Simultaneous Linear Equations — Exercise 5.2

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

$\dfrac{3}{4}x - \dfrac{2}{3}y = 1$

$\dfrac{3}{8}x - \dfrac{1}{6}y = 1$

Answer

Given,

$\dfrac{3}{4}x - \dfrac{2}{3}y = 1$ ........(i)

$\dfrac{3}{8}x - \dfrac{1}{6}y = 1$ .........(ii)

Multiplying eq. (ii) by 2 we get,

$\dfrac{3}{4}x - \dfrac{1}{3}y = 2$ ........(iii)

Subtracting eq. (i) from (iii) we get,

$\Rightarrow \dfrac{3}{4}x - \dfrac{1}{3}y - \Big(\dfrac{3}{4}x - \dfrac{2}{3}y\Big) = 2 - 1 \\[1em] \Rightarrow \dfrac{3}{4}x - \dfrac{3}{4}x - \dfrac{1}{3}y + \dfrac{2}{3}y = 1 \\[1em] \Rightarrow \dfrac{1}{3}y = 1 \\[1em] \Rightarrow y = 3.$

Substituting value of y in eq. (i) we get,

$\Rightarrow \dfrac{3}{4}x - \dfrac{2}{3} \times 3 = 1 \\[1em] \Rightarrow \dfrac{3}{4}x - 2 = 1 \\[1em] \Rightarrow \dfrac{3}{4}x = 3 \\[1em] \Rightarrow x = \dfrac{3 \times 4}{3} \\[1em] \Rightarrow x = 4.$

Hence, x = 4 and y = 3.

Solve the following systems of simultaneous linear equations by the elimination method

2x - 3y - 3 = 0

$\dfrac{2x}{3} + 4y + \dfrac{1}{2} = 0$

Answer

Given,

2x - 3y - 3 = 0 ...........(i)

$\dfrac{2x}{3} + 4y + \dfrac{1}{2} = 0$ ........(ii)

Multiplying eq. (ii) by 3 we get,

$2x + 12y + \dfrac{3}{2} = 0$ .........(iii)

Subtracting eq. (i) from (iii) we get,

$\Rightarrow 2x + 12y + \dfrac{3}{2} - (2x - 3y - 3) = 0 \\[1em] \Rightarrow 2x - 2x + 12y + 3y + \dfrac{3}{2} + 3 = 0 \\[1em] \Rightarrow 15y + \dfrac{9}{2} = 0 \\[1em] \Rightarrow 15y = -\dfrac{9}{2} \\[1em] \Rightarrow y = \dfrac{-9}{2 \times 15} \\[1em] \Rightarrow y = -\dfrac{3}{10}.$

Substituting value of y in eq. (i) we get,

$\Rightarrow 2x - 3y - 3 = 0 \\[1em] \Rightarrow 2x - 3 \times -\dfrac{3}{10} - 3 = 0 \\[1em] \Rightarrow 2x + \dfrac{9}{10} - 3 = 0 \\[1em] \Rightarrow 2x + \dfrac{9 - 30}{10} = 0 \\[1em] \Rightarrow 2x - \dfrac{21}{10} = 0 \\[1em] \Rightarrow x = \dfrac{21}{20}.$

Hence, $x = \dfrac{21}{20} \text{and y} = -\dfrac{3}{10}.$

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

$\dfrac{x}{6} = y - 6$

$\dfrac{3x}{4} = 1 + y$

Answer

Given,

$\dfrac{x}{6} = y - 6$ .......(i)

$\dfrac{3x}{4} = 1 + y$ ......(ii)

Subtracting eq. (i) from (ii) we get,

$\Rightarrow \dfrac{3x}{4} - \dfrac{x}{6} = 1 + y - (y - 6) \\[1em] \Rightarrow \dfrac{9x - 2x}{12} = 7 \\[1em] \Rightarrow \dfrac{7x}{12} = 7 \\[1em] \Rightarrow x = \dfrac{7 \times 12}{7} \\[1em] \Rightarrow x = 12.$

Substituting value of x in eq. (i) we get,

$\Rightarrow \dfrac{12}{6} = y - 6 \\[1em] \Rightarrow 2 = y - 6 \\[1em] \Rightarrow y = 2 + 6 = 8.$

Hence, x = 12 and y = 8.

