Simultaneous (Linear) Equations — Exercise 6(A)

Solution of equations x - 2y = 7 and 2x - y = 8 is :

1. x = 3, y = 2

2. x = -3, y = 2

3. x = 3, y = -2

4. x = -3, y = -2

Answer

Given,

Equations : x - 2y = 7 and 2x - y = 8.

⇒ x - 2y = 7

⇒ x = 7 + 2y ........(1)

Substituting value of x from equation (1) in 2x - y = 8, we get :

⇒ 2(7 + 2y) - y = 8

⇒ 14 + 4y - y = 8

⇒ 3y = 8 - 14

⇒ 3y = -6

⇒ y = $-\dfrac{6}{3}$ = -2.

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

⇒ x = 7 + 2 × -2 = 7 - 4 = 3.

Hence, Option 3 is the correct option.

Solution of equations x + y = 2.1 and x - y = 0.3 is :

1. x = 1.2, y = 0.9

2. x = 1.2, y = -0.9

3. x = -1.2, y = 0.9

4. x = -1.2, y = -0.9

Answer

Given,

Equations :

⇒ x + y = 2.1 ..........(1)

⇒ x - y = 0.3 ............(2)

Subtracting equation (2) from (1), we get :

⇒ x + y - (x - y) = 2.1 - 0.3

⇒ x - x + y - (-y) = 1.8

⇒ 2y = 1.8

⇒ y = $\dfrac{1.8}{2}$ = 0.9

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

⇒ x + 0.9 = 2.1

⇒ x = 2.1 - 0.9 = 1.2

Hence, Option 1 is the correct option.

Solution of equations $\dfrac{1}{x} + \dfrac{1}{y} = 21 \text{ and } \dfrac{1}{x} - \dfrac{1}{y} + 9 = 0$ is :

1. $x = \dfrac{1}{6}, y = -\dfrac{1}{15}$

2. $x = -\dfrac{1}{6}, y = \dfrac{1}{15}$

3. $x = -\dfrac{1}{6}, y = -\dfrac{1}{15}$

4. $x = \dfrac{1}{6}, y = \dfrac{1}{15}$

Answer

Given,

Equations :

$\Rightarrow \dfrac{1}{x} + \dfrac{1}{y} = 21 .......(1) \\[1em] \Rightarrow \dfrac{1}{x} - \dfrac{1}{y} + 9 = 0 \\[1em] \Rightarrow \dfrac{1}{x} - \dfrac{1}{y} = -9 .......(2)$

Adding equations (1) and (2), we get :

$\Rightarrow \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{x} - \dfrac{1}{y} = 21 + (-9) \\[1em] \Rightarrow \dfrac{2}{x} = 12 \\[1em] \Rightarrow x = \dfrac{2}{12} = \dfrac{1}{6}.$

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

$\Rightarrow \dfrac{1}{x} + \dfrac{1}{y} = 21 \\[1em] \Rightarrow \dfrac{1}{\dfrac{1}{6}} + \dfrac{1}{y} = 21 \\[1em] \Rightarrow 6 + \dfrac{1}{y} = 21 \\[1em] \Rightarrow \dfrac{1}{y} = 21 - 6 \\[1em] \Rightarrow \dfrac{1}{y} = 15 \\[1em] \Rightarrow y = \dfrac{1}{15}.$

Hence, Option 4 is the correct option.

Solution of equations $\dfrac{x}{2} - \dfrac{y}{3} = 0 \text{ and } \dfrac{x}{2} + \dfrac{y}{3} = 6$ is :

1. x = 6, y = -9

2. x = 6, y = 9

3. x = -6, y = -9

4. x = -6, y = 9

Answer

Given,

Equations :

$\Rightarrow \dfrac{x}{2} - \dfrac{y}{3} = 0 ..........(1) \\[1em] \Rightarrow \dfrac{x}{2} + \dfrac{y}{3} = 6 ..........(2)$

Adding equations (1) and (2), we get :

