -27
⇒ y = − 9 4 -\dfrac{9}{4} − 4 9
Substituting value of y in eq (ii) we get,
⇒ x − ( − 9 4 ) = 3 ⇒ x + 9 4 = 3 ⇒ x = 3 − 9 4 ⇒ x = 12 − 9 4 ⇒ x = 3 4 . \Rightarrow x - \Big(-\dfrac{9}{4}\Big) = 3 \\[1em] \Rightarrow x + \dfrac{9}{4} = 3 \\[1em] \Rightarrow x = 3 - \dfrac{9}{4} \\[1em] \Rightarrow x = \dfrac{12 - 9}{4} \\[1em] \Rightarrow x = \dfrac{3}{4}. ⇒ x − ( − 4 9 ) = 3 ⇒ x + 4 9 = 3 ⇒ x = 3 − 4 9 ⇒ x = 4 12 − 9 ⇒ x = 4 3 .
Hence, x = 3 4 and y = − 9 4 . x = \dfrac{3}{4} \text{ and } y = -\dfrac{9}{4}. x = 4 3 and y = − 4 9 .
Solve the following systems of simultaneous linear equations by the elimination method
x - 3y = 3x - 1 = 2x - y.
Answer
x - 3y = 3x - 1
⇒ 3x - x + 3y = 1
⇒ 2x + 3y = 1 .......(i)
3x - 1 = 2x - y
⇒ 3x - 2x + y = 1
⇒ x + y = 1 .......(ii)
Multiplying eq. (ii) by 2 we get,
2x + 2y = 2 ........(iii)
Subtracting eq. (iii) from (i) we get,
⇒ 2x + 3y - (2x + 2y) = 1 - 2
⇒ 2x - 2x + 3y - 2y = -1
⇒ y = -1.
Substituting value of y in eq. (ii) we get,
⇒ x + (-1) = 1
⇒ x = 1 + 1
⇒ x = 2.
Hence, x = 2 and y = -1.
Solve the following systems of simultaneous linear equations by the elimination method
4 x + x − y 8 4x + \dfrac{x - y}{8} 4 x + 8 x − y
2 y + x − 5 y + 2 3 = 2 2y + x - \dfrac{5y + 2}{3} = 2 2 y + x − 3 5 y + 2 = 2
Answer
Solving 1st equation,
⇒ 4 x + x − y 8 = 17 ⇒ 32 x + x − y 8 = 17 ⇒ 33 x − y = 136....... ( i ) \Rightarrow 4x + \dfrac{x - y}{8} = 17 \\[1em] \Rightarrow \dfrac{32x + x - y}{8} = 17 \\[1em] \Rightarrow 33x - y = 136 .......(i) ⇒ 4 x + 8 x − y = 17 ⇒ 8 32 x + x − y = 17 ⇒ 33 x − y = 136....... ( i )
Solving 2nd equation,
⇒ 2 y + x − 5 y + 2 3 = 2 ⇒ 6 y + 3 x − 5 y − 2 3 = 2 ⇒ y + 3 x − 2 3 = 2 ⇒ y + 3 x − 2 = 6 ⇒ y + 3 x = 8........ ( i i ) \Rightarrow 2y + x - \dfrac{5y + 2}{3} = 2 \\[1em] \Rightarrow \dfrac{6y + 3x - 5y - 2}{3} = 2 \\[1em] \Rightarrow \dfrac{y + 3x - 2}{3} = 2 \\[1em] \Rightarrow y + 3x - 2 = 6 \\[1em] \Rightarrow y + 3x = 8 ........(ii) ⇒ 2 y + x − 3 5 y + 2 = 2 ⇒ 3 6 y + 3 x − 5 y − 2 = 2 ⇒ 3 y + 3 x − 2 = 2 ⇒ y + 3 x − 2 = 6 ⇒ y + 3 x = 8........ ( ii )
Adding equation (i) and (ii) we get,
⇒ 33x - y + y + 3x = 136 + 8
⇒ 36x = 144
⇒ x = 4.
Substituting value of x in equation (ii) we get,
⇒ y + 3(4) = 8
⇒ y + 12 = 8
⇒ y = -4.
Hence, x = 4 and y = -4.
Solve the following systems of simultaneous linear equations by the elimination method
x + 1 2 + y − 1 3 = 8 \dfrac{x + 1}{2} + \dfrac{y - 1}{3} = 8 2 x + 1 + 3 y − 1 = 8
x − 1 3 + y + 1 2 = 9 \dfrac{x - 1}{3} + \dfrac{y + 1}{2} = 9 3 x − 1 + 2 y + 1 = 9
Answer
Solving 1st equation,
⇒ x + 1 2 + y − 1 3 = 8 ⇒ 3 ( x + 1 ) + 2 ( y − 1 ) 6 = 8 ⇒ 3 x + 3 + 2 y − 2 6 = 8 ⇒ 3 x + 2 y + 1 6 = 8 ⇒ 3 x + 2 y + 1 = 48 ⇒ 3 x + 2 y = 47...... ( i ) \Rightarrow \dfrac{x + 1}{2} + \dfrac{y - 1}{3} = 8 \\[1em] \Rightarrow \dfrac{3(x + 1) + 2(y - 1)}{6} = 8 \\[1em] \Rightarrow \dfrac{3x + 3 + 2y - 2}{6} = 8 \\[1em] \Rightarrow \dfrac{3x + 2y + 1}{6} = 8 \\[1em] \Rightarrow 3x + 2y + 1 = 48 \\[1em] \Rightarrow 3x + 2y = 47 ......(i) ⇒ 2 x + 1 + 3 y − 1 = 8 ⇒ 6 3 ( x + 1 ) + 2 ( y − 1 ) = 8 ⇒ 6 3 x + 3 + 2 y − 2 = 8 ⇒ 6 3 x + 2 y + 1 = 8 ⇒ 3 x + 2 y + 1 = 48 ⇒ 3 x + 2 y = 47...... ( i )
Solving 2nd equation,
⇒ x − 1 3 + y + 1 2 = 9 ⇒ 2 ( x − 1 ) + 3 ( y + 1 ) 6 = 9 ⇒ 2 x − 2 + 3 y + 3 6 = 9 ⇒ 2 x + 3 y + 1 6 = 9 ⇒ 2 x + 3 y + 1 = 54 ⇒ 2 x + 3 y = 53....... ( i i ) \Rightarrow \dfrac{x - 1}{3} + \dfrac{y + 1}{2} = 9 \\[1em] \Rightarrow \dfrac{2(x - 1) + 3(y + 1)}{6} = 9 \\[1em] \Rightarrow \dfrac{2x - 2 + 3y + 3}{6} = 9 \\[1em] \Rightarrow \dfrac{2x + 3y + 1}{6} = 9 \\[1em] \Rightarrow 2x + 3y + 1 = 54 \\[1em] \Rightarrow 2x + 3y = 53 .......(ii) ⇒ 3 x − 1 + 2 y + 1 = 9 ⇒ 6 2 ( x − 1 ) + 3 ( y + 1 ) = 9 ⇒ 6 2 x − 2 + 3 y + 3 = 9 ⇒ 6 2 x + 3 y + 1 = 9 ⇒ 2 x + 3 y + 1 = 54 ⇒ 2 x + 3 y = 53....... ( ii )
Multiplying eq. (i) by 2 and eq. (ii) by 3 we get,
6x + 4y = 94 ......(iii)
6x + 9y = 159 .......(iv)
Subtracting eq. (iii) from (iv) we get,
⇒ 6x + 9y - (6x + 4y) = 159 - 94
⇒ 5y = 65
⇒ y = 13.
Substituting value of y in eq. (iii) we get,
⇒ 6x + 4(13) = 94
⇒ 6x + 52 = 94
⇒ 6x = 42
⇒ x = 7.
Hence, x = 7 and y = 13.
Solve the following systems of simultaneous linear equations by the elimination method
3 x + 4 y = 7 \dfrac{3}{x} + 4y = 7 x 3 + 4 y = 7
5 x + 6 y = 13 \dfrac{5}{x} + 6y = 13 x 5 + 6 y = 13
Answer
Given,
3 x + 4 y = 7 \dfrac{3}{x} + 4y = 7 x 3 + 4 y = 7
5 x + 6 y = 13 \dfrac{5}{x} + 6y = 13 x 5 + 6 y = 13
Multiplying eq. (i) by 3 and eq. (ii) by 2 we get,
9 x + 12 y = 21 \dfrac{9}{x} + 12y = 21 x 9 + 12 y = 21
10 x + 12 y = 26 \dfrac{10}{x} + 12y = 26 x 10 + 12 y = 26
Subtracting eq. (iii) from (iv) we get,
⇒ 10 x − 9 x + 12 y − 12 y = 26 − 21 ⇒ 1 x = 5 ⇒ x = 1 5 . \Rightarrow \dfrac{10}{x} - \dfrac{9}{x} + 12y - 12y = 26 - 21 \\[1em] \Rightarrow \dfrac{1}{x} = 5 \\[1em] \Rightarrow x = \dfrac{1}{5}. ⇒ x 10 − x 9 + 12 y − 12 y = 26 − 21 ⇒ x 1 = 5 ⇒ x = 5 1 .
Substituting value of x in eq. (i) we get,
⇒ 3 x + 4 y = 7 ⇒ 3 1 5 + 4 y = 7 ⇒ 15 + 4 y = 7 ⇒ 4 y = − 8 ⇒ y = − 2. \Rightarrow \dfrac{3}{x} + 4y = 7 \\[1em] \Rightarrow \dfrac{3}{\dfrac{1}{5}} + 4y = 7 \\[1em] \Rightarrow 15 + 4y = 7 \\[1em] \Rightarrow 4y = -8 \\[1em] \Rightarrow y = -2. ⇒ x 3 + 4 y = 7 ⇒ 5 1 3 + 4 y = 7 ⇒ 15 + 4 y = 7 ⇒ 4 y = − 8 ⇒ y = − 2.
Hence, x = 1 5 \dfrac{1}{5} 5 1
Solve the following systems of simultaneous linear equations by the elimination method
5 x − 9 = 1 y 5x - 9 = \dfrac{1}{y} 5 x − 9 = y 1
x + 1 y = 3 x + \dfrac{1}{y} = 3 x + y 1 = 3
Answer
5 x − 9 = 1 y or 5 x − 1 y = 9 5x - 9 = \dfrac{1}{y} \text{ or } 5x - \dfrac{1}{y} = 9 5 x − 9 = y 1 or 5 x − y 1 = 9
x + 1 y = 3 x + \dfrac{1}{y} = 3 x + y 1 = 3
Adding eq. (i) and (ii) we get,
⇒ 5 x − 1 y + x + 1 y = 9 + 3 ⇒ 6 x = 12 ⇒ x = 2. \Rightarrow 5x - \dfrac{1}{y} + x + \dfrac{1}{y} = 9 + 3 \\[1em] \Rightarrow 6x = 12 \\[1em] \Rightarrow x = 2. ⇒ 5 x − y 1 + x + y 1 = 9 + 3 ⇒ 6 x = 12 ⇒ x = 2.
Substituting value of x in eq. (ii) we get,
⇒ x + 1 y = 3 ⇒ 2 + 1 y = 3 ⇒ 1 y = 1 ⇒ y = 1. \Rightarrow x + \dfrac{1}{y} = 3 \\[1em] \Rightarrow 2 + \dfrac{1}{y} = 3 \\[1em] \Rightarrow \dfrac{1}{y} = 1 \\[1em] \Rightarrow y = 1. ⇒ x + y 1 = 3 ⇒ 2 + y 1 = 3 ⇒ y 1 = 1 ⇒ y = 1.
Hence, x = 2 and y = 1.
Solve the following systems of simultaneous linear equations by the elimination method
px + qy = p - q
qx - py = p + q
Answer
Given,
px + qy = p - q .......(i)
qx - py = p + q .......(ii)
Multiplying eq. (i) by q and eq. (ii) by p we get,
pqx + q2 y = pq - q2 .......(iii)
pqx - p2 y = p2 + pq .......(iv)
Subtracting eq. (iv) from (iii) we get,
⇒ pqx + q2 y - (pqx - p2 y) = pq - q2 - (p2 + pq)
⇒ pqx - pqx + q2 y + p2 y = pq - pq - q2 - p2
⇒ q2 y + p2 y = -q2 - p2
⇒ y(q2 + p2 ) = -(q2 + p2 )
⇒ y = − ( q 2 + p 2 ) q 2 + p 2 \dfrac{-(\text{q}^2 + \text{p}^2)}{\text{q}^2 + \text{p}^2} q 2 + p 2 − ( q 2 + p 2 )
Substituting value of y in eq. (i) we get,
⇒ px + q(-1) = p - q
⇒ px - q = p - q
⇒ px = p - q + q
⇒ px = p
⇒ x = 1.
