Simultaneous Linear Equations — Exercise 5.4

Solve the following pairs of linear equations:

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

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

Answer

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

$2a + \dfrac{2}{3}b = \dfrac{1}{3}$

Multiply by 3:

6a + 2b = 1 .......(i)

2a - b = 2 .......(ii)

⇒ -b = 2 - 2a

⇒ b = 2a - 2

Substituting this value of b in equation (i) we get,

6a + 2(2a - 2) = 1

⇒ 6a + 4a - 4 = 1

⇒ 10a - 4 = 1

⇒ 10a = 1 + 4

⇒ 10a = 5

⇒ a = $\dfrac{5}{10}$

⇒ a = $\dfrac{1}{2}$

⇒ $\dfrac{1}{x} = \dfrac{1}{2}$

⇒ $x = 2$

From (ii) we have,

b = 2a - 2

Substituting value of a in above equation we get,

b = 2 x $\dfrac{1}{2}$ - 2

⇒ b = 1 - 2

⇒ b = -1

⇒ $\dfrac{1}{y}$ = -1

⇒ $y$ = -1

Hence, x = 2 and y = -1.

Solve the following pairs of linear equations:

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

$\dfrac{5}{x} - \dfrac{3}{y} = 1$

Answer

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

$\dfrac{3}{2}a + \dfrac{2}{3}b = 5$ ......(i)

5a - 3b = 1 .......(ii)

Solving eq (i) we get,

$\Rightarrow \dfrac{3}{2}a + \dfrac{2}{3}b = 5 \\[1em] \Rightarrow \dfrac{9a + 4b}{6} = 5 \\[1em] \Rightarrow 9a + 4b = 30 .......(iii)$

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.

$\therefore \dfrac{1}{x} = 2 \\[1em] \Rightarrow x = \dfrac{1}{2}.$

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

⇒ 5(2) - 3b = 1

⇒ 10 - 3b = 1

⇒ 3b = 10 - 1

⇒ 3b = 9

⇒ b = 3.

$\therefore \dfrac{1}{y} = 3 \\[1em] \Rightarrow y = \dfrac{1}{3}.$

Hence, $x = \dfrac{1}{2} \text{ and } y = \dfrac{1}{3}.$

Solve the following pairs of linear equations:

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

$\dfrac{8x + 7y}{xy} = 15$

Answer

Given,

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

$\dfrac{8x + 7y}{xy} = 15 \text{ or } \dfrac{8}{y} + \dfrac{7}{x} = 15$

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

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.

$\therefore \dfrac{1}{y} = 1 \\[1em] \Rightarrow y = 1.$

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

⇒ 7(1) - 2a = 5

⇒ 7 - 2a = 5

⇒ 2a = 7 - 5

⇒ 2a = 2

⇒ a = 1.

$\therefore \dfrac{1}{x} = 1 \\[1em] \Rightarrow x = 1.$

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,

$\dfrac{99}{y} + \dfrac{101}{x} = 499$ .......(i)

$\dfrac{101}{y} + \dfrac{99}{x} = 501$ .......(ii)

Substituting $\dfrac{1}{x} = p \text{ and } \dfrac{1}{y} = q$ in both equations and multiplying eq. (i) by 101 and (ii) by 99 we get,

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.

$\therefore \dfrac{1}{x} = 2 \text{ or } x = \dfrac{1}{2}.$

Substituting value of p in (iii) we get,

⇒ 9999q + 10201(2) = 50399

⇒ 9999q + 20402 = 50399

⇒ 9999q = 29997

⇒ q = 3.