$\Rightarrow \dfrac{x}{2} - \dfrac{y}{3} + \dfrac{x}{2} + \dfrac{y}{3} = 0 + 6 \\[1em] \Rightarrow \dfrac{2x}{2} = 6 \\[1em] \Rightarrow x = 6.$

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

$\Rightarrow \dfrac{6}{2} - \dfrac{y}{3} = 0 ..........(1) \\[1em] \Rightarrow \dfrac{6}{2} = \dfrac{y}{3} \\[1em] \Rightarrow \dfrac{y}{3} = 3 \\[1em] \Rightarrow y = 9.$

Hence, Option 2 is the correct option.

Solution of equations $\dfrac{x + 8}{5} - \dfrac{2y - 4}{3} = 0$ and y - x = 3 is :

1. x = 2, y = 5

2. x = -2, y = 5

3. x = 2, y = -5

4. x = -2, y = -5

Answer

Given,

Equations :

$\Rightarrow \dfrac{x + 8}{5} - \dfrac{2y - 4}{3} = 0$ ......(1)

⇒ y - x = 3

⇒ y = x + 3 .........(2)

Substituting value of y from equation (2) in (1), we get :

$\Rightarrow \dfrac{x + 8}{5} - \dfrac{2(x + 3) - 4}{3} = 0 \\[1em] \Rightarrow \dfrac{x + 8}{5} - \dfrac{2x + 6 - 4}{3} = 0 \\[1em] \Rightarrow \dfrac{x + 8}{5} - \dfrac{2x + 2}{3} = 0 \\[1em] \Rightarrow \dfrac{3(x + 8) - 5(2x + 2)}{15} = 0 \\[1em] \Rightarrow 3x + 24 - 10x - 10 = 0 \\[1em] \Rightarrow -7x + 14 = 0 \\[1em] \Rightarrow 7x = 14 \\[1em] \Rightarrow x = \dfrac{14}{7} = 2.$

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

⇒ y = x + 3 = 2 + 3 = 5.

Hence, Option 1 is the correct option.

Solve the following pairs of linear (simultaneously) equations using method of elimination by substitution:

2x + 3y = 8
2x = 2 + 3y

Answer

Given,

Equations : 2x + 3y = 8 and 2x = 2 + 3y

⇒ 2x + 3y = 8

⇒ 2x = 8 - 3y

⇒ x = $\dfrac{8 - 3y}{2}$ ............(1)

Substituting value of x from equation (1) in 2x = 2 + 3y, we get :

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

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

$\Rightarrow x = \dfrac{8 - 3 \times 1}{2} \\[1em] = \dfrac{8 - 3}{2} \\[1em] = \dfrac{5}{2} = 2.5$

Hence, x = 2.5 and y = 1.

0.2x + 0.1y = 25
2(x - 2) - 1.6y = 116

Answer

Given,

Equations : 0.2x + 0.1y = 25 and 2(x - 2) - 1.6y = 116

⇒ 0.2x + 0.1y = 25

⇒ 0.2x = 25 - 0.1y

⇒ x = $\dfrac{25 - 0.1y}{0.2}$ ..........(1)

$\Rightarrow 2(x - 2) - 1.6y = 116 \\[1em] \Rightarrow 2x - 4 - 1.6y = 116 \\[1em] \Rightarrow 2x - 1.6y = 116 + 4 \\[1em] \Rightarrow 2x - 1.6y = 120 \\[1em]$

Substituting value of x from equation (1) in above equation, we get :

$\Rightarrow 2 \times \Big(\dfrac{25 - 0.1y}{0.2}\Big) - 1.6y = 120 \\[1em] \Rightarrow \dfrac{25 - 0.1y}{0.1} - 1.6y = 120 \\[1em] \Rightarrow \dfrac{25 - 0.1y - 0.16y}{0.1} = 120 \\[1em] \Rightarrow 25 - 0.26y = 120 \times 0.1 \\[1em] \Rightarrow 25 - 0.26y = 12 \\[1em] \Rightarrow 0.26y = 25 - 12 \\[1em] \Rightarrow 0.26y = 13 \\[1em] \Rightarrow y = \dfrac{13}{0.26} = \dfrac{1300}{26} = 50.$

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

$\Rightarrow x = \dfrac{25 - 0.1 \times 50}{0.2} \\[1em] \Rightarrow x = \dfrac{25 - 5}{0.2} \\[1em] \Rightarrow x = \dfrac{20}{0.2} = 100.$

Hence, x = 100 and y = 50.

