Simultaneous Linear Equations — Exercise 5.1

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
⟹ 4 + 2y = 14
⟹ 2y = 14 - 4
⟹ 2y = 10
⟹ y = 5.

∴ 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

$\dfrac{s}{3} + \dfrac{t}{2} = 6$

Answer

Given,

s - t = 3 ........(i)

$\dfrac{s}{3} + \dfrac{t}{2} = 6$ ........(ii)

From eqn. (i) we get,

s = 3 + t.

Substituting above value of s in eqn. (ii) we get,

$\Rightarrow \dfrac{s}{3} + \dfrac{t}{2} = 6 \\[1em] \Rightarrow \dfrac{3 + t}{3} + \dfrac{t}{2} = 6 \\[1em] \Rightarrow \dfrac{2(3 + t) + 3t}{6} = 6 \\[1em] \Rightarrow 6 + 2t + 3t = 36 \\[1em] \Rightarrow 5t = 36 - 6 \\[1em] \Rightarrow 5t = 30 \\[1em] \Rightarrow t = 6.$

∴ 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 = $\dfrac{9 - 3y}{2}$

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

$\Rightarrow 3\Big(\dfrac{9 - 3y}{2}\Big) + 4y = 5 \\[1em] \Rightarrow \dfrac{27 - 9y}{2} + 4y = 5 \\[1em] \Rightarrow \dfrac{27 - 9y + 8y}{2} = 5 \\[1em] \Rightarrow 27 - y = 10 \\[1em] \Rightarrow y = 27 - 10 = 17.$

Solving for x,

$x = \dfrac{9 - 3y}{2} = \dfrac{9 - 3(17)}{2} \\[1em] = \dfrac{9 - 51}{2} \\[1em] = \dfrac{-42}{2} \\[1em] = -21.$

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 = $\dfrac{4 + 5y}{3}$.

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

$\Rightarrow 9x - 2y = 7 \\[1em] \Rightarrow 9\Big(\dfrac{4 + 5y}{3}\Big) - 2y = 7 \\[1em] \Rightarrow 3(4 + 5y) - 2y = 7 \\[1em] \Rightarrow 12 + 15y - 2y = 7 \\[1em] \Rightarrow 12 + 13y = 7 \\[1em] \Rightarrow 13y = 7 - 12 \\[1em] \Rightarrow 13y = -5 \\[1em] \Rightarrow y = -\dfrac{5}{13}.$

Solving for x by substituting value of y,

$\Rightarrow x = \dfrac{4 + 5y}{3} \\[1em] = \dfrac{4 + 5 \times \dfrac{-5}{13}}{3} \\[1em] = \dfrac{4 - \dfrac{25}{13}}{3} \\[1em] = \dfrac{\dfrac{52 - 25}{13}}{3} \\[1em] = \dfrac{27}{39} \\[1em] = \dfrac{9}{13}.$

Hence, x = $\dfrac{9}{13}\text{ and y } = -\dfrac{5}{13}.$

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

Substituting above value of x in equation (2), we get :

$\Rightarrow 7\Big(\dfrac{-2 + 5y}{3}\Big) - 3y = -9\\[1em] \Rightarrow \dfrac{7(-2 + 5y) - 3y \times 3}{3} = -9\\[1em] \Rightarrow \dfrac{7(-2 + 5y) - 9y}{3} = -9\\[1em] \Rightarrow 7({-2 + 5y}) - 9y = -27\\[1em] \Rightarrow -14 + 35y - 9y = -27\\[1em] \Rightarrow -14 + 26y = -27\\[1em] \Rightarrow 26y = -27 + 14\\[1em] \Rightarrow 26y = -13\\[1em] \Rightarrow y = -\dfrac{13}{26}\\[1em] \Rightarrow y = -\dfrac{1}{2}.$

Substituting value of y in x = $\dfrac{-2 + 5y}{3}$, we get :

$\Rightarrow x = \dfrac{-2 + 5 \times \dfrac{-1}{2}}{3}\\[1em] \Rightarrow x = \dfrac{-2 - \dfrac{5}{2}}{3} \\[1em] \Rightarrow x = \dfrac{\dfrac{-4 - 5}{2}}{3} \\[1em] \Rightarrow x = \dfrac{-4 - 5}{6}\\[1em] \Rightarrow x = \dfrac{-9}{6}\\[1em] \Rightarrow x = -\dfrac{3}{2}.$

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

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 = $-\dfrac{96}{64} = -\dfrac{3}{2}$.

