Simultaneous Linear Equations — Chapter Test

Solve the following simultaneous linear equations:

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

5x - 2y = 7

Answer

Given,

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

5x - 2y = 7 ..........(ii)

Multiplying eq. (i) by 5 and (ii) by 2 we get,

10x - $\dfrac{15}{4}y$ = 15 .......(iii)

10x - 4y = 14 ........(iv)

Subtracting eq. (iv) from (iii),

$\Rightarrow 10x - \dfrac{15}{4}y - (10x - 4y) = 15 - 14 \\[1em] \Rightarrow 4y - \dfrac{15}{4}y = 1 \\[1em] \Rightarrow \dfrac{16y - 15y}{4} = 1 \\[1em] \Rightarrow \dfrac{y}{4} = 1 \\[1em] \Rightarrow y = 4.$

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 = $\dfrac{46200}{6600}$ = 7.

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 = $-\dfrac{1125}{1125}$ = -1.

Substituting value of y in equation (1), we get :

⇒ 67x - 58(-1) = 192

⇒ 67x + 58 = 192

⇒ 67x = 192 - 58

⇒ 67x = 134

⇒ x = $\dfrac{134}{67}$ = 2.

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,

$\Rightarrow \dfrac{x}{xy} + \dfrac{y}{xy} = \dfrac{7xy}{xy} \\[1em] \Rightarrow \dfrac{1}{y} + \dfrac{1}{x} = 7 ......(i)$

Dividing 2x - 3y + xy = 0 by xy,

$\Rightarrow \dfrac{2x}{xy} - \dfrac{3y}{xy} + \dfrac{xy}{xy} = 0 \\[1em] \Rightarrow \dfrac{2}{y} - \dfrac{3}{x} + 1 = 0 \\[1em] \Rightarrow \dfrac{2}{y} - \dfrac{3}{x} = -1 ......(ii)$

Substituting $\dfrac{1}{x}$ as p and $\dfrac{1}{y}$ as q we get,

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 = $\dfrac{20}{5}$

⇒ q = 4.

$\therefore \dfrac{1}{y} = 4 \text{ or } y = \dfrac{1}{4}$.

Substituting value of q in (iii) we get,

⇒ 4 + p = 7

⇒ p = 3.

$\therefore \dfrac{1}{x} = 3 \text{ or } x = \dfrac{1}{3}$.

Hence, x = 0, y = 0 and x = $\dfrac{1}{3}, y = \dfrac{1}{4}$.

Solve the following simultaneous linear equations:

$\dfrac{30}{x - y} + \dfrac{44}{x + y} = 10$

$\dfrac{40}{x - y} + \dfrac{55}{x + y} = 13$

Answer

Substituting $\dfrac{1}{x - y} = a \text{ and } \dfrac{1}{x + y} = b$ in above equations we get,

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 = $\dfrac{1}{11}$.

$\therefore \dfrac{1}{x + y} = \dfrac{1}{11} \\[1em] \Rightarrow x + y = 11 .........(v)$

Substituting value of b in (iv),

⇒ 120a + $165 \times \dfrac{1}{11}$ = 39

⇒ 120a + 15 = 39

⇒ 120a = 24

⇒ a = $\dfrac{24}{120} = \dfrac{1}{5}$.

$\therefore \dfrac{1}{x - y} = \dfrac{1}{5} \\[1em] \Rightarrow x - y = 5 ........(vi)$

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 + b²y = ab - b² ......(iii)

abx - a²y = a² + ab .......(iv)

Subtracting eq. (iv) from (iii),

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

⇒ b²y + a²y = -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

$\dfrac{1}{x} + \dfrac{2}{y} = 1\dfrac{1}{6}$

Answer

Given,

3x + 2y = 2xy .......(i)

$\dfrac{1}{x} + \dfrac{2}{y} = 1\dfrac{1}{6}$ ......(ii)

Dividing eq. (i) by xy we get,

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

Substituting $\dfrac{1}{x}$ as p and $\dfrac{1}{y}$ as q in (ii) and (iii) we get,

p + 2q = $\dfrac{7}{6}$ ......(iv)

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 = $\dfrac{1}{3}$.