6x = 7y + 7
7y - x = 8

Answer

Given,

Equations : 6x = 7y + 7 and 7y - x = 8

⇒ 7y - x = 8

⇒ x = 7y - 8 .........(1)

Substituting value of x from equation (1) in 6x = 7y + 7, we get :

⇒ 6(7y - 8) = 7y + 7

⇒ 42y - 48 = 7y + 7

⇒ 42y - 7y = 7 + 48

⇒ 35y = 55

⇒ y = $\dfrac{55}{35} = \dfrac{11}{7}$.

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

$\Rightarrow x = 7 \times \dfrac{11}{7} - 8 \\[1em] \Rightarrow x = 11 - 8 = 3.$

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

y = 4x - 7
16x - 5y = 25

Answer

Given,

Equations :

⇒ y = 4x - 7 .........(1)

⇒ 16x - 5y = 25 .......(2)

Substituting value of y from equation (1) in (2), we get :

⇒ 16x - 5(4x - 7) = 25

⇒ 16x - 20x + 35 = 25

⇒ -4x = 25 - 35

⇒ -4x = -10

⇒ 4x = 10

⇒ x = $\dfrac{10}{4} = \dfrac{5}{2}$.

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

$\Rightarrow y = 4 \times \dfrac{5}{2} - 7 \\[1em] \Rightarrow y = 10 - 7 = 3.$

Hence, x = $\dfrac{5}{2}$ and y = 3.

1.5x + 0.1y = 6.2
3x - 0.4y = 11.2

Answer

Given,

Equations : 1.5x + 0.1y = 6.2 and 3x - 0.4y = 11.2

⇒ 1.5x + 0.1y = 6.2

⇒ 1.5x = 6.2 - 0.1y

⇒ x = $\dfrac{6.2 - 0.1y}{1.5}$ ........(1)

Substituting value of x from equation (1) in 3x - 0.4y = 11.2, we get :

$\Rightarrow 3 \times \Big(\dfrac{6.2 - 0.1y}{1.5}\Big) - 0.4y = 11.2 \\[1em] \Rightarrow \dfrac{6.2 - 0.1y}{0.5} - 0.4y = 11.2 \\[1em] \Rightarrow \dfrac{6.2 - 0.1y - 0.2y}{0.5} = 11.2 \\[1em] \Rightarrow 6.2 - 0.3y = 11.2 \times 0.5 \\[1em] \Rightarrow 6.2 - 0.3y = 5.6 \\[1em] \Rightarrow 0.3y = 6.2 - 5.6 \\[1em] \Rightarrow 0.3y = 0.6 \\[1em] \Rightarrow y = \dfrac{0.6}{0.3} = 2.$

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

$\Rightarrow x = \dfrac{6.2 - 0.1 \times 2}{1.5} \\[1em] = \dfrac{6.2 - 0.2}{1.5} \\[1em] = \dfrac{6}{1.5} \\[1em] = 4.$

Hence, x = 4 and y = 2.