Hence, x = 1 and y = -1.
Solve the following systems of simultaneous linear equations by the elimination method
x a − y b = 0 \dfrac{x}{a} - \dfrac{y}{b} = 0 a x − b y = 0
ax + by = a2 + b2
Answer
⇒ x a − y b = 0 ⇒ b x − a y a b = 0 ⇒ b x − a y = 0. \phantom{\Rightarrow} \dfrac{x}{a} - \dfrac{y}{b} = 0 \\[1em] \Rightarrow \dfrac{bx - ay}{ab} = 0 \\[1em] \Rightarrow bx - ay = 0. ⇒ a x − b y = 0 ⇒ ab b x − a y = 0 ⇒ b x − a y = 0.
Given equations can be written as,
bx - ay = 0 .......(i)
ax + by = (a2 + b2 ) .......(ii)
Multiplying eq. (i) by a and (ii) by b we get,
abx - a2 y = 0 .........(iii)
abx + b2 y = b(a2 + b2 ) .......(iv)
Subtracting eq. (iii) from (iv) we get,
⇒ abx + b2 y - (abx - a2 y) = b(a2 + b2 )
⇒ abx - abx + b2 y + a2 y = b(a2 + b2 )
⇒ y(b2 + a2 ) = b(a2 + b2 )
⇒ y = b.
Substituting value of y in eq. (i) we get,
⇒ bx - ay = 0
⇒ bx - ab = 0
⇒ bx = ab
⇒ x = a.
Hence, x = a and y = b.
Solve 2x + y = 23, 4x - y = 19. Hence, find the values of x - 3y and 5y - 2x.
Answer
Given,
2x + y = 23 .......(i)
4x - y = 19 .......(ii)
Multiplying eq. (i) by 2 we get,
4x + 2y = 46 ......(iii)
Subtracting eq. (ii) from (iii) we get,
⇒ 4x + 2y - (4x - y) = 46 - 19
⇒ 4x - 4x + 2y + y = 27
⇒ 3y = 27
⇒ y = 9.
Substituting value of y in eq. (ii) we get,
⇒ 4x - 9 = 19
⇒ 4x = 28
⇒ x = 7.
Substituting value of x and y in x - 3y,
⇒ x - 3y = 7 - 3(9) = 7 - 27 = -20.
Substituting value of x and y in 5y - 2x,
⇒ 5y - 2x = 5(9) - 2(7) = 45 - 14 = 31.
Hence, x = 7, y = 9, x - 3y = -20 and 5y - 2x = 31.
The expression ax + by has value 7 when x = 2 and y = 1. When x = -1, y = 1, it has value 1, find a and b.
Answer
Given,
ax + by = 7, when x = 2, y = 1.
⇒ 2a + b = 7 ......(i)
ax + by = 1 when x = -1, y = 1.
⇒ -a + b = 1 .......(ii)
Multiplying eq. (ii) by 2 we get,
⇒ -2a + 2b = 2 .......(iii)
Adding eq. (i) and (iii) we get,
⇒ 2a + b + (-2a + 2b) = 7 + 2
⇒ 2a - 2a + b + 2b = 9
⇒ 3b = 9
⇒ b = 3.
Substituting value of b in eq. (ii) we get,
⇒ -a + 3 = 1
⇒ -a = -2
⇒ a = 2.
Hence, a = 2 and b = 3.
Can the following equations hold simultaneously?
3x - 7y = 7
11x + 5y = 87
5x + 4y = 43.
If so, find x and y.
Answer
Given,
3x - 7y = 7 ........(i)
11x + 5y = 87 .......(ii)
5x + 4y = 43 .......(iii)
Solving first two equations simultaneously,
Multiplying eq. (i) by 11 and eq. (ii) by 3 we get,
33x - 77y = 77 ......(iv)
33x + 15y = 261 ......(v)
Subtracting eq. (iv) from (v) we get,
⇒ 33x + 15y - (33x - 77y) = 261 - 77
⇒ 33x - 33x + 15y + 77y = 184
⇒ 92y = 184
⇒ y = 2.
Substituting value of y in eq. (i) we get,
⇒ 3x - 7y = 7
⇒ 3x - 7(2) = 7
⇒ 3x - 14 = 7
⇒ 3x = 21
⇒ x = 7.
Substituting x = 7 and y = 2 in L.H.S. of eq. (iii),
5x + 4y = 43 .......(iii)
⇒ 5(7) + 4(2) = 35 + 8 = 43.
Since, L.H.S. = R.H.S. hence following equations can be held simultaneously.
Hence, x = 7 and y = 2.
Solve the following systems of simultaneous linear equations by cross-multiplication method :
3x + 2y = 4
8x + 5y = 9
Answer
Given equations can be written as,
3x + 2y - 4 = 0
8x + 5y - 9 = 0
By cross multiplication method,
∴ x 2 × ( − 9 ) − 5 × ( − 4 ) = y ( − 4 ) × 8 − ( − 9 ) × 3 = 1 3 × 5 − 8 × 2 ⇒ x − 18 + 20 = y − 32 + 27 = 1 15 − 16 ⇒ x 2 = y − 5 = 1 − 1 ∴ x 2 = 1 − 1 and y − 5 = 1 − 1 ⇒ x = − 2 and y = 5. \therefore \dfrac{x}{2 \times (-9) - 5 \times (-4)} = \dfrac{y}{(-4) \times 8 - (-9) \times 3} = \dfrac{1}{3 \times 5 - 8 \times 2} \\[1em] \Rightarrow \dfrac{x}{-18 + 20} = \dfrac{y}{-32 + 27} = \dfrac{1}{15 - 16} \\[1em] \Rightarrow \dfrac{x}{2} = \dfrac{y}{-5} = \dfrac{1}{-1} \\[1em] \therefore \dfrac{x}{2} = \dfrac{1}{-1} \text{ and } \dfrac{y}{-5} = \dfrac{1}{-1} \\[1em] \Rightarrow x = -2 \text{ and } y = 5. ∴ 2 × ( − 9 ) − 5 × ( − 4 ) x = ( − 4 ) × 8 − ( − 9 ) × 3 y = 3 × 5 − 8 × 2 1 ⇒ − 18 + 20 x = − 32 + 27 y = 15 − 16 1 ⇒ 2 x = − 5 y = − 1 1 ∴ 2 x = − 1 1 and − 5 y = − 1 1 ⇒ x = − 2 and y = 5.
Hence, x = -2 and y = 5.
Solve the following systems of simultaneous linear equations by cross-multiplication method :
3x - 7y + 10 = 0
y - 2x = 3
Answer
Given equations can be written as,
3x - 7y + 10 = 0
-2x + y - 3 = 0
By cross multiplication method,
∴ x ( − 7 ) × ( − 3 ) − 1 × 10 = y 10 × ( − 2 ) − ( − 3 ) × 3 = 1 3 × 1 − ( − 2 ) × ( − 7 ) ⇒ x 21 − 10 = y − 20 + 9 = 1 3 − 14 ⇒ x 11 = y − 11 = 1 − 11 ∴ x 11 = 1 − 11 and y − 11 = 1 − 11 ⇒ x = − 1 and y = 1. \therefore \dfrac{x}{(-7) \times (-3) - 1 \times 10} = \dfrac{y}{10 \times (-2) - (-3) \times 3} = \dfrac{1}{3 \times 1 - (-2) \times (-7)} \\[1em] \Rightarrow \dfrac{x}{21 - 10} = \dfrac{y}{-20 + 9} = \dfrac{1}{3 - 14} \\[1em] \Rightarrow \dfrac{x}{11} = \dfrac{y}{-11} = \dfrac{1}{-11} \\[1em] \therefore \dfrac{x}{11} = \dfrac{1}{-11} \text{ and } \dfrac{y}{-11} = \dfrac{1}{-11} \\[1em] \Rightarrow x = -1 \text{ and } y = 1. ∴ ( − 7 ) × ( − 3 ) − 1 × 10 x = 10 × ( − 2 ) − ( − 3 ) × 3 y = 3 × 1 − ( − 2 ) × ( − 7 ) 1 ⇒ 21 − 10 x = − 20 + 9 y = 3 − 14 1 ⇒ 11 x = − 11 y = − 11 1 ∴ 11 x = − 11 1 and − 11 y = − 11 1 ⇒ x = − 1 and y = 1.
Hence, x = -1 and y = 1.
Solve the following system of simultaneous linear equations by cross - multiplication method :
2x - 5y = -1
3x + y = 7
Answer
Given equations can be written as,
2x - 5y + 1 = 0
3x + y - 7 = 0
By cross multiplication method,
∴ x − 5 × ( − 7 ) − 1 × 1 = y 1 × 3 − ( − 7 ) × 2 = 1 2 × 1 − 3 × − 5 ⇒ x 35 − 1 = y 3 − ( − 14 ) = 1 2 − ( − 15 ) ⇒ x 34 = y 3 + 14 = 1 2 + 15 ⇒ x 34 = y 17 = 1 17 ⇒ x 34 = 1 17 and y 17 = 1 17 ⇒ x = 34 17 and y = 17 17 ⇒ x = 2 and y = 1. \therefore \dfrac{x}{-5 \times (-7) - 1 \times 1} = \dfrac{y}{1 \times 3 - (-7) \times 2} = \dfrac{1}{2 \times 1 - 3 \times -5}\\[1em] \Rightarrow \dfrac{x}{35 - 1} = \dfrac{y}{3 - (-14)} = \dfrac{1}{2 - (-15)}\\[1em] \Rightarrow \dfrac{x}{34} = \dfrac{y}{3 + 14} = \dfrac{1}{2 + 15}\\[1em] \Rightarrow \dfrac{x}{34} = \dfrac{y}{17} = \dfrac{1}{17}\\[1em] \Rightarrow \dfrac{x}{34} = \dfrac{1}{17} \text{ and } \dfrac{y}{17} = \dfrac{1}{17}\\[1em] \Rightarrow x = \dfrac{34}{17} \text{ and }y = \dfrac{17}{17}\\[1em] \Rightarrow x = 2 \text{ and }y = 1. ∴ − 5 × ( − 7 ) − 1 × 1 x = 1 × 3 − ( − 7 ) × 2 y = 2 × 1 − 3 × − 5 1 ⇒ 35 − 1 x = 3 − ( − 14 ) y = 2 − ( − 15 ) 1 ⇒ 34 x = 3 + 14 y = 2 + 15 1 ⇒ 34 x = 17 y = 17 1 ⇒ 34 x = 17 1 and 17 y = 17 1 ⇒ x = 17 34 and y = 17 17 ⇒ x = 2 and y = 1.
Hence, x = 2 and y = 1.
Solve the following system of simultaneous linear equations by cross-multiplication method :
x + 3y + 4 = 0
3x - y = -2
Answer
Given equations can be written as,
x + 3y + 4 = 0
3x - y + 2 = 0
By cross multiplication method,
∴ x 3 × 2 − ( − 1 ) × 4 = y 4 × 3 − 2 × 1 = 1 1 × ( − 1 ) − 3 × 3 ⇒ x 6 + 4 = y 12 − 2 = 1 − 1 − 9 ⇒ x 10 = y 10 = 1 − 10 ⇒ x 10 = 1 − 10 and y 10 = 1 − 10 ⇒ x = − 10 10 and y = − 10 10 ⇒ x = − 1 and y = − 1 \therefore \dfrac{x}{3 \times 2 - (-1) \times 4} = \dfrac{y}{4 \times 3 - 2 \times 1} = \dfrac{1}{1 \times (-1) - 3 \times 3}\\[1em] \Rightarrow \dfrac{x}{6 + 4} = \dfrac{y}{12 - 2} = \dfrac{1}{-1 - 9}\\[1em] \Rightarrow \dfrac{x}{10} = \dfrac{y}{10} = \dfrac{1}{-10}\\[1em] \Rightarrow \dfrac{x}{10} = \dfrac{1}{-10} \text{ and }\dfrac{y}{10} = \dfrac{1}{-10}\\[1em] \Rightarrow x = -\dfrac{10}{10} \text{ and }y = -\dfrac{10}{10}\\[1em] \Rightarrow x = -1 \text{ and }y = -1 ∴ 3 × 2 − ( − 1 ) × 4 x = 4 × 3 − 2 × 1 y = 1 × ( − 1 ) − 3 × 3 1 ⇒ 6 + 4 x = 12 − 2 y = − 1 − 9 1 ⇒ 10 x = 10 y = − 10 1 ⇒ 10 x = − 10 1 and 10 y = − 10 1 ⇒ x = − 10 10 and y = − 10 10 ⇒ x = − 1 and y = − 1
Hence, x = -1 and y = -1.