Solve the following systems of simultaneous linear equations by the elimination method

$x - \dfrac{2}{3}y = \dfrac{8}{3}$

$\dfrac{2x}{5} - y = \dfrac{7}{5}$

Answer

Given,

$x - \dfrac{2}{3}y = \dfrac{8}{3}$ ........(i)

$\dfrac{2x}{5} - y = \dfrac{7}{5}$ ........(ii)

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 = $-\dfrac{5}{11}$.

Substituting value of y in eq. (iv) we get,

$\Rightarrow 6x - 15 \times -\dfrac{5}{11} = 21 \\[1em] \Rightarrow 6x + \dfrac{75}{11} = 21 \\[1em] \Rightarrow 6x = 21 - \dfrac{75}{11} \\[1em] \Rightarrow 6x = \dfrac{231 - 75}{11} \\[1em] \Rightarrow 6x = \dfrac{156}{11} \\[1em] \Rightarrow x = \dfrac{156}{66} \\[1em] \Rightarrow x = \dfrac{26}{11}.$

Hence, $x = \dfrac{26}{11} \text{ and } y = -\dfrac{5}{11}.$

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 = $-\dfrac{8}{3}$.

Substituting value of y in eq. (i) we get,

$\Rightarrow x + \Big(-\dfrac{8}{3}\Big) = 6 \\[1em] \Rightarrow x = 6 + \dfrac{8}{3} \\[1em] \Rightarrow x = \dfrac{26}{3}.$

Hence, $x = \dfrac{26}{3} \text{ and } y = -\dfrac{8}{3}$.

Solve the following systems of simultaneous linear equations by the elimination method

$2x + \dfrac{x - y}{6} = 2$

$x - \dfrac{2x + y}{3} = 1$

Answer

Solving 1st equation,

$\Rightarrow 2x + \dfrac{x - y}{6} = 2 \\[1em] \Rightarrow \dfrac{12x + x - y}{6} = 2 \\[1em] \Rightarrow 13x - y = 12 .......(i)$

Solving 2nd equation,

$\Rightarrow x - \dfrac{2x + y}{3} = 1 \\[1em] \Rightarrow \dfrac{3x - 2x - y}{3} = 1 \\[1em] \Rightarrow x - y = 3 ........(ii)$

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 = $-\dfrac{9}{4}$.

Substituting value of y in eq (ii) we get,

$\Rightarrow x - \Big(-\dfrac{9}{4}\Big) = 3 \\[1em] \Rightarrow x + \dfrac{9}{4} = 3 \\[1em] \Rightarrow x = 3 - \dfrac{9}{4} \\[1em] \Rightarrow x = \dfrac{12 - 9}{4} \\[1em] \Rightarrow x = \dfrac{3}{4}.$

Hence, $x = \dfrac{3}{4} \text{ and } y = -\dfrac{9}{4}.$

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

$4x + \dfrac{x - y}{8}$ = 17

$2y + x - \dfrac{5y + 2}{3} = 2$

Answer

Solving 1st equation,

$\Rightarrow 4x + \dfrac{x - y}{8} = 17 \\[1em] \Rightarrow \dfrac{32x + x - y}{8} = 17 \\[1em] \Rightarrow 33x - y = 136 .......(i)$

Solving 2nd equation,

$\Rightarrow 2y + x - \dfrac{5y + 2}{3} = 2 \\[1em] \Rightarrow \dfrac{6y + 3x - 5y - 2}{3} = 2 \\[1em] \Rightarrow \dfrac{y + 3x - 2}{3} = 2 \\[1em] \Rightarrow y + 3x - 2 = 6 \\[1em] \Rightarrow y + 3x = 8 ........(ii)$

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

$\dfrac{x + 1}{2} + \dfrac{y - 1}{3} = 8$

$\dfrac{x - 1}{3} + \dfrac{y + 1}{2} = 9$

Answer

Solving 1st equation,

$\Rightarrow \dfrac{x + 1}{2} + \dfrac{y - 1}{3} = 8 \\[1em] \Rightarrow \dfrac{3(x + 1) + 2(y - 1)}{6} = 8 \\[1em] \Rightarrow \dfrac{3x + 3 + 2y - 2}{6} = 8 \\[1em] \Rightarrow \dfrac{3x + 2y + 1}{6} = 8 \\[1em] \Rightarrow 3x + 2y + 1 = 48 \\[1em] \Rightarrow 3x + 2y = 47 ......(i)$