2(x - 3) + 3(y - 5) = 0
5(x - 1) + 4(y - 4) = 0

Answer

Given,

Equations : 2(x - 3) + 3(y - 5) = 0 and 5(x - 1) + 4(y - 4) = 0

⇒ 2(x - 3) + 3(y - 5) = 0

⇒ 2x - 6 + 3y - 15 = 0

⇒ 2x + 3y - 21 = 0

⇒ 2x = 21 - 3y

⇒ x = $\dfrac{21 - 3y}{2}$ ..........(1)

⇒ 5(x - 1) + 4(y - 4) = 0

⇒ 5x - 5 + 4y - 16 = 0

⇒ 5x + 4y - 21 = 0

Substituting value of x from equation (1) in above equation, we get :

$\Rightarrow 5 \times \Big(\dfrac{21 - 3y}{2}\Big) + 4y - 21 = 0 \\[1em] \Rightarrow \dfrac{105 - 15y}{2} + 4y - 21 = 0 \\[1em] \Rightarrow \dfrac{105 - 15y + 8y - 42}{2} = 0 \\[1em] \Rightarrow 63 - 7y = 0 \\[1em] \Rightarrow 7y = 63 \\[1em] \Rightarrow y = \dfrac{63}{7} = 9.$

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

$\Rightarrow x = \dfrac{21 - 3 \times 9}{2} \\[1em] = \dfrac{21 - 27}{2} \\[1em] = -\dfrac{6}{2} \\[1em] = -3.$

Hence, x = -3 and y = 9.

$\dfrac{2x + 1}{7} + \dfrac{5y - 3}{3} = 12$

$\dfrac{3x + 2}{2} - \dfrac{4y + 3}{9} = 13$

Answer

Simplifying first equation :

$\Rightarrow \dfrac{2x + 1}{7} + \dfrac{5y - 3}{3} = 12 \\[1em] \Rightarrow \dfrac{3(2x + 1) + 7(5y - 3)}{21} = 12 \\[1em] \Rightarrow 6x + 3 + 35y - 21 = 12 \times 21 \\[1em] \Rightarrow 6x + 35y - 18 = 252 \\[1em] \Rightarrow 6x + 35y = 252 + 18 \\[1em] \Rightarrow 6x + 35y = 270 \\[1em] \Rightarrow 6x = 270 - 35y \\[1em] \Rightarrow x = \dfrac{270 - 35y}{6} \text{ ........(1)}$

Simplifying second equation :

$\Rightarrow \dfrac{3x + 2}{2} - \dfrac{4y + 3}{9} = 13 \\[1em] \Rightarrow \dfrac{9(3x + 2) - 2(4y + 3)}{18} = 13 \\[1em] \Rightarrow 27x + 18 - 8y - 6 = 18 \times 13 \\[1em] \Rightarrow 27x - 8y + 12 = 234 \\[1em] \Rightarrow 27x - 8y = 234 -12 \\[1em] \Rightarrow 27x - 8y = 222 \text{ .......(2)}$

Substituting value of x from equation (1) in (2), we get :

$\Rightarrow 27 \times \Big(\dfrac{270 - 35y}{6}\Big) - 8y = 222 \\[1em] \Rightarrow \dfrac{9(270 - 35y)}{2} - 8y = 222 \\[1em] \Rightarrow \dfrac{2430 - 315y - 16y}{2} = 222 \\[1em] \Rightarrow 2430 - 331y = 444 \\[1em] \Rightarrow 331y = 2430 - 444 \\[1em] \Rightarrow 331y = 1986 \\[1em] \Rightarrow y = \dfrac{1986}{331} = 6.$

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

$\Rightarrow x = \dfrac{270 - 35 \times 6}{6} \\[1em] = \dfrac{270 - 210}{6} \\[1em] = \dfrac{60}{6} \\[1em] = 10.$

Hence, x = 10 and y = 6.

For solving each pair of equations, use the method of elimination by equating coefficients :

3x - y = 23

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

Answer

Given equations :

⇒ 3x - y = 23 .........(1)

⇒ $\dfrac{x}{3} + \dfrac{y}{4}$ = 4 ......(2)

Multiplying equation (1) by 3, we get :

⇒ 3(3x - y) = 3 × 23

⇒ 9x - 3y = 69 .........(3)

Multiplying equation (2) by 12, we get :

⇒ $12\