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

Hence, x = 0, y = 0 and x = $\dfrac{1}{2} \text{ and } y = \dfrac{1}{3}.$

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,

$\Rightarrow 3x + 14y = 5xy \\[1em] \Rightarrow \dfrac{3x}{xy} + \dfrac{14y}{xy} = \dfrac{5xy}{xy} \\[1em] \Rightarrow \dfrac{3}{y} + \dfrac{14}{x} = 5 ......(iii)$

Dividing eq. (ii) by xy we get,

$\Rightarrow 21y - x = 2xy \\[1em] \Rightarrow \dfrac{21y}{xy} - \dfrac{x}{xy} = 2 \\[1em] \Rightarrow \dfrac{21}{x} - \dfrac{1}{y} = 2 .......(iv)$

Substituting $\dfrac{1}{x} = a \text{ and } \dfrac{1}{y} = b$ in eq. (iii) and (iv) we get,

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 + 6
⇒ 77a = 11
⇒ a = $\dfrac{11}{77} = \dfrac{1}{7}$.

$\therefore \dfrac{1}{x} = \dfrac{1}{7} \\[1em] \Rightarrow x = 7.$

Substituting value of a in eq. (v) we get,

⇒ 3b + $14\times \dfrac{1}{7}$ = 5

⇒ 3b + 2 = 5

⇒ 3b = 3

⇒ b = 1.

$\therefore \dfrac{1}{y} = 1 \\[1em] \Rightarrow y = 1.$

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,

$\dfrac{3}{y} + \dfrac{5}{x} = 4$ .......(iii)

$\dfrac{2}{x} - \dfrac{1}{y} = 1$ .......(iv)

Substituting $\dfrac{1}{x} = a \text{ and } \dfrac{1}{y} = b$ in eq. (iii) and (iv) we get,

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 = $\dfrac{7}{11}$.

$\therefore \dfrac{1}{x} = \dfrac{7}{11} \\[1em] \Rightarrow x = \dfrac{11}{7}.$

Substituting value of a in eq. (vi) we get,

$\Rightarrow 2 \times \dfrac{7}{11} - b = 1 \\[1em] \Rightarrow \dfrac{14}{11} - b = 1 \\[1em] \Rightarrow b = \dfrac{14}{11} - 1 \\[1em] \Rightarrow b = \dfrac{3}{11} \\[1em] \therefore \dfrac{1}{y} = \dfrac{3}{11} \\[1em] \Rightarrow y = \dfrac{11}{3}.$

Hence, x = 0, y = 0 and x = $\dfrac{11}{7}, y = \dfrac{11}{3}.$

Solve the following pairs of linear equations:

$\dfrac{20}{x + 1} + \dfrac{4}{y - 1} = 5$

$\dfrac{10}{x + 1} - \dfrac{4}{y - 1} = 1$

Answer

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

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 = $\dfrac{3}{12} = \dfrac{1}{4}$.

$\therefore \dfrac{1}{y - 1} = \dfrac{1}{4} \\[1em] \Rightarrow y - 1 = 4 \\[1em] \Rightarrow y = 5.$

Substituting value of b in (iii) we get,

⇒ 20a - $8 \times \dfrac{1}{4}$ = 2

⇒ 20a - 2 = 2

⇒ 20a = 2 + 2

⇒ a = $\dfrac{4}{20}$

⇒ a = $\dfrac{1}{5}$

$\therefore \dfrac{1}{x + 1} = \dfrac{1}{5} \\[1em] \Rightarrow x + 1 = 5 \\[1em] \Rightarrow x = 4.$

Hence, x = 4 and y = 5.

Solve the following pairs of linear equations:

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

$\dfrac{2}{x + y} + \dfrac{3}{x - y} = \dfrac{11}{3}$

Answer

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

3a + 2b = 3 .......(i)

2a + 3b = $\dfrac{11}{3}$ .......(ii)

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.

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

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

⇒ 6a + 4(1) = 6

⇒ 6a + 4 = 6

⇒ 6a = 2

⇒ a = $\dfrac{2}{6} = \dfrac{1}{3}.$

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

⇒ 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:

$\dfrac{1}{2(2x + 3y)} + \dfrac{12}{7(3x - 2y)} = \dfrac{1}{2}$

$\dfrac{7}{2x + 3y} + \dfrac{4}{3x - 2y} = 2$

Answer

Substituting $\dfrac{1}{2x + 3y} = a \text{ and } \dfrac{1}{3x - 2y} = b$ in above eq. we get,

$\dfrac{1}{2}a + \dfrac{12}{7}b = \dfrac{1}{2}$ .....(i)

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 = $\dfrac{5}{20} = \dfrac{1}{4}$.