Solve the following pairs of linear equations by cross-multiplication method:
x - y = a + b
ax + by = a2 - b2
Answer
Given equations can be written as,
x - y - (a + b) = 0
ax + by - (a2 - b2 ) = 0
By cross multiplication method,
∴ x ( − 1 ) × [ − ( a 2 − b 2 ) ] − b × [ − ( a + b ) ] = y [ − ( a + b ) × a ] − [ − ( a 2 − b 2 ) × 1 ] = 1 1 × b − a × ( − 1 ) ⇒ x a 2 − b 2 + a b + b 2 = y − a 2 − a b + a 2 − b 2 = 1 b + a ⇒ x a 2 + a b = y − a b − b 2 = 1 b + a ∴ x a 2 + a b = 1 b + a and y − b ( a + b ) = 1 b + a ⇒ x a ( a + b ) = 1 b + a and y − b ( a + b ) = 1 b + a ⇒ x = a ( a + b ) b + a and y = − b ( a + b ) b + a ⇒ x = a and y = − b . \therefore \dfrac{x}{(-1) \times [-(a^2 - b^2)] - b \times [-(a + b)]} = \dfrac{y}{[-(a + b) \times a] - [-(a^2 - b^2) \times 1] } = \dfrac{1}{1 \times b - a \times (-1)} \\[1em] \Rightarrow \dfrac{x}{a^2 - b^2 + ab + b^2} = \dfrac{y}{-a^2 - ab + a^2 - b^2} = \dfrac{1}{b + a} \\[1em] \Rightarrow \dfrac{x}{a^2 + ab} = \dfrac{y}{-ab - b^2} = \dfrac{1}{b + a} \\[1em] \therefore \dfrac{x}{a^2 + ab} = \dfrac{1}{b + a} \text{ and } \dfrac{y}{-b(a + b)} = \dfrac{1}{b + a} \\[1em] \Rightarrow \dfrac{x}{a(a + b)} = \dfrac{1}{b + a} \text{ and } \dfrac{y}{-b(a + b)} = \dfrac{1}{b + a} \\[1em] \Rightarrow x = \dfrac{a(a + b)}{b + a} \text{ and } y = \dfrac{-b(a + b)}{b + a} \\[1em] \Rightarrow x = a \text{ and } y = -b. ∴ ( − 1 ) × [ − ( a 2 − b 2 )] − b × [ − ( a + b )] x = [ − ( a + b ) × a ] − [ − ( a 2 − b 2 ) × 1 ] y = 1 × b − a × ( − 1 ) 1 ⇒ a 2 − b 2 + ab + b 2 x = − a 2 − ab + a 2 − b 2 y = b + a 1 ⇒ a 2 + ab x = − ab − b 2 y = b + a 1 ∴ a 2 + ab x = b + a 1 and − b ( a + b ) y = b + a 1 ⇒ a ( a + b ) x = b + a 1 and − b ( a + b ) y = b + a 1 ⇒ x = b + a a ( a + b ) and y = b + a − b ( a + b ) ⇒ x = a and y = − b .
Hence, x = a and y = -b.
Solve the following pairs of linear equations by cross-multiplication method:
2bx + ay = 2ab
bx - ay = 4ab.
Answer
Given equations can be written as,
2bx + ay - 2ab = 0
bx - ay - 4ab = 0
By cross multiplication method,
∴ x a × − ( 4 a b ) − ( − a ) × [ − 2 a b ] = y [ − ( 2 a b ) × b ] − [ − ( 4 a b ) × 2 b ] = 1 2 b × ( − a ) − b × a ⇒ x − 4 a 2 b − 2 a 2 b = y − 2 a b 2 + 8 a b 2 = 1 − 2 a b − a b ⇒ − x 6 a 2 b = y 6 a b 2 = − 1 3 a b ∴ − x 6 a 2 b = − 1 3 a b and y 6 a b 2 = − 1 3 a b ⇒ x = 6 a 2 b 3 a b and y = − 6 a b 2 3 a b ⇒ x = 2 a and y = − 2 b . \therefore \dfrac{x}{a \times -(4ab) - (-a) \times [-2ab]} = \dfrac{y}{[-(2ab) \times b] - [-(4ab) \times 2b] } = \dfrac{1}{2b \times (-a) - b \times a} \\[1em] \Rightarrow \dfrac{x}{-4a^2b - 2a^2b} = \dfrac{y}{-2ab^2 + 8ab^2} = \dfrac{1}{-2ab - ab} \\[1em] \Rightarrow -\dfrac{x}{6a^2b} = \dfrac{y}{6ab^2} = -\dfrac{1}{3ab} \\[1em] \therefore -\dfrac{x}{6a^2b} = -\dfrac{1}{3ab} \text{ and } \dfrac{y}{6ab^2} = -\dfrac{1}{3ab} \\[1em] \Rightarrow x = \dfrac{6a^2b}{3ab} \text{ and } y = -\dfrac{6ab^2}{3ab} \\[1em] \Rightarrow x = 2a \text{ and } y = -2b. ∴ a × − ( 4 ab ) − ( − a ) × [ − 2 ab ] x = [ − ( 2 ab ) × b ] − [ − ( 4 ab ) × 2 b ] y = 2 b × ( − a ) − b × a 1 ⇒ − 4 a 2 b − 2 a 2 b x = − 2 a b 2 + 8 a b 2 y = − 2 ab − ab 1 ⇒ − 6 a 2 b x = 6 a b 2 y = − 3 ab 1 ∴ − 6 a 2 b x = − 3 ab 1 and 6 a b 2 y = − 3 ab 1 ⇒ x = 3 ab 6 a 2 b and y = − 3 ab 6 a b 2 ⇒ x = 2 a and y = − 2 b .
Hence, x = 2a and y = -2b.
Solve the following pairs of linear equations:
2 x + 2 3 y = 1 6 \dfrac{2}{x} + \dfrac{2}{3y} = \dfrac{1}{6} x 2 + 3 y 2 = 6 1
2 x − 1 y = 1 \dfrac{2}{x} - \dfrac{1}{y} = 1 x 2 − y 1 = 1
Answer
Substituting 1 x = a and 1 y = b \dfrac{1}{x} = a \text{ and } \dfrac{1}{y} = b x 1 = a and y 1 = b
2 a + 2 3 b = 1 6 2a + \dfrac{2}{3}b = \dfrac{1}{6} 2 a + 3 2 b = 6 1
2a - b = 1 .......(ii)
Subtracting eq. (ii) from (i) we get,
⇒ 2 a + 2 3 b − ( 2 a − b ) = 1 6 − 1 ⇒ 2 a − 2 a + 2 3 b + b = 1 − 6 6 ⇒ 2 b + 3 b 3 = − 5 6 ⇒ 5 b 3 = − 5 6 ⇒ b = − 5 × 3 6 × 5 ⇒ b = − 1 2 ∴ 1 y = − 1 2 ⇒ y = − 2. \Rightarrow 2a + \dfrac{2}{3}b - (2a - b) = \dfrac{1}{6} - 1 \\[1em] \Rightarrow 2a - 2a + \dfrac{2}{3}b + b = \dfrac{1 - 6}{6} \\[1em] \Rightarrow \dfrac{2b + 3b}{3} = -\dfrac{5}{6} \\[1em] \Rightarrow \dfrac{5b}{3} = -\dfrac{5}{6} \\[1em] \Rightarrow b = -\dfrac{5 \times 3}{6 \times 5} \\[1em] \Rightarrow b = -\dfrac{1}{2} \\[1em] \therefore \dfrac{1}{y} = -\dfrac{1}{2} \\[1em] \Rightarrow y = -2. ⇒ 2 a + 3 2 b − ( 2 a − b ) = 6 1 − 1 ⇒ 2 a − 2 a + 3 2 b + b = 6 1 − 6 ⇒ 3 2 b + 3 b = − 6 5 ⇒ 3 5 b = − 6 5 ⇒ b = − 6 × 5 5 × 3 ⇒ b = − 2 1 ∴ y 1 = − 2 1 ⇒ y = − 2.
Substituting value of b in eq (ii),
⇒ 2 a − ( − 1 2 ) = 1 ⇒ 2 a + 1 2 = 1 ⇒ 2 a = 1 − 1 2 ⇒ 2 a = 1 2 ⇒ a = 1 4 ∴ 1 x = 1 4 ⇒ x = 4. \Rightarrow 2a - \Big(-\dfrac{1}{2}\Big) = 1 \\[1em] \Rightarrow 2a + \dfrac{1}{2} = 1 \\[1em] \Rightarrow 2a = 1 - \dfrac{1}{2} \\[1em] \Rightarrow 2a = \dfrac{1}{2} \\[1em] \Rightarrow a = \dfrac{1}{4} \\[1em] \therefore \dfrac{1}{x} = \dfrac{1}{4} \\[1em] \Rightarrow x = 4. ⇒ 2 a − ( − 2 1 ) = 1 ⇒ 2 a + 2 1 = 1 ⇒ 2 a = 1 − 2 1 ⇒ 2 a = 2 1 ⇒ a = 4 1 ∴ x 1 = 4 1 ⇒ x = 4.
Hence, x = 4 and y = -2.
Solve the following pairs of linear equations:
3 2 x + 2 3 y = 5 \dfrac{3}{2x} + \dfrac{2}{3y} = 5 2 x 3 + 3 y 2 = 5
5 x − 3 y = 1 \dfrac{5}{x} - \dfrac{3}{y} = 1 x 5 − y 3 = 1
Answer
Substituting 1 x = a and 1 y = b \dfrac{1}{x} = a \text{ and } \dfrac{1}{y} = b x 1 = a and y 1 = b
3 2 a + 2 3 b = 5 \dfrac{3}{2}a + \dfrac{2}{3}b = 5 2 3 a + 3 2 b = 5
5a - 3b = 1 .......(ii)
Solving eq (i) we get,
⇒ 3 2 a + 2 3 b = 5 ⇒ 9 a + 4 b 6 = 5 ⇒ 9 a + 4 b = 30....... ( i i i ) \Rightarrow \dfrac{3}{2}a + \dfrac{2}{3}b = 5 \\[1em] \Rightarrow \dfrac{9a + 4b}{6} = 5 \\[1em] \Rightarrow 9a + 4b = 30 .......(iii) ⇒ 2 3 a + 3 2 b = 5 ⇒ 6 9 a + 4 b = 5 ⇒ 9 a + 4 b = 30....... ( iii )
Multiplying eq. (ii) by 4 we get,
20a - 12b = 4 .......(iv)
Multiplying eq. (iii) by 3 we get,
27a + 12b = 90 .......(v)
Adding eq. (iv) and (v) we get,
⇒ 20a - 12b + 27a + 12b = 4 + 90
⇒ 47a = 94
⇒ a = 2.
∴ 1 x = 2 ⇒ x = 1 2 . \therefore \dfrac{1}{x} = 2 \\[1em] \Rightarrow x = \dfrac{1}{2}. ∴ x 1 = 2 ⇒ x = 2 1 .
Substituting value of a in eq. (ii) we get,
⇒ 5(2) - 3b = 1
⇒ 10 - 3b = 1
⇒ 3b = 10 - 1
⇒ 3b = 9
⇒ b = 3.
∴ 1 y = 3 ⇒ y = 1 3 . \therefore \dfrac{1}{y} = 3 \\[1em] \Rightarrow y = \dfrac{1}{3}. ∴ y 1 = 3 ⇒ y = 3 1 .
Hence, x = 1 2 and y = 1 3 . x = \dfrac{1}{2} \text{ and } y = \dfrac{1}{3}. x = 2 1 and y = 3 1 .