$\therefore \dfrac{1}{y} = \dfrac{1}{3} \text{ or } y = 3.$

Substituting value of q in (iv) we get,

$\Rightarrow p + 2 \times \dfrac{1}{3} = \dfrac{7}{6} \\[1em] \Rightarrow p + \dfrac{2}{3} = \dfrac{7}{6} \\[1em] \Rightarrow p = \dfrac{7}{6} - \dfrac{2}{3} \\[1em] \Rightarrow p = \dfrac{7 - 4}{6} \\[1em] \Rightarrow p = \dfrac{3}{6} \\[1em] \therefore \dfrac{1}{x} = \dfrac{3}{6} \\[1em] \Rightarrow x = \dfrac{6}{3} = 2.$

Hence, x = 2 and y = 3.

$\dfrac{2}{3(2x - y)} + \dfrac{1}{2(x + 2y)} = \dfrac{5}{12}, \dfrac{1}{2x - y} - \dfrac{2}{x + 2y} = \dfrac{1}{6}$

Answer

Substituting $\dfrac{1}{2x - y}$ = a and $\dfrac{1}{x + 2y}$ = b in above equations, we get :

⇒ $\dfrac{2}{3}a + \dfrac{1}{2}b = \dfrac{5}{12}$ ..........(1)

⇒ a - 2b = $\dfrac{1}{6}$ .........(2)

Multiplying equation (1) by $\dfrac{3}{2}$ we get :

$\Rightarrow \dfrac{3}{2}\Big(\dfrac{2}{3}a + \dfrac{1}{2}b\Big) = \dfrac{3}{2} \times \dfrac{5}{12} \\[1em] \Rightarrow a + \dfrac{3}{4}b = \dfrac{5}{8} \text{ ...........(3)}$

Subtracting equation (2) from (3), we get :

$\Rightarrow a + \dfrac{3}{4}b - (a - 2b) = \dfrac{5}{8} - \dfrac{1}{6} \\[1em] \Rightarrow a - a + \dfrac{3}{4}b + 2b = \dfrac{15 - 4}{24} \\[1em] \Rightarrow \dfrac{3b + 8b}{4} = \dfrac{11}{24} \\[1em] \Rightarrow \dfrac{11b}{4} = \dfrac{11}{24} \\[1em] \Rightarrow b = \dfrac{11}{24} \times \dfrac{4}{11} = \dfrac{1}{6}. \\[1em] \therefore \dfrac{1}{x + 2y} = \dfrac{1}{6} \\[1em] \Rightarrow x + 2y = 6 \\[1em] \Rightarrow x = 6 - 2y \text{ ............(4)}$

Substituting b = $\dfrac{1}{6}$ in equation (2), we get :

$\Rightarrow a - 2 \times \dfrac{1}{6} = \dfrac{1}{6} \\[1em] \Rightarrow a - \dfrac{1}{3} = \dfrac{1}{6} \\[1em] \Rightarrow a = \dfrac{1}{6} + \dfrac{1}{3} \\[1em] \Rightarrow a = \dfrac{1 + 2}{6} \\[1em] \Rightarrow a = \dfrac{3}{6} \\[1em] \Rightarrow a = \dfrac{1}{2} \\[1em] \therefore \dfrac{1}{2x - y} = \dfrac{1}{2}\\[1em] \Rightarrow 2x - y = 2 \text{ ......(5)}$

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 = $\dfrac{-10}{-5}$ = 2.

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 $2x - \dfrac{3}{y} = 9, 3x + \dfrac{7}{y} = 2.$ Hence, find the value of k if x = ky + 5.

Answer

Given,

$2x - \dfrac{3}{y} = 9$ .......(i)

$3x + \dfrac{7}{y} = 2$ .......(ii)

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

$6x - \dfrac{9}{y} = 27$ .......(iii)

$6x + \dfrac{14}{y} = 4$ .......(iv)

Subtracting (iii) from (iv) we get,

$\Rightarrow 6x + \dfrac{14}{y} - \Big(6x - \dfrac{9}{y}\Big) = 4 - 27 \\[1em] \Rightarrow \dfrac{14}{y} + \dfrac{9}{y} = -23 \\[1em] \Rightarrow \dfrac{23}{y} = -23 \\[1em] \Rightarrow y = -1.$

Substituting value of y in (iv),

$\Rightarrow 6x + \dfrac{14}{-1} = 4 \\[1em] \Rightarrow 6x - 14 = 4 \\[1em] \Rightarrow 6x = 18 \\[1em] \Rightarrow x = 3.$

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 $\dfrac{1}{x + y} - \dfrac{1}{2x} = \dfrac{1}{30}, \dfrac{5}{x + y} + \dfrac{1}{x} = \dfrac{4}{3}.$ Hence, find the value of 2x² - y².