Solving 2nd equation,

$\Rightarrow \dfrac{x - 1}{3} + \dfrac{y + 1}{2} = 9 \\[1em] \Rightarrow \dfrac{2(x - 1) + 3(y + 1)}{6} = 9 \\[1em] \Rightarrow \dfrac{2x - 2 + 3y + 3}{6} = 9 \\[1em] \Rightarrow \dfrac{2x + 3y + 1}{6} = 9 \\[1em] \Rightarrow 2x + 3y + 1 = 54 \\[1em] \Rightarrow 2x + 3y = 53 .......(ii)$

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

$\dfrac{3}{x} + 4y = 7$

$\dfrac{5}{x} + 6y = 13$

Answer

Given,

$\dfrac{3}{x} + 4y = 7$ .......(i)

$\dfrac{5}{x} + 6y = 13$ .......(ii)

Multiplying eq. (i) by 3 and eq. (ii) by 2 we get,

$\dfrac{9}{x} + 12y = 21$ ......(iii)

$\dfrac{10}{x} + 12y = 26$ .......(iv)

Subtracting eq. (iii) from (iv) we get,

$\Rightarrow \dfrac{10}{x} - \dfrac{9}{x} + 12y - 12y = 26 - 21 \\[1em] \Rightarrow \dfrac{1}{x} = 5 \\[1em] \Rightarrow x = \dfrac{1}{5}.$

Substituting value of x in eq. (i) we get,

$\Rightarrow \dfrac{3}{x} + 4y = 7 \\[1em] \Rightarrow \dfrac{3}{\dfrac{1}{5}} + 4y = 7 \\[1em] \Rightarrow 15 + 4y = 7 \\[1em] \Rightarrow 4y = -8 \\[1em] \Rightarrow y = -2.$

Hence, x = $\dfrac{1}{5}$ and y = -2.

Solve the following systems of simultaneous linear equations by the elimination method

$5x - 9 = \dfrac{1}{y}$

$x + \dfrac{1}{y} = 3$

Answer

$5x - 9 = \dfrac{1}{y} \text{ or } 5x - \dfrac{1}{y} = 9$ ........(i)

$x + \dfrac{1}{y} = 3$ .......(ii)

Adding eq. (i) and (ii) we get,

$\Rightarrow 5x - \dfrac{1}{y} + x + \dfrac{1}{y} = 9 + 3 \\[1em] \Rightarrow 6x = 12 \\[1em] \Rightarrow x = 2.$

Substituting value of x in eq. (ii) we get,

$\Rightarrow x + \dfrac{1}{y} = 3 \\[1em] \Rightarrow 2 + \dfrac{1}{y} = 3 \\[1em] \Rightarrow \dfrac{1}{y} = 1 \\[1em] \Rightarrow y = 1.$

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 + q²y = pq - q² .......(iii)

pqx - p²y = p² + pq .......(iv)

Subtracting eq. (iv) from (iii) we get,

⇒ pqx + q²y - (pqx - p²y) = pq - q² - (p² + pq)

⇒ pqx - pqx + q²y + p²y = pq - pq - q² - p²

⇒ q²y + p²y = -q² - p²

⇒ y(q² + p²) = -(q² + p²)

⇒ y = $\dfrac{-(\text{q}^2 + \text{p}^2)}{\text{q}^2 + \text{p}^2}$ = -1.

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

$\dfrac{x}{a} - \dfrac{y}{b} = 0$

ax + by = a² + b²

Answer

$\phantom{\Rightarrow} \dfrac{x}{a} - \dfrac{y}{b} = 0 \\[1em] \Rightarrow \dfrac{bx - ay}{ab} = 0 \\[1em] \Rightarrow bx - ay = 0.$

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 - a²y = 0 .........(iii)

abx + b²y = b(a² + b²) .......(iv)

Subtracting eq. (iii) from (iv) we get,

⇒ abx + b²y - (abx - a²y) = b(a² + b²)

⇒ abx - abx + b²y + a²y = 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.