Solve the following pairs of linear equations:
7 x − 2 y x y = 5 \dfrac{7x - 2y}{xy} = 5 x y 7 x − 2 y = 5
8 x + 7 y x y = 15 \dfrac{8x + 7y}{xy} = 15 x y 8 x + 7 y = 15
Answer
Given,
7 x − 2 y x y = 5 or 7 y − 2 x = 5 \dfrac{7x - 2y}{xy} = 5 \text{ or } \dfrac{7}{y} - \dfrac{2}{x} = 5 x y 7 x − 2 y = 5 or y 7 − x 2 = 5
8 x + 7 y x y = 15 or 8 y + 7 x = 15 \dfrac{8x + 7y}{xy} = 15 \text{ or } \dfrac{8}{y} + \dfrac{7}{x} = 15 x y 8 x + 7 y = 15 or y 8 + x 7 = 15
Substituting 1 x = a and 1 y = b \dfrac{1}{x} = a \text{ and } \dfrac{1}{y} = b x 1 = a and y 1 = b
7b - 2a = 5 ......(i)
8b + 7a = 15 ......(ii)
Multiplying eq. (i) by 7 and eq. (ii) by 2 we get,
49b - 14a = 35 ......(iii)
16b + 14a = 30 .......(iv)
Adding equations (iii) and (iv) we get,
⇒ 49b - 14a + 16b + 14a = 35 + 30
⇒ 65b = 65
⇒ b = 1.
∴ 1 y = 1 ⇒ y = 1. \therefore \dfrac{1}{y} = 1 \\[1em] \Rightarrow y = 1. ∴ y 1 = 1 ⇒ y = 1.
Substituting value of b in eq (i) we get,
⇒ 7(1) - 2a = 5
⇒ 7 - 2a = 5
⇒ 2a = 7 - 5
⇒ 2a = 2
⇒ a = 1.
∴ 1 x = 1 ⇒ x = 1. \therefore \dfrac{1}{x} = 1 \\[1em] \Rightarrow x = 1. ∴ x 1 = 1 ⇒ x = 1.
Hence, x = 1 and y = 1.
Solve the following pairs of linear equations:
99x + 101y = 499xy
101x + 99y = 501xy
Answer
Given,
99x + 101y = 499xy
101x + 99y = 501xy
First we note that x = 0, y = 0 is a solution of equations.
Now when x ≠ 0 and y ≠ 0.
Dividing the above equations by xy we get,
99 y + 101 x = 499 \dfrac{99}{y} + \dfrac{101}{x} = 499 y 99 + x 101 = 499
101 y + 99 x = 501 \dfrac{101}{y} + \dfrac{99}{x} = 501 y 101 + x 99 = 501
Substituting 1 x = p and 1 y = q \dfrac{1}{x} = p \text{ and } \dfrac{1}{y} = q x 1 = p and y 1 = q
9999q + 10201p = 50399 ......(iii)
9999q + 9801p = 49599 ......(iv)
Subtracting (iv) from (iii) we get,
⇒ 9999q + 10201p - (9999q + 9801p) = 50399 - 49599
⇒ 9999q - 9999q + 10201p - 9801p = 800
⇒ 400p = 800
⇒ p = 2.
∴ 1 x = 2 or x = 1 2 . \therefore \dfrac{1}{x} = 2 \text{ or } x = \dfrac{1}{2}. ∴ x 1 = 2 or x = 2 1 .
Substituting value of p in (iii) we get,
⇒ 9999q + 10201(2) = 50399
⇒ 9999q + 20402 = 50399
⇒ 9999q = 29997
⇒ q = 3.
∴ 1 y = 3 or y = 1 3 \therefore \dfrac{1}{y} = 3 \text{ or } y = \dfrac{1}{3} ∴ y 1 = 3 or y = 3 1
Hence, x = 0, y = 0 and x = 1 2 and y = 1 3 . \dfrac{1}{2} \text{ and } y = \dfrac{1}{3}. 2 1 and y = 3 1 .
Solve the following pairs of linear equations:
3x + 14y = 5xy
21y - x = 2xy
Answer
Given,
3x + 14y = 5xy .......(i)
21y - x = 2xy ........(ii)
First we note that x = 0, y = 0 is a solution of the equations.
Now when x ≠ 0 and y ≠ 0.
Dividing eq. (i) by xy we get,
⇒ 3 x + 14 y = 5 x y ⇒ 3 x x y + 14 y x y = 5 x y x y ⇒ 3 y + 14 x = 5...... ( i i i ) \Rightarrow 3x + 14y = 5xy \\[1em] \Rightarrow \dfrac{3x}{xy} + \dfrac{14y}{xy} = \dfrac{5xy}{xy} \\[1em] \Rightarrow \dfrac{3}{y} + \dfrac{14}{x} = 5 ......(iii) ⇒ 3 x + 14 y = 5 x y ⇒ x y 3 x + x y 14 y = x y 5 x y ⇒ y 3 + x 14 = 5...... ( iii )
Dividing eq. (ii) by xy we get,
⇒ 21 y − x = 2 x y ⇒ 21 y x y − x x y = 2 ⇒ 21 x − 1 y = 2....... ( i v ) \Rightarrow 21y - x = 2xy \\[1em] \Rightarrow \dfrac{21y}{xy} - \dfrac{x}{xy} = 2 \\[1em] \Rightarrow \dfrac{21}{x} - \dfrac{1}{y} = 2 .......(iv) ⇒ 21 y − x = 2 x y ⇒ x y 21 y − x y x = 2 ⇒ x 21 − y 1 = 2....... ( i v )
Substituting 1 x = a and 1 y = b \dfrac{1}{x} = a \text{ and } \dfrac{1}{y} = b x 1 = a and y 1 = b
3b + 14a = 5 ........(v)
21a - b = 2 .........(vi)
Multiplying eq. (vi) by 3 we get,
63a - 3b = 6 .......(vii)
Adding eq. (v) and (vii) we get,
⇒ 3b + 14a + (63a - 3b) = 5 + 611 77 = 1 7 \dfrac{11}{77} = \dfrac{1}{7} 77 11 = 7 1
∴ 1 x = 1 7 ⇒ x = 7. \therefore \dfrac{1}{x} = \dfrac{1}{7} \\[1em] \Rightarrow x = 7. ∴ x 1 = 7 1 ⇒ x = 7.
Substituting value of a in eq. (v) we get,
⇒ 3b + 14 × 1 7 14\times \dfrac{1}{7} 14 × 7 1
⇒ 3b + 2 = 5
⇒ 3b = 3
⇒ b = 1.
∴ 1 y = 1 ⇒ y = 1. \therefore \dfrac{1}{y} = 1 \\[1em] \Rightarrow y = 1. ∴ y 1 = 1 ⇒ y = 1.
Hence, x = 0, y = 0 and x = 7, y = 1.
Solve the following pairs of linear equations:
3x + 5y = 4xy
2y - x = xy.
Answer
Given,
3x + 5y = 4xy ........(i)
2y - x = xy .......(ii)
First we note that x = 0, y = 0 is a solution of equations.
Now when x ≠ 0 and y ≠ 0.
Dividing above equations by xy we get,
3 y + 5 x = 4 \dfrac{3}{y} + \dfrac{5}{x} = 4 y 3 + x 5 = 4
2 x − 1 y = 1 \dfrac{2}{x} - \dfrac{1}{y} = 1 x 2 − y 1 = 1
Substituting 1 x = a and 1 y = b \dfrac{1}{x} = a \text{ and } \dfrac{1}{y} = b x 1 = a and y 1 = b
3b + 5a = 4 .......(v)
2a - b = 1 .......(vi)
Multiplying eq. (vi) by 3 we get,
6a - 3b = 3 ........(vii)
Adding eq. (v) and (vii) we get,
⇒ 3b + 5a + 6a - 3b = 4 + 3
⇒ 11a = 7
⇒ a = 7 11 \dfrac{7}{11} 11 7
∴ 1 x = 7 11 ⇒ x = 11 7 . \therefore \dfrac{1}{x} = \dfrac{7}{11} \\[1em] \Rightarrow x = \dfrac{11}{7}. ∴ x 1 = 11 7 ⇒ x = 7 11 .
Substituting value of a in eq. (vi) we get,
⇒ 2 × 7 11 − b = 1 ⇒ 14 11 − b = 1 ⇒ b = 14 11 − 1 ⇒ b = 3 11 ∴ 1 y = 3 11 ⇒ y = 11 3 . \Rightarrow 2 \times \dfrac{7}{11} - b = 1 \\[1em] \Rightarrow \dfrac{14}{11} - b = 1 \\[1em] \Rightarrow b = \dfrac{14}{11} - 1 \\[1em] \Rightarrow b = \dfrac{3}{11} \\[1em] \therefore \dfrac{1}{y} = \dfrac{3}{11} \\[1em] \Rightarrow y = \dfrac{11}{3}. ⇒ 2 × 11 7 − b = 1 ⇒ 11 14 − b = 1 ⇒ b = 11 14 − 1 ⇒ b = 11 3 ∴ y 1 = 11 3 ⇒ y = 3 11 .
Hence, x = 0, y = 0 and x = 11 7 , y = 11 3 . \dfrac{11}{7}, y = \dfrac{11}{3}. 7 11 , y = 3 11 .
Solve the following pairs of linear equations:
20 x + 1 + 4 y − 1 = 5 \dfrac{20}{x + 1} + \dfrac{4}{y - 1} = 5 x + 1 20 + y − 1 4 = 5
10 x + 1 − 4 y − 1 = 1 \dfrac{10}{x + 1} - \dfrac{4}{y - 1} = 1 x + 1 10 − y − 1 4 = 1
Answer
Substituting 1 x + 1 = a and 1 y − 1 = b \dfrac{1}{x + 1} = a \text{ and } \dfrac{1}{y - 1} = b x + 1 1 = a and y − 1 1 = b
20a + 4b = 5 .......(i)
10a - 4b = 1 .......(ii)
Multiplying equation (ii) by 2 we get,
20a - 8b = 2 .......(iii)
Subtracting eq. (iii) from (i) we get,
⇒ 20a + 4b - (20a - 8b) = 5 - 2
⇒ 12b = 3
⇒ b = 3 12 = 1 4 \dfrac{3}{12} = \dfrac{1}{4} 12 3 = 4 1
∴ 1 y − 1 = 1 4 ⇒ y − 1 = 4 ⇒ y = 5. \therefore \dfrac{1}{y - 1} = \dfrac{1}{4} \\[1em] \Rightarrow y - 1 = 4 \\[1em] \Rightarrow y = 5. ∴ y − 1 1 = 4 1 ⇒ y − 1 = 4 ⇒ y = 5.
Substituting value of b in (iii) we get,
⇒ 20a - 8 × 1 4 8 \times \dfrac{1}{4} 8 × 4 1
⇒ 20a - 2 = 2
⇒ 20a = 2 + 2
⇒ a = 4 20 \dfrac{4}{20} 20 4
⇒ a = 1 5 \dfrac{1}{5} 5 1
∴ 1 x + 1 = 1 5 ⇒ x + 1 = 5 ⇒ x = 4. \therefore \dfrac{1}{x + 1} = \dfrac{1}{5} \\[1em] \Rightarrow x + 1 = 5 \\[1em] \Rightarrow x = 4. ∴ x + 1 1 = 5 1 ⇒ x + 1 = 5 ⇒ x = 4.
Hence, x = 4 and y = 5.
Solve the following pairs of linear equations:
3 x + y + 2 x − y = 3 \dfrac{3}{x + y} + \dfrac{2}{x - y} = 3 x + y 3 + x − y 2 = 3
2 x + y + 3 x − y = 11 3 \dfrac{2}{x + y} + \dfrac{3}{x - y} = \dfrac{11}{3} x + y 2 + x − y 3 = 3 11
Answer
Substituting 1 x + y = a and 1 x − y = b \dfrac{1}{x + y} = a \text{ and } \dfrac{1}{x - y} = b x + y 1 = a and x − y 1 = b
3a + 2b = 3 .......(i)
2a + 3b = 11 3 \dfrac{11}{3} 3 11
Multiplying eq. (i) by 2 and (ii) by 3 we get,
6a + 4b = 6 .......(iii)
6a + 9b = 11 .......(iv)
Subtracting eq. (iii) from iv we get,
⇒ 6a + 9b - (6a + 4b) = 11 - 6
⇒ 6a + 9b - 6a - 4b = 5
⇒ 5b = 5
⇒ b = 1.