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

⇒ 3x - 2y = 4 ........(iv)

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

⇒ 7a + $24 \times \dfrac{1}{4}$ = 7

⇒ 7a + 6 = 7

⇒ 7a = 1

⇒ a = $\dfrac{1}{7}$.

$\therefore \dfrac{1}{2x + 3y} = \dfrac{1}{7}$

⇒ 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:

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

$\dfrac{5}{4(x + 2y)} - \dfrac{3}{5(3x - 2y)} = \dfrac{61}{60}$

Answer

Substitute $\dfrac{1}{x + 2y} = p \text{ and } \dfrac{1}{3x - 2y} = q$ in above equations,

$\dfrac{1}{2}p + \dfrac{5}{3}q = -\dfrac{3}{2}$ ........(i)

$\dfrac{5}{4}p - \dfrac{3}{5}q = \dfrac{61}{60}$ ........(ii)

Multiplying (i) by $\dfrac{3}{5}$ and (ii) by $\dfrac{5}{3}$ we get,

$\dfrac{3}{10}p + q = -\dfrac{9}{10}$ .......(iii)

$\dfrac{25}{12}p - q = \dfrac{61}{36}$ .......(iv)

Adding (iii) and (iv) we get,

$\Rightarrow \dfrac{3}{10}p + q + \dfrac{25}{12}p - q = -\dfrac{9}{10} + \dfrac{61}{36} \\[1em] \Rightarrow \dfrac{18p + 125p}{60} = \dfrac{-162 + 305}{180} \\[1em] \Rightarrow \dfrac{143}{60}p = \dfrac{143}{180} \\[1em] \Rightarrow p = \dfrac{143 \times 60}{180 \times 143} \\[1em] \Rightarrow p = \dfrac{1}{3} \\[1em] \therefore \dfrac{1}{x + 2y} = \dfrac{1}{3} \\[1em] \Rightarrow x + 2y = 3 .......(v)$

Substituting value of p in (i) we get,

$\Rightarrow \dfrac{1}{2} \times \dfrac{1}{3} + \dfrac{5}{3}q = -\dfrac{3}{2} \\[1em] \Rightarrow \dfrac{1}{6} + \dfrac{5}{3}q = -\dfrac{3}{2} \\[1em] \Rightarrow \dfrac{5}{3}q = -\dfrac{3}{2} - \dfrac{1}{6} \\[1em] \Rightarrow \dfrac{5}{3}q = \dfrac{-9 - 1}{6} \\[1em] \Rightarrow \dfrac{5}{3}q = -\dfrac{10}{6} \\[1em] \Rightarrow q = -\dfrac{10 \times 3}{6 \times 5} \\[1em] \Rightarrow q = -1 \\[1em] \therefore \dfrac{1}{3x - 2y} = -1 \\[1em] \Rightarrow 3x - 2y = -1 \\[1em] \Rightarrow 2y - 3x = 1 ......(vi)$

Subtracting (vi) from (v) we get,

⇒ x + 2y - (2y - 3x) = 3 - 1

⇒ x + 3x = 2

⇒ 4x = 2

⇒ x = $\dfrac{1}{2}$.

Substituting value of x from (v) we get,

$\Rightarrow \dfrac{1}{2} + 2y = 3 \\[1em] \Rightarrow 2y = 3 - \dfrac{1}{2} \\[1em] \Rightarrow 2y = \dfrac{6 - 1}{2} \\[1em] \Rightarrow 2y = \dfrac{5}{2} \\[1em] \Rightarrow y = \dfrac{5}{4}.$

Hence, x = $\dfrac{1}{2} \text{ and } y = \dfrac{5}{4}.$