Answer

Given,

$\dfrac{1}{x + y} - \dfrac{1}{2x} = \dfrac{1}{30}$

$\dfrac{5}{x + y} + \dfrac{1}{x} = \dfrac{4}{3}$

Substituting $\dfrac{1}{x + y} = a \text{ and } \dfrac{1}{x} = b$ in above equations,

$a - \dfrac{b}{2} = \dfrac{1}{30}$ ........(i)

$5a + b = \dfrac{4}{3}$ ........(ii)

Multiplying eq. (i) by 5 we get,

$5a - \dfrac{5b}{2} = \dfrac{1}{6}$ .......(iii)

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

$\Rightarrow 5a + b - \Big(5a - \dfrac{5b}{2}\Big) = \dfrac{4}{3} - \dfrac{1}{6} \\[1em] \Rightarrow b + \dfrac{5b}{2} = \dfrac{8 - 1}{6} \\[1em] \Rightarrow \dfrac{2b + 5b}{2} = \dfrac{7}{6} \\[1em] \Rightarrow \dfrac{7b}{2} = \dfrac{7}{6} \\[1em] \Rightarrow b = \dfrac{1}{3} \\[1em] \therefore \dfrac{1}{x} = \dfrac{1}{3} \\[1em] \Rightarrow x = 3.$

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

$\Rightarrow a - \dfrac{\dfrac{1}{3}}{2} = \dfrac{1}{30} \\[1em] \Rightarrow a - \dfrac{1}{6} = \dfrac{1}{30} \\[1em] \Rightarrow a = \dfrac{1}{30} + \dfrac{1}{6} \\[1em] \Rightarrow a = \dfrac{1 + 5}{30} \\[1em] \Rightarrow a = \dfrac{6}{30} \\[1em] \Rightarrow a = \dfrac{1}{5} \\[1em] \therefore \dfrac{1}{x + y} = \dfrac{1}{5} \\[1em] \Rightarrow x + y = 5 \\[1em] \Rightarrow 3 + y = 5 \\[1em] \Rightarrow y = 2.$

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?

$\dfrac{2}{y} + \dfrac{5}{x} = 19, \dfrac{5}{y} - \dfrac{3}{x} = 1$, 3x + 8y = 5.

If so, find them.

Answer

Given,

$\dfrac{2}{y} + \dfrac{5}{x} = 19$ ......(i)

$\dfrac{5}{y} - \dfrac{3}{x} = 1$ .......(ii)

3x + 8y = 5 .......(iii)

Multiplying eq. (i) by 5 and eq. (ii) by 2 we get,

$\dfrac{10}{y} + \dfrac{25}{x} = 95$ ......(iii)

$\dfrac{10}{y} - \dfrac{6}{x} = 2$ .......(iv)

Subtracting (iv) from (iii) we get,

$\Rightarrow \dfrac{10}{y} + \dfrac{25}{x} - \Big(\dfrac{10}{y} - \dfrac{6}{x}\Big) = 95 - 2 \\[1em] \Rightarrow \dfrac{25}{x} + \dfrac{6}{x} = 93 \\[1em] \Rightarrow \dfrac{31}{x} = 93 \\[1em] \Rightarrow x = \dfrac{31}{93} = \dfrac{1}{3}.$

Substituting value of x in (i) we get,

$\Rightarrow \dfrac{2}{y} + \dfrac{5}{\dfrac{1}{3}} = 19 \\[1em] \Rightarrow \dfrac{2}{y} + 15 = 19 \\[1em] \Rightarrow \dfrac{2}{y} = 4 \\[1em] \Rightarrow y = \dfrac{1}{2}.$

Substituting values of x and y in eq. (iii),

$\Rightarrow 3 \times \dfrac{1}{3} + 8 \times \dfrac{1}{2} = 5 \\[1em] \Rightarrow 1 + 4 = 5 \\[1em] \Rightarrow 5 = 5.$

Since, L.H.S. = R.H.S. hence, $x = \dfrac{1}{3} \text{ and } y = \dfrac{1}{2}$ satisfies the equation.

Hence, equations can be satisfied simultaneously with $x = \dfrac{1}{3} \text{ and } y = \dfrac{1}{2}$.