∴ 1 x − y = 1 ⇒ x − y = 1....... ( v ) \therefore \dfrac{1}{x - y} = 1 \\[1em] \Rightarrow x - y = 1 .......(v) ∴ x − y 1 = 1 ⇒ x − y = 1....... ( v )
Substituting value of b in eq. (iii) we get,
⇒ 6a + 4(1) = 6
⇒ 6a + 4 = 6
⇒ 6a = 2
⇒ a = 2 6 = 1 3 . \dfrac{2}{6} = \dfrac{1}{3}. 6 2 = 3 1 .
∴ 1 x + y = 1 3 \therefore \dfrac{1}{x + y} = \dfrac{1}{3} ∴ x + y 1 = 3 1
⇒ x + y = 3
⇒ x = 3 - y.
Putting above value of x in eq. (v) we get,
⇒ (3 - y) - y = 1
⇒ 3 - 2y = 1
⇒ 2y = 3 - 1
⇒ 2y = 2
⇒ y = 1.
⇒ x = 3 - y = 3 - 1 = 2.
Hence, x = 2 and y = 1.
Solve the following pairs of linear equations:
1 2 ( 2 x + 3 y ) + 12 7 ( 3 x − 2 y ) = 1 2 \dfrac{1}{2(2x + 3y)} + \dfrac{12}{7(3x - 2y)} = \dfrac{1}{2} 2 ( 2 x + 3 y ) 1 + 7 ( 3 x − 2 y ) 12 = 2 1
7 2 x + 3 y + 4 3 x − 2 y = 2 \dfrac{7}{2x + 3y} + \dfrac{4}{3x - 2y} = 2 2 x + 3 y 7 + 3 x − 2 y 4 = 2
Answer
Substituting 1 2 x + 3 y = a and 1 3 x − 2 y = b \dfrac{1}{2x + 3y} = a \text{ and } \dfrac{1}{3x - 2y} = b 2 x + 3 y 1 = a and 3 x − 2 y 1 = b
1 2 a + 12 7 b = 1 2 \dfrac{1}{2}a + \dfrac{12}{7}b = \dfrac{1}{2} 2 1 a + 7 12 b = 2 1
7a + 4b = 2 .......(ii)
Multiplying eq. (i) by 14 we get,
7a + 24b = 7 ......(iii)
Subtracting eq. (ii) from (iii) we get,
⇒ 7a + 24b - (7a + 4b) = 7 - 2
⇒ 7a + 24b - 7a - 4b = 7 - 2
⇒ 20b = 5
⇒ b = 5 20 = 1 4 \dfrac{5}{20} = \dfrac{1}{4} 20 5 = 4 1
∴ 1 3 x − 2 y = 1 4 \therefore \dfrac{1}{3x - 2y} = \dfrac{1}{4} ∴ 3 x − 2 y 1 = 4 1
⇒ 3x - 2y = 4 ........(iv)
Substituting value of b in eq. (iii) we get,
⇒ 7a + 24 × 1 4 24 \times \dfrac{1}{4} 24 × 4 1
⇒ 7a + 6 = 7
⇒ 7a = 1
⇒ a = 1 7 \dfrac{1}{7} 7 1
∴ 1 2 x + 3 y = 1 7 \therefore \dfrac{1}{2x + 3y} = \dfrac{1}{7} ∴ 2 x + 3 y 1 = 7 1
⇒ 2x + 3y = 7 .......(v)
Multiplying eq. (iv) by 2 and eq. (v) by 3 we get,
⇒ 6x - 4y = 8 .......(vi)
⇒ 6x + 9y = 21 .......(vii)
Subtracting eq. (vi) from (vii) we get,
⇒ (6x + 9y) - (6x - 4y) = 21 - 8
⇒ 13y = 13
⇒ y = 1.
Substituting value of y in eq. (vi) we get,
⇒ 6x - 4(1) = 8
⇒ 6x = 8 + 4
⇒ 6x = 12
⇒ x = 2.
Hence, x = 2 and y = 1.
Solve the following pairs of linear equations:
1 2 ( x + 2 y ) + 5 3 ( 3 x − 2 y ) = − 3 2 \dfrac{1}{2(x + 2y)} + \dfrac{5}{3(3x - 2y)} = -\dfrac{3}{2} 2 ( x + 2 y ) 1 + 3 ( 3 x − 2 y ) 5 = − 2 3
5 4 ( x + 2 y ) − 3 5 ( 3 x − 2 y ) = 61 60 \dfrac{5}{4(x + 2y)} - \dfrac{3}{5(3x - 2y)} = \dfrac{61}{60} 4 ( x + 2 y ) 5 − 5 ( 3 x − 2 y ) 3 = 60 61
Answer
Substitute 1 x + 2 y = p and 1 3 x − 2 y = q \dfrac{1}{x + 2y} = p \text{ and } \dfrac{1}{3x - 2y} = q x + 2 y 1 = p and 3 x − 2 y 1 = q
1 2 p + 5 3 q = − 3 2 \dfrac{1}{2}p + \dfrac{5}{3}q = -\dfrac{3}{2} 2 1 p + 3 5 q = − 2 3
5 4 p − 3 5 q = 61 60 \dfrac{5}{4}p - \dfrac{3}{5}q = \dfrac{61}{60} 4 5 p − 5 3 q = 60 61
Multiplying (i) by 3 5 \dfrac{3}{5} 5 3 5 3 \dfrac{5}{3} 3 5
3 10 p + q = − 9 10 \dfrac{3}{10}p + q = -\dfrac{9}{10} 10 3 p + q = − 10 9
25 12 p − q = 61 36 \dfrac{25}{12}p - q = \dfrac{61}{36} 12 25 p − q = 36 61
Adding (iii) and (iv) we get,
⇒ 3 10 p + q + 25 12 p − q = − 9 10 + 61 36 ⇒ 18 p + 125 p 60 = − 162 + 305 180 ⇒ 143 60 p = 143 180 ⇒ p = 143 × 60 180 × 143 ⇒ p = 1 3 ∴ 1 x + 2 y = 1 3 ⇒ x + 2 y = 3....... ( v ) \Rightarrow \dfrac{3}{10}p + q + \dfrac{25}{12}p - q = -\dfrac{9}{10} + \dfrac{61}{36} \\[1em] \Rightarrow \dfrac{18p + 125p}{60} = \dfrac{-162 + 305}{180} \\[1em] \Rightarrow \dfrac{143}{60}p = \dfrac{143}{180} \\[1em] \Rightarrow p = \dfrac{143 \times 60}{180 \times 143} \\[1em] \Rightarrow p = \dfrac{1}{3} \\[1em] \therefore \dfrac{1}{x + 2y} = \dfrac{1}{3} \\[1em] \Rightarrow x + 2y = 3 .......(v) ⇒ 10 3 p + q + 12 25 p − q = − 10 9 + 36 61 ⇒ 60 18 p + 125 p = 180 − 162 + 305 ⇒ 60 143 p = 180 143 ⇒ p = 180 × 143 143 × 60 ⇒ p = 3 1 ∴ x + 2 y 1 = 3 1 ⇒ x + 2 y = 3....... ( v )
Substituting value of p in (i) we get,
⇒ 1 2 × 1 3 + 5 3 q = − 3 2 ⇒ 1 6 + 5 3 q = − 3 2 ⇒ 5 3 q = − 3 2 − 1 6 ⇒ 5 3 q = − 9 − 1 6 ⇒ 5 3 q = − 10 6 ⇒ q = − 10 × 3 6 × 5 ⇒ q = − 1 ∴ 1 3 x − 2 y = − 1 ⇒ 3 x − 2 y = − 1 ⇒ 2 y − 3 x = 1...... ( v i ) \Rightarrow \dfrac{1}{2} \times \dfrac{1}{3} + \dfrac{5}{3}q = -\dfrac{3}{2} \\[1em] \Rightarrow \dfrac{1}{6} + \dfrac{5}{3}q = -\dfrac{3}{2} \\[1em] \Rightarrow \dfrac{5}{3}q = -\dfrac{3}{2} - \dfrac{1}{6} \\[1em] \Rightarrow \dfrac{5}{3}q = \dfrac{-9 - 1}{6} \\[1em] \Rightarrow \dfrac{5}{3}q = -\dfrac{10}{6} \\[1em] \Rightarrow q = -\dfrac{10 \times 3}{6 \times 5} \\[1em] \Rightarrow q = -1 \\[1em] \therefore \dfrac{1}{3x - 2y} = -1 \\[1em] \Rightarrow 3x - 2y = -1 \\[1em] \Rightarrow 2y - 3x = 1 ......(vi) ⇒ 2 1 × 3 1 + 3 5 q = − 2 3 ⇒ 6 1 + 3 5 q = − 2 3 ⇒ 3 5 q = − 2 3 − 6 1 ⇒ 3 5 q = 6 − 9 − 1 ⇒ 3 5 q = − 6 10 ⇒ q = − 6 × 5 10 × 3 ⇒ q = − 1 ∴ 3 x − 2 y 1 = − 1 ⇒ 3 x − 2 y = − 1 ⇒ 2 y − 3 x = 1...... ( v i )
Subtracting (vi) from (v) we get,
⇒ x + 2y - (2y - 3x) = 3 - 1
⇒ x + 3x = 2
⇒ 4x = 2
⇒ x = 1 2 \dfrac{1}{2} 2 1
Substituting value of x from (v) we get,
⇒ 1 2 + 2 y = 3 ⇒ 2 y = 3 − 1 2 ⇒ 2 y = 6 − 1 2 ⇒ 2 y = 5 2 ⇒ y = 5 4 . \Rightarrow \dfrac{1}{2} + 2y = 3 \\[1em] \Rightarrow 2y = 3 - \dfrac{1}{2} \\[1em] \Rightarrow 2y = \dfrac{6 - 1}{2} \\[1em] \Rightarrow 2y = \dfrac{5}{2} \\[1em] \Rightarrow y = \dfrac{5}{4}. ⇒ 2 1 + 2 y = 3 ⇒ 2 y = 3 − 2 1 ⇒ 2 y = 2 6 − 1 ⇒ 2 y = 2 5 ⇒ y = 4 5 .
Hence, x = 1 2 and y = 5 4 . \dfrac{1}{2} \text{ and } y = \dfrac{5}{4}. 2 1 and y = 4 5 .
Multiple Choice Questions
If x = 3, y = k is a solution of the equation 3x - 4y + 7 = 0, then the value of k is
16
-16
4
-4
Answer
Substituting x = 3 and y = k in 3x - 4y + 7 = 0 we get,
⇒ 3(3) - 4(k) + 7 = 0
⇒ 9 - 4k + 7 = 0
⇒ 4k = 16
⇒ k = 4.
Hence, Option 3 is the correct option.
The solution of the pair of linear equations 2x - y = 5 and 5x - y = 11 is
x = -1, y = 2
x = 2, y = -1
x = 0, y = -5
x = 5 2 \dfrac{5}{2} 2 5
Answer
Given,
2x - y = 5 ......(i)
5x - y = 11 .....(ii)
Subtracting eq. (i) from (ii) we get,
⇒ 5x - y - (2x - y) = 11 - 5
⇒ 5x - y - 2x + y = 6
⇒ 3x = 6
⇒ x = 2.
Substituting value of x in eq. (i),
⇒ 2(2) - y = 5
⇒ 4 - y = 5
⇒ y = 4 - 5
⇒ y = -1.
Hence, Option 2 is the correct option.
If x = a, y = b is the solution of the equations x - y = 2 and x + y = 4, then the values of a and b are respectively,
3 and 5
5 and 3
3 and 1
-1 and -3
Answer
Given,
x - y = 2 ......(i)
x + y = 4 ......(ii)
Adding both the equations we get,
⇒ x - y + x + y = 2 + 4
⇒ 2x = 6
⇒ x = 3.
Substituting value of x in (i),
⇒ 3 - y = 2
⇒ y = 3 - 2
⇒ y = 1.
Hence, Option 3 is the correct option.