Substituting value of y in eqn. (iii) we get,

$x = 12 \times \dfrac{-3}{2} + 20 \\[1em] = 6 \times -3 + 20 \\[1em] = -18 + 20 \\[1em] = 2.$

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

Solve the following system of simultaneous linear equations by the substitution method:

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

5x - 2y - 7 = 0

Answer

Given,

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

5x - 2y - 7 = 0 .......(ii)

Multiplying eqn. (i) by 4 we get,

$\Rightarrow 4\Big(2x - \dfrac{3y}{4}\Big) = 3 \times 4 \\[1em] \Rightarrow 8x - 3y = 12 \\[1em] \Rightarrow 8x = 12 + 3y \\[1em] \Rightarrow x = \dfrac{12 + 3y}{8} ......(iii)$

Putting value of x from eqn. (iii) in eqn. (ii),

$\Rightarrow 5\Big(\dfrac{12 + 3y}{8}\Big) - 2y - 7 = 0 \\[1em] \Rightarrow \dfrac{60 + 15y}{8} - 2y - 7 = 0 \\[1em] \Rightarrow \dfrac{60 + 15y - 16y - 56}{8} = 0 \\[1em] \Rightarrow 4 - y = 0 \\[1em] \Rightarrow y = 4.$

Substituting value of y in eqn. (iii) we get,

$\Rightarrow x = \dfrac{12 + 3y}{8} \\[1em] = \dfrac{12 + 3(4)}{8} \\[1em] = \dfrac{12 + 12}{8} \\[1em] = \dfrac{24}{8} \\[1em] = 3.$

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 = $\dfrac{23 - 3y}{2}$ .....(iii)

Substituting value of x from eqn. (iii) in eqn. (ii) we get,

$\Rightarrow 5\Big(\dfrac{23 - 3y}{2}\Big) - 20 = 8y \\[1em] \Rightarrow \dfrac{115 - 15y}{2} - 20 = 8y \\[1em] \Rightarrow \dfrac{115 - 15y - 40}{2} = 8y \\[1em] \Rightarrow 115 - 15y - 40 = 16y \\[1em] \Rightarrow 75 - 15y = 16y \\[1em] \Rightarrow 31y = 75 \\[1em] \Rightarrow y = \dfrac{75}{31} = 2\dfrac{13}{31}.$

Substituting value of y in eqn.(iii) we get,

$\Rightarrow x = \dfrac{23 - 3y}{2} \\[1em] = \dfrac{23 - 3 \times \dfrac{75}{31}}{2} \\[1em] = \dfrac{\dfrac{23 \times 31 - 3 \times 75}{31}}{2} \\[1em] = \dfrac{713 - 225}{31 \times 2} \\[1em] = \dfrac{488}{62} \\[1em] = \dfrac{244}{31} = 7\dfrac{27}{31}.$

Hence, x = $7\dfrac{27}{31}\text{ and y =} 2\dfrac{13}{31}$.

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 = $\dfrac{\text{m}^2 - \text{n}^2}{\text{m} + \text{n}} = \dfrac{(\text{m} - \text{n})(\text{m} + \text{n})}{\text{m} + \text{n}}$ = (m - n).

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:

$\dfrac{2x}{a} + \dfrac{y}{b} = 2$

$\dfrac{x}{a} - \dfrac{y}{b} = 4$

Answer

Given,

$\dfrac{2x}{a} + \dfrac{y}{b} = 2$ .......(i)

$\dfrac{x}{a} - \dfrac{y}{b} = 4$ ........(ii)

Multiplying both side of eqn (i) by ab we get,

$\Rightarrow ab\Big(\dfrac{2x}{a} + \dfrac{y}{b}\Big) = 2ab \\[1em] \Rightarrow 2bx + ay = 2ab ......(iii)$

Multiplying both side of eqn (ii) by ab we get,

$\Rightarrow ab\Big(\dfrac{x}{a} - \dfrac{y}{b}\Big) = 4ab \\[1em] \Rightarrow bx - ay = 4ab \\[1em] \Rightarrow bx = 4ab + ay ......(iv)$

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 = $\dfrac{-6ab}{3a}$ = -2b.

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 $\dfrac{x}{y}.$

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.

⟹ $\dfrac{x}{y} = \dfrac{15}{5} = 3.$

Hence, x = 15, y = 5 and $\dfrac{x}{y} = 3.$

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 = $\dfrac{16}{5}$.

Substituting value of x in eqn. (iii) we get,

$\Rightarrow y = 3 \times \dfrac{16}{5} - 5 \\[1em] = \dfrac{48}{5} - 5 \\[1em] = \dfrac{48 - 25}{5} \\[1em] = \dfrac{23}{5}$

Given, y = px - 3. Substituting value of x and y in equation,

$\Rightarrow \dfrac{23}{5} = \dfrac{16}{5}p - 3 \\[1em] \Rightarrow \dfrac{23}{5} = \dfrac{16p - 15}{5} \\[1em] \Rightarrow 23 = 16p - 15 \\[1em] \Rightarrow 16p = 38 \\[1em] \Rightarrow p = \dfrac{38}{16} = \dfrac{19}{8}.$

Hence, x =$\dfrac{16}{5},\text{ y }= \dfrac{23}{5}\text{ and p} = \dfrac{19}{8}.$