The solution of the system of equations 4 x + 5 y = 7 and 3 x + 4 y = 5 \dfrac{4}{x} + 5y = 7 \text{ and } \dfrac{3}{x} + 4y = 5 x 4 + 5 y = 7 and x 3 + 4 y = 5
x = 1 3 , y = − 1 x = \dfrac{1}{3}, y = -1 x = 3 1 , y = − 1
x = − 1 3 , y = 1 x = -\dfrac{1}{3}, y = 1 x = − 3 1 , y = 1
x = 3, y = -1
x = -3, y = 1
Answer
Given,
4 x + 5 y = 7 \dfrac{4}{x} + 5y = 7 x 4 + 5 y = 7
3 x + 4 y = 5 \dfrac{3}{x} + 4y = 5 x 3 + 4 y = 5
Multiplying eq. (i) by 4 and eq. (ii) by 5 we get,
16 x + 20 y = 28 \dfrac{16}{x} + 20y = 28 x 16 + 20 y = 28
15 x + 20 y = 25 \dfrac{15}{x} + 20y = 25 x 15 + 20 y = 25
Subtracting eq. (iv) from (iii) we get,
⇒ 16 x + 20 y − ( 15 x + 20 y ) = 28 − 25 ⇒ 16 x − 15 x = 3 ⇒ 1 x = 3 ⇒ x = 1 3 . \Rightarrow \dfrac{16}{x} + 20y - \Big(\dfrac{15}{x} + 20y\Big) = 28 - 25 \\[1em] \Rightarrow \dfrac{16}{x} - \dfrac{15}{x} = 3 \\[1em] \Rightarrow \dfrac{1}{x} = 3 \\[1em] \Rightarrow x = \dfrac{1}{3}. ⇒ x 16 + 20 y − ( x 15 + 20 y ) = 28 − 25 ⇒ x 16 − x 15 = 3 ⇒ x 1 = 3 ⇒ x = 3 1 .
Substituting value of x in (ii),
⇒ 3 1 3 + 4 y = 5 ⇒ 9 + 4 y = 5 ⇒ 4 y = 5 − 9 ⇒ 4 y = − 4 ⇒ y = − 1. \Rightarrow \dfrac{3}{\dfrac{1}{3}} + 4y = 5 \\[1em] \Rightarrow 9 + 4y = 5 \\[1em] \Rightarrow 4y = 5 - 9 \\[1em] \Rightarrow 4y = -4 \\[1em] \Rightarrow y = -1. ⇒ 3 1 3 + 4 y = 5 ⇒ 9 + 4 y = 5 ⇒ 4 y = 5 − 9 ⇒ 4 y = − 4 ⇒ y = − 1.
Hence, Option 1 is the correct option.
A pair of linear equations which has a unique solution x = 2, y = -3 is
x + y = -1
2x + 5y = -11
2x - y = 1
x - 4y - 14 = 0
Answer
Substituting x = 2, y = -3 in x - 4y - 14 = 0 we get,
= 2 - 4(-3) - 14
= 2 + 12 - 14
= 0.
Substituting x = 2, y = -3 in 5x - y - 13 = 0 we get,
= 5(2) - (-3) - 13
= 10 + 3 - 13
= 13 - 13
= 0.
Hence, Option 4 is the correct option.
Consider the following two statements.
Statement 1: A solution to linear equation 5x - 2y = 1 is x = 3, y = 7.
Statement 2: The linear equation 5x - 2y = 1 has a unique solution.
Which of the following is valid?
Both the statements are true.
Both the statements are false.
Statement 1 is true, and Statement 2 is false.
Statement 1 is false, and Statement 2 is true.
Answer
If we substitute x = 3 and y = 7 into the equation 5x - 2y = 1, we get
Taking L.H.S.
⇒ 5.(3) - 2.(7)
⇒ 15 - 14
⇒ 1.
Since, L.H.S. = R.H.S.
This confirms that (3, 7) is indeed a solution.
∴ Statement 1 is true.
A single linear equation with two variables (like x and y) represents a line in a coordinate plane.
A line has infinitely many points on it, and each point corresponds to a solution of the equation.
Therefore, a single linear equation has infinitely many solutions, not just one.
∴ Statement 2 is false.
∴ Statement 1 is true, and Statement 2 is false.
Hence, option 3 is the correct option.
Assertion Reason Type Questions
Assertion (A): A solution of x - y = 1, 2x + y = 7 2 \dfrac{7}{2} 2 7 3 2 \dfrac{3}{2} 2 3 1 2 \dfrac{1}{2} 2 1
Reason (R): One of the methods of solving a pair of linear equations is elimination method.
Assertion (A) is true, Reason (R) is false.
Assertion (A) is false, Reason (R) is true.
Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer
One of the methods of solving a pair of linear equations is elimination method.
∴ Reason (R) is true.
Given, x - y = 1 .................(1)
2x + y = 7 2 \dfrac{7}{2} 2 7
Adding equation (1) and (2) we get
⇒ ( x − y ) + ( 2 x + y ) = 1 + 7 2 ⇒ x − y + 2 x + y = 2 2 + 7 2 ⇒ 3 x = 9 2 ⇒ x = 9 2 × 3 ⇒ x = 3 2 \Rightarrow (x - y) + (2x + y) = 1 + \dfrac{7}{2}\\[1em] \Rightarrow x - y + 2x + y = \dfrac{2}{2} + \dfrac{7}{2}\\[1em] \Rightarrow 3x = \dfrac{9}{2}\\[1em] \Rightarrow x = \dfrac{9}{2 \times 3}\\[1em] \Rightarrow x = \dfrac{3}{2} ⇒ ( x − y ) + ( 2 x + y ) = 1 + 2 7 ⇒ x − y + 2 x + y = 2 2 + 2 7 ⇒ 3 x = 2 9 ⇒ x = 2 × 3 9 ⇒ x = 2 3
Substituting the value of x in equation (1), we get :
⇒ x - y = 1
⇒ 3 2 − y = 1 ⇒ 3 2 − 1 = y ⇒ 3 − 2 2 = y ⇒ y = 1 2 . \Rightarrow \dfrac{3}{2} - y = 1\\[1em] \Rightarrow \dfrac{3}{2} - 1 = y\\[1em] \Rightarrow \dfrac{3 - 2}{2} = y\\[1em] \Rightarrow y = \dfrac{1}{2}. ⇒ 2 3 − y = 1 ⇒ 2 3 − 1 = y ⇒ 2 3 − 2 = y ⇒ y = 2 1 .
So, x = 3 2 \dfrac{3}{2} 2 3 1 2 \dfrac{1}{2} 2 1
∴ Assertion (A) is true.
∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason (or explanation) for Assertion (A).
Hence, option 3 is the correct option.
Assertion (A): Solving 2 x − 3 y \sqrt{2}x - \sqrt{3}y 2 x − 3 y 3 x + 2 y \sqrt{3}x + \sqrt{2}y 3 x + 2 y 3 \sqrt{3} 3 2 \sqrt{2} 2
Reason (R): We can use cross-multiplication method to solve a pair of linear equations.
Assertion (A) is true, Reason (R) is false.
Assertion (A) is false, Reason (R) is true.
Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer
We can use cross-multiplication method to solve a pair of linear equations.
∴ Reason (R) is true.
Given, equations can be written as :
2 x − 3 y − 0 \sqrt{2}x - \sqrt{3}y - 0 2 x − 3 y − 0
3 x + 2 y \sqrt{3}x + \sqrt{2}y 3 x + 2 y
By cross multiplication method,
∴ x − 3 × ( − 5 ) − 2 × 0 = y 0 × 3 − ( − 5 ) × 2 = 1 2 × 2 − 3 × ( − 3 ) ⇒ x 5 3 − 0 = y 0 + 5 2 = 1 2 + 3 ⇒ x 5 3 = y 5 2 = 1 5 ⇒ x 5 3 = 1 5 and y 5 2 = 1 5 ⇒ x = 5 3 5 and y = 5 2 5 ⇒ x = 3 and y = 2 . \therefore \dfrac{x}{-\sqrt{3} \times (-5) - \sqrt{2} \times 0} = \dfrac{y}{0 \times \sqrt{3} - (-5) \times \sqrt{2}} = \dfrac{1}{\sqrt{2} \times \sqrt{2} - \sqrt{3} \times (-\sqrt{3})}\\[1em] \Rightarrow \dfrac{x}{5\sqrt{3} - 0} = \dfrac{y}{0 + 5 \sqrt{2}} = \dfrac{1}{2 + 3}\\[1em] \Rightarrow \dfrac{x}{5\sqrt{3}} = \dfrac{y}{5 \sqrt{2}} = \dfrac{1}{5}\\[1em] \Rightarrow \dfrac{x}{5\sqrt{3}} = \dfrac{1}{5} \text{ and } \dfrac{y}{5 \sqrt{2}} = \dfrac{1}{5} \\[1em] \Rightarrow x = \dfrac{5\sqrt{3}}{5} \text{ and } y = \dfrac{5\sqrt{2}}{5}\\[1em] \Rightarrow x = \sqrt{3} \text{ and } y = \sqrt{2}. ∴ − 3 × ( − 5 ) − 2 × 0 x = 0 × 3 − ( − 5 ) × 2 y = 2 × 2 − 3 × ( − 3 ) 1 ⇒ 5 3 − 0 x = 0 + 5 2 y = 2 + 3 1 ⇒ 5 3 x = 5 2 y = 5 1 ⇒ 5 3 x = 5 1 and 5 2 y = 5 1 ⇒ x = 5 5 3 and y = 5 5 2 ⇒ x = 3 and y = 2 .
So, the solution are x = 3 \sqrt{3} 3 2 \sqrt{2} 2
∴ Assertion (A) is true.
∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
Hence, option 3 is the correct option.
Solve the following simultaneous linear equations:
2x - 3 4 y \dfrac{3}{4}y 4 3 y
5x - 2y = 7
Answer
Given,
2x - 3 4 y \dfrac{3}{4}y 4 3 y
5x - 2y = 7 ..........(ii)
Multiplying eq. (i) by 5 and (ii) by 2 we get,
10x - 15 4 y \dfrac{15}{4}y 4 15 y
10x - 4y = 14 ........(iv)
Subtracting eq. (iv) from (iii),
⇒ 10 x − 15 4 y − ( 10 x − 4 y ) = 15 − 14 ⇒ 4 y − 15 4 y = 1 ⇒ 16 y − 15 y 4 = 1 ⇒ y 4 = 1 ⇒ y = 4. \Rightarrow 10x - \dfrac{15}{4}y - (10x - 4y) = 15 - 14 \\[1em] \Rightarrow 4y - \dfrac{15}{4}y = 1 \\[1em] \Rightarrow \dfrac{16y - 15y}{4} = 1 \\[1em] \Rightarrow \dfrac{y}{4} = 1 \\[1em] \Rightarrow y = 4. ⇒ 10 x − 4 15 y − ( 10 x − 4 y ) = 15 − 14 ⇒ 4 y − 4 15 y = 1 ⇒ 4 16 y − 15 y = 1 ⇒ 4 y = 1 ⇒ y = 4.
Substituting value of y in eq. (ii) we get,
⇒ 5x - 2(4) = 7
⇒ 5x - 8 = 7
⇒ 5x = 15
⇒ x = 3.
Hence, x = 3 and y = 4.
Solve the following simultaneous linear equations:
2(x - 4) = 9y + 2
x - 6y = 2.
Answer
Given,
2(x - 4) = 9y + 2 ......(i)
x - 6y = 2 or x = 2 + 6y......(ii)
Substituting value of x from eq. (ii) in (i),
⇒ 2[(2 + 6y) - 4] = 9y + 2
⇒ 2[6y - 2] = 9y + 2
⇒ 12y - 4 = 9y + 2
⇒ 12y - 9y = 2 + 4
⇒ 3y = 6
⇒ y = 2.
x = 2 + 6y = 2 + 6(2) = 2 + 12 = 14.
Hence, x = 14 and y = 2.
Solve the following simultaneous linear equations:
97x + 53y = 177
53x + 97y = 573
Answer
Given,
97x + 53y = 177 .......(i)
53x + 97y = 573 .......(ii)
Multiplying eq. (i) by 53 and (ii) by 97,
5141x + 2809y = 9381 .......(iii)
5141x + 9409y = 55581 ......(iv)
Subtracting eq. (iii) from (iv) we get,
⇒ 5141x + 9409y - (5141x + 2809y) = 55581 - 9381
⇒ 6600y = 46200
⇒ y = 46200 6600 \dfrac{46200}{6600} 6600 46200
Substituting value of y in (ii) we get,
⇒ 53x + 97(7) = 573
⇒ 53x + 679 = 573
⇒ 53x = 573 - 679
⇒ 53x = -106
⇒ x = -2.
Hence, x = -2 and y = 7.
67x - 58y = 192
58x - 67y = 183
Answer
Given,
Equations :
67x - 58y = 192 ........(1)
58x - 67y = 183 ........(2)
Multiplying equation (1) by 58, we get :
⇒ 58(67x - 58y) = 192 × 58
⇒ 3886x - 3364y = 11136 ........(3)
Multiplying equation (2) by 67, we get :
⇒ 67(58x - 67y) = 183 × 67
⇒ 3886x - 4489y = 12261 .........(4)
Subtracting equation (3) from (4), we get :
⇒ 3886x - 4489y - (3886x - 3364y) = 12261 - 11136
⇒ 3886x - 3886x - 4489y + 3364y = 1125
⇒ -1125y = 1125
⇒ y = − 1125 1125 -\dfrac{1125}{1125} − 1125 1125
Substituting value of y in equation (1), we get :
⇒ 67x - 58(-1) = 192
⇒ 67x + 58 = 192
⇒ 67x = 192 - 58
⇒ 67x = 134
⇒ x = 134 67 \dfrac{134}{67} 67 134
Hence, x = 2 and y = -1.
Solve the following simultaneous linear equations:
x + y = 7xy
2x - 3y + xy = 0
Answer
Given,
x + y = 7xy
2x - 3y + xy = 0
First we note that x = 0, y = 0 is a solution of equations.
Now when x ≠ 0 and y ≠ 0.
Dividing x + y = 7xy by xy,
⇒ x x y + y x y = 7 x y x y ⇒ 1 y + 1 x = 7...... ( i ) \Rightarrow \dfrac{x}{xy} + \dfrac{y}{xy} = \dfrac{7xy}{xy} \\[1em] \Rightarrow \dfrac{1}{y} + \dfrac{1}{x} = 7 ......(i) ⇒ x y x + x y y = x y 7 x y ⇒ y 1 + x 1 = 7...... ( i )
Dividing 2x - 3y + xy = 0 by xy,
⇒ 2 x x y − 3 y x y + x y x y = 0 ⇒ 2 y − 3 x + 1 = 0 ⇒ 2 y − 3 x = − 1...... ( i i ) \Rightarrow \dfrac{2x}{xy} - \dfrac{3y}{xy} + \dfrac{xy}{xy} = 0 \\[1em] \Rightarrow \dfrac{2}{y} - \dfrac{3}{x} + 1 = 0 \\[1em] \Rightarrow \dfrac{2}{y} - \dfrac{3}{x} = -1 ......(ii) ⇒ x y 2 x − x y 3 y + x y x y = 0 ⇒ y 2 − x 3 + 1 = 0 ⇒ y 2 − x 3 = − 1...... ( ii )
Substituting 1 x \dfrac{1}{x} x 1 1 y \dfrac{1}{y} y 1
q + p = 7 .......(iii)
2q - 3p = -1 ......(iv)
Multiplying (iii) by 3 we get,
3q + 3p = 21 .......(v)
Adding (iv) and (v) we get,
⇒ 2q - 3p + (3q + 3p) = -1 + 21
⇒ 2q + 3q = 20
⇒ 5q = 20
⇒ q = 20 5 \dfrac{20}{5} 5 20
⇒ q = 4.
∴ 1 y = 4 or y = 1 4 \therefore \dfrac{1}{y} = 4 \text{ or } y = \dfrac{1}{4} ∴ y 1 = 4 or y = 4 1
Substituting value of q in (iii) we get,
⇒ 4 + p = 7
⇒ p = 3.
∴ 1 x = 3 or x = 1 3 \therefore \dfrac{1}{x} = 3 \text{ or } x = \dfrac{1}{3} ∴ x 1 = 3 or x = 3 1
Hence, x = 0, y = 0 and x = 1 3 , y = 1 4 \dfrac{1}{3}, y = \dfrac{1}{4} 3 1 , y = 4 1
Solve the following simultaneous linear equations:
30 x − y + 44 x + y = 10 \dfrac{30}{x - y} + \dfrac{44}{x + y} = 10 x − y 30 + x + y 44 = 10
40 x − y + 55 x + y = 13 \dfrac{40}{x - y} + \dfrac{55}{x + y} = 13 x − y 40 + x + y 55 = 13
Answer
Substituting 1 x − y = a and 1 x + y = b \dfrac{1}{x - y} = a \text{ and } \dfrac{1}{x + y} = b x − y 1 = a and x + y 1 = b
30a + 44b = 10 .......(i)
40a + 55b = 13 .......(ii)
Multiplying (i) by 4 and (ii) by 3 we get,
120a + 176b = 40 .......(iii)
120a + 165b = 39 .......(iv)
Subtracting eq. (iii) from (iv) we get,
⇒ 120a + 165b - (120a + 176b) = 39 - 40
⇒ 165b - 176b = -1
⇒ -11b = -1
⇒ b = 1 11 \dfrac{1}{11} 11 1
∴ 1 x + y = 1 11 ⇒ x + y = 11......... ( v ) \therefore \dfrac{1}{x + y} = \dfrac{1}{11} \\[1em] \Rightarrow x + y = 11 .........(v) ∴ x + y 1 = 11 1 ⇒ x + y = 11......... ( v )
Substituting value of b in (iv),
⇒ 120a + 165 × 1 11 165 \times \dfrac{1}{11} 165 × 11 1
⇒ 120a + 15 = 39
⇒ 120a = 24
⇒ a = 24 120 = 1 5 \dfrac{24}{120} = \dfrac{1}{5} 120 24 = 5 1
∴ 1 x − y = 1 5 ⇒ x − y = 5........ ( v i ) \therefore \dfrac{1}{x - y} = \dfrac{1}{5} \\[1em] \Rightarrow x - y = 5 ........(vi) ∴ x − y 1 = 5 1 ⇒ x − y = 5........ ( v i )
Adding eq. (v) and (vi) we get,
⇒ x + y + (x - y) = 11 + 5
⇒ 2x = 16
⇒ x = 8.
Substituting value of x in (v)
⇒ x + y = 11
⇒ 8 + y = 11
⇒ y = 11 - 8 = 3.
Hence, x = 8 and y = 3.
Solve the following simultaneous linear equations:
ax + by = a - b
bx - ay = a + b
Answer
Given,
ax + by = a - b .......(i)
bx - ay = a + b .......(ii)
Multiplying eq. (i) by b and (ii) by a we get,
abx + b2 y = ab - b2 ......(iii)
abx - a2 y = a2 + ab .......(iv)
Subtracting eq. (iv) from (iii),
⇒ abx + b2 y - (abx - a2 y) = ab - b2 - (a2 + ab)
⇒ b2 y + a2 y = -b2 - a2
⇒ y(b2 + a2 ) = -(b2 + a2 )
⇒ y = -1.
Substituting value of y in (i),
⇒ ax + b(-1) = a - b
⇒ ax - b = a - b
⇒ ax = a - b + b
⇒ ax = a
⇒ x = 1.
Hence, x = 1 and y = -1.
Solve the following simultaneous linear equations:
3x + 2y = 2xy
1 x + 2 y = 1 1 6 \dfrac{1}{x} + \dfrac{2}{y} = 1\dfrac{1}{6} x 1 + y 2 = 1 6 1
Answer
Given,
3x + 2y = 2xy .......(i)
1 x + 2 y = 1 1 6 \dfrac{1}{x} + \dfrac{2}{y} = 1\dfrac{1}{6} x 1 + y 2 = 1 6 1
Dividing eq. (i) by xy we get,
3 y + 2 x = 2 \dfrac{3}{y} + \dfrac{2}{x} = 2 y 3 + x 2 = 2
Substituting 1 x \dfrac{1}{x} x 1 1 y \dfrac{1}{y} y 1
p + 2q = 7 6 \dfrac{7}{6} 6 7
3q + 2p = 2 ........(v)
Multiplying (iv) by 6 and (v) by 3 we get,
6p + 12q = 7 ........(vi)
9q + 6p = 6 .......(vii)
Subtracting (vii) from (vi) we get,
6p + 12q - (9q + 6p) = 7 - 6
3q = 1
q = 1 3 \dfrac{1}{3} 3 1
∴ 1 y = 1 3 or y = 3. \therefore \dfrac{1}{y} = \dfrac{1}{3} \text{ or } y = 3. ∴ y 1 = 3 1 or y = 3.
Substituting value of q in (iv) we get,
⇒ p + 2 × 1 3 = 7 6 ⇒ p + 2 3 = 7 6 ⇒ p = 7 6 − 2 3 ⇒ p = 7 − 4 6 ⇒ p = 3 6 ∴ 1 x = 3 6 ⇒ x = 6 3 = 2. \Rightarrow p + 2 \times \dfrac{1}{3} = \dfrac{7}{6} \\[1em] \Rightarrow p + \dfrac{2}{3} = \dfrac{7}{6} \\[1em] \Rightarrow p = \dfrac{7}{6} - \dfrac{2}{3} \\[1em] \Rightarrow p = \dfrac{7 - 4}{6} \\[1em] \Rightarrow p = \dfrac{3}{6} \\[1em] \therefore \dfrac{1}{x} = \dfrac{3}{6} \\[1em] \Rightarrow x = \dfrac{6}{3} = 2. ⇒ p + 2 × 3 1 = 6 7 ⇒ p + 3 2 = 6 7 ⇒ p = 6 7 − 3 2 ⇒ p = 6 7 − 4 ⇒ p = 6 3 ∴ x 1 = 6 3 ⇒ x = 3 6 = 2.
Hence, x = 2 and y = 3.
2 3 ( 2 x − y ) + 1 2 ( x + 2 y ) = 5 12 , 1 2 x − y − 2 x + 2 y = 1 6 \dfrac{2}{3(2x - y)} + \dfrac{1}{2(x + 2y)} = \dfrac{5}{12}, \dfrac{1}{2x - y} - \dfrac{2}{x + 2y} = \dfrac{1}{6} 3 ( 2 x − y ) 2 + 2 ( x + 2 y ) 1 = 12 5 , 2 x − y 1 − x + 2 y 2 = 6 1
Answer
Substituting 1 2 x − y \dfrac{1}{2x - y} 2 x − y 1 1 x + 2 y \dfrac{1}{x + 2y} x + 2 y 1
⇒ 2 3 a + 1 2 b = 5 12 \dfrac{2}{3}a + \dfrac{1}{2}b = \dfrac{5}{12} 3 2 a + 2 1 b = 12 5
⇒ a - 2b = 1 6 \dfrac{1}{6} 6 1
Multiplying equation (1) by 3 2 \dfrac{3}{2} 2 3
⇒ 3 2 ( 2 3 a + 1 2 b ) = 3 2 × 5 12 ⇒ a + 3 4 b = 5 8 ...........(3) \Rightarrow \dfrac{3}{2}\Big(\dfrac{2}{3}a + \dfrac{1}{2}b\Big) = \dfrac{3}{2} \times \dfrac{5}{12} \\[1em] \Rightarrow a + \dfrac{3}{4}b = \dfrac{5}{8} \text{ ...........(3)} ⇒ 2 3 ( 3 2 a + 2 1 b ) = 2 3 × 12 5 ⇒ a + 4 3 b = 8 5 ...........(3)
Subtracting equation (2) from (3), we get :
⇒ a + 3 4 b − ( a − 2 b ) = 5 8 − 1 6 ⇒ a − a + 3 4 b + 2 b = 15 − 4 24 ⇒ 3 b + 8 b 4 = 11 24 ⇒ 11 b 4 = 11 24 ⇒ b = 11 24 × 4 11 = 1 6 . ∴ 1 x + 2 y = 1 6 ⇒ x + 2 y = 6 ⇒ x = 6 − 2 y ............(4) \Rightarrow a + \dfrac{3}{4}b - (a - 2b) = \dfrac{5}{8} - \dfrac{1}{6} \\[1em] \Rightarrow a - a + \dfrac{3}{4}b + 2b = \dfrac{15 - 4}{24} \\[1em] \Rightarrow \dfrac{3b + 8b}{4} = \dfrac{11}{24} \\[1em] \Rightarrow \dfrac{11b}{4} = \dfrac{11}{24} \\[1em] \Rightarrow b = \dfrac{11}{24} \times \dfrac{4}{11} = \dfrac{1}{6}. \\[1em] \therefore \dfrac{1}{x + 2y} = \dfrac{1}{6} \\[1em] \Rightarrow x + 2y = 6 \\[1em] \Rightarrow x = 6 - 2y \text{ ............(4)} ⇒ a + 4 3 b − ( a − 2 b ) = 8 5 − 6 1 ⇒ a − a + 4 3 b + 2 b = 24 15 − 4 ⇒ 4 3 b + 8 b = 24 11 ⇒ 4 11 b = 24 11 ⇒ b = 24 11 × 11 4 = 6 1 . ∴ x + 2 y 1 = 6 1 ⇒ x + 2 y = 6 ⇒ x = 6 − 2 y ............(4)
Substituting b = 1 6 \dfrac{1}{6} 6 1
⇒ a − 2 × 1 6 = 1 6 ⇒ a − 1 3 = 1 6 ⇒ a = 1 6 + 1 3 ⇒ a = 1 + 2 6 ⇒ a = 3 6 ⇒ a = 1 2 ∴ 1 2 x − y = 1 2 ⇒ 2 x − y = 2 ......(5) \Rightarrow a - 2 \times \dfrac{1}{6} = \dfrac{1}{6} \\[1em] \Rightarrow a - \dfrac{1}{3} = \dfrac{1}{6} \\[1em] \Rightarrow a = \dfrac{1}{6} + \dfrac{1}{3} \\[1em] \Rightarrow a = \dfrac{1 + 2}{6} \\[1em] \Rightarrow a = \dfrac{3}{6} \\[1em] \Rightarrow a = \dfrac{1}{2} \\[1em] \therefore \dfrac{1}{2x - y} = \dfrac{1}{2}\\[1em] \Rightarrow 2x - y = 2 \text{ ......(5)} ⇒ a − 2 × 6 1 = 6 1 ⇒ a − 3 1 = 6 1 ⇒ a = 6 1 + 3 1 ⇒ a = 6 1 + 2 ⇒ a = 6 3 ⇒ a = 2 1 ∴ 2 x − y 1 = 2 1 ⇒ 2 x − y = 2 ......(5)
Substituting value of x from equation (4) in (5), we get :
⇒ 2(6 - 2y) - y = 2
⇒ 12 - 4y - y = 2
⇒ 12 - 5y = 2
⇒ -5y = 2 - 12
⇒ -5y = -10
⇒ y = − 10 − 5 \dfrac{-10}{-5} − 5 − 10
Substituting y = 2 in equation (4), we get :
⇒ x = 6 - 2y
⇒ x = 6 - 2(2) = 6 - 4 = 2.
Hence, x = 2 and y = 2.
Solve 2 x − 3 y = 9 , 3 x + 7 y = 2. 2x - \dfrac{3}{y} = 9, 3x + \dfrac{7}{y} = 2. 2 x − y 3 = 9 , 3 x + y 7 = 2.
Answer
Given,
2 x − 3 y = 9 2x - \dfrac{3}{y} = 9 2 x − y 3 = 9
3 x + 7 y = 2 3x + \dfrac{7}{y} = 2 3 x + y 7 = 2
Multiplying (i) by 3 and (ii) by 2 we get,
6 x − 9 y = 27 6x - \dfrac{9}{y} = 27 6 x − y 9 = 27
6 x + 14 y = 4 6x + \dfrac{14}{y} = 4 6 x + y 14 = 4
Subtracting (iii) from (iv) we get,
⇒ 6 x + 14 y − ( 6 x − 9 y ) = 4 − 27 ⇒ 14 y + 9 y = − 23 ⇒ 23 y = − 23 ⇒ y = − 1. \Rightarrow 6x + \dfrac{14}{y} - \Big(6x - \dfrac{9}{y}\Big) = 4 - 27 \\[1em] \Rightarrow \dfrac{14}{y} + \dfrac{9}{y} = -23 \\[1em] \Rightarrow \dfrac{23}{y} = -23 \\[1em] \Rightarrow y = -1. ⇒ 6 x + y 14 − ( 6 x − y 9 ) = 4 − 27 ⇒ y 14 + y 9 = − 23 ⇒ y 23 = − 23 ⇒ y = − 1.
Substituting value of y in (iv),
⇒ 6 x + 14 − 1 = 4 ⇒ 6 x − 14 = 4 ⇒ 6 x = 18 ⇒ x = 3. \Rightarrow 6x + \dfrac{14}{-1} = 4 \\[1em] \Rightarrow 6x - 14 = 4 \\[1em] \Rightarrow 6x = 18 \\[1em] \Rightarrow x = 3. ⇒ 6 x + − 1 14 = 4 ⇒ 6 x − 14 = 4 ⇒ 6 x = 18 ⇒ x = 3.
Substituting values of x and y in x = ky + 5 we get,
⇒ 3 = k(-1) + 5
⇒ 3 = -k + 5
⇒ k = 5 - 3 = 2.
Hence, x = 3, y = -1 and k = 2.
Solve 1 x + y − 1 2 x = 1 30 , 5 x + y + 1 x = 4 3 . \dfrac{1}{x + y} - \dfrac{1}{2x} = \dfrac{1}{30}, \dfrac{5}{x + y} + \dfrac{1}{x} = \dfrac{4}{3}. x + y 1 − 2 x 1 = 30 1 , x + y 5 + x 1 = 3 4 . 2 - y2 .
Answer
Given,
1 x + y − 1 2 x = 1 30 \dfrac{1}{x + y} - \dfrac{1}{2x} = \dfrac{1}{30} x + y 1 − 2 x 1 = 30 1
5 x + y + 1 x = 4 3 \dfrac{5}{x + y} + \dfrac{1}{x} = \dfrac{4}{3} x + y 5 + x 1 = 3 4
Substituting 1 x + y = a and 1 x = b \dfrac{1}{x + y} = a \text{ and } \dfrac{1}{x} = b x + y 1 = a and x 1 = b
a − b 2 = 1 30 a - \dfrac{b}{2} = \dfrac{1}{30} a − 2 b = 30 1
5 a + b = 4 3 5a + b = \dfrac{4}{3} 5 a + b = 3 4
Multiplying eq. (i) by 5 we get,
5 a − 5 b 2 = 1 6 5a - \dfrac{5b}{2} = \dfrac{1}{6} 5 a − 2 5 b = 6 1
Subtracting eq. (iii) from (ii) we get,
⇒ 5 a + b − ( 5 a − 5 b 2 ) = 4 3 − 1 6 ⇒ b + 5 b 2 = 8 − 1 6 ⇒ 2 b + 5 b 2 = 7 6 ⇒ 7 b 2 = 7 6 ⇒ b = 1 3 ∴ 1 x = 1 3 ⇒ x = 3. \Rightarrow 5a + b - \Big(5a - \dfrac{5b}{2}\Big) = \dfrac{4}{3} - \dfrac{1}{6} \\[1em] \Rightarrow b + \dfrac{5b}{2} = \dfrac{8 - 1}{6} \\[1em] \Rightarrow \dfrac{2b + 5b}{2} = \dfrac{7}{6} \\[1em] \Rightarrow \dfrac{7b}{2} = \dfrac{7}{6} \\[1em] \Rightarrow b = \dfrac{1}{3} \\[1em] \therefore \dfrac{1}{x} = \dfrac{1}{3} \\[1em] \Rightarrow x = 3. ⇒ 5 a + b − ( 5 a − 2 5 b ) = 3 4 − 6 1 ⇒ b + 2 5 b = 6 8 − 1 ⇒ 2 2 b + 5 b = 6 7 ⇒ 2 7 b = 6 7 ⇒ b = 3 1 ∴ x 1 = 3 1 ⇒ x = 3.
Substituting value of b in eq. (i) we get,
⇒ a − 1 3 2 = 1 30 ⇒ a − 1 6 = 1 30 ⇒ a = 1 30 + 1 6 ⇒ a = 1 + 5 30 ⇒ a = 6 30 ⇒ a = 1 5 ∴ 1 x + y = 1 5 ⇒ x + y = 5 ⇒ 3 + y = 5 ⇒ y = 2. \Rightarrow a - \dfrac{\dfrac{1}{3}}{2} = \dfrac{1}{30} \\[1em] \Rightarrow a - \dfrac{1}{6} = \dfrac{1}{30} \\[1em] \Rightarrow a = \dfrac{1}{30} + \dfrac{1}{6} \\[1em] \Rightarrow a = \dfrac{1 + 5}{30} \\[1em] \Rightarrow a = \dfrac{6}{30} \\[1em] \Rightarrow a = \dfrac{1}{5} \\[1em] \therefore \dfrac{1}{x + y} = \dfrac{1}{5} \\[1em] \Rightarrow x + y = 5 \\[1em] \Rightarrow 3 + y = 5 \\[1em] \Rightarrow y = 2. ⇒ a − 2 3 1 = 30 1 ⇒ a − 6 1 = 30 1 ⇒ a = 30 1 + 6 1 ⇒ a = 30 1 + 5 ⇒ a = 30 6 ⇒ a = 5 1 ∴ x + y 1 = 5 1 ⇒ x + y = 5 ⇒ 3 + y = 5 ⇒ y = 2.
Substituting values of x and y in 2x2 - y2 we get,
⇒ 2x2 - y2
= 2(3)2 - (2)2
= 2(9) - 4
= 18 - 4 = 14.
Hence, x = 3, y = 2 and 2x2 - y2 = 14.
Can x, y be found to satisfy the following equations simultaneously?
2 y + 5 x = 19 , 5 y − 3 x = 1 \dfrac{2}{y} + \dfrac{5}{x} = 19, \dfrac{5}{y} - \dfrac{3}{x} = 1 y 2 + x 5 = 19 , y 5 − x 3 = 1
If so, find them.
Answer
Given,
2 y + 5 x = 19 \dfrac{2}{y} + \dfrac{5}{x} = 19 y 2 + x 5 = 19
5 y − 3 x = 1 \dfrac{5}{y} - \dfrac{3}{x} = 1 y 5 − x 3 = 1
3x + 8y = 5 .......(iii)
Multiplying eq. (i) by 5 and eq. (ii) by 2 we get,
10 y + 25 x = 95 \dfrac{10}{y} + \dfrac{25}{x} = 95 y 10 + x 25 = 95
10 y − 6 x = 2 \dfrac{10}{y} - \dfrac{6}{x} = 2 y 10 − x 6 = 2
Subtracting (iv) from (iii) we get,
⇒ 10 y + 25 x − ( 10 y − 6 x ) = 95 − 2 ⇒ 25 x + 6 x = 93 ⇒ 31 x = 93 ⇒ x = 31 93 = 1 3 . \Rightarrow \dfrac{10}{y} + \dfrac{25}{x} - \Big(\dfrac{10}{y} - \dfrac{6}{x}\Big) = 95 - 2 \\[1em] \Rightarrow \dfrac{25}{x} + \dfrac{6}{x} = 93 \\[1em] \Rightarrow \dfrac{31}{x} = 93 \\[1em] \Rightarrow x = \dfrac{31}{93} = \dfrac{1}{3}. ⇒ y 10 + x 25 − ( y 10 − x 6 ) = 95 − 2 ⇒ x 25 + x 6 = 93 ⇒ x 31 = 93 ⇒ x = 93 31 = 3 1 .
Substituting value of x in (i) we get,
⇒ 2 y + 5 1 3 = 19 ⇒ 2 y + 15 = 19 ⇒ 2 y = 4 ⇒ y = 1 2 . \Rightarrow \dfrac{2}{y} + \dfrac{5}{\dfrac{1}{3}} = 19 \\[1em] \Rightarrow \dfrac{2}{y} + 15 = 19 \\[1em] \Rightarrow \dfrac{2}{y} = 4 \\[1em] \Rightarrow y = \dfrac{1}{2}. ⇒ y 2 + 3 1 5 = 19 ⇒ y 2 + 15 = 19 ⇒ y 2 = 4 ⇒ y = 2 1 .
Substituting values of x and y in eq. (iii),
⇒ 3 × 1 3 + 8 × 1 2 = 5 ⇒ 1 + 4 = 5 ⇒ 5 = 5. \Rightarrow 3 \times \dfrac{1}{3} + 8 \times \dfrac{1}{2} = 5 \\[1em] \Rightarrow 1 + 4 = 5 \\[1em] \Rightarrow 5 = 5. ⇒ 3 × 3 1 + 8 × 2 1 = 5 ⇒ 1 + 4 = 5 ⇒ 5 = 5.
Since, L.H.S. = R.H.S. hence, x = 1 3 and y = 1 2 x = \dfrac{1}{3} \text{ and } y = \dfrac{1}{2} x = 3 1 and y = 2 1
Hence, equations can be satisfied simultaneously with x = 1 3 and y = 1 2 x = \dfrac{1}{3} \text{ and } y = \dfrac{1}{2} x = 3 1 and y = 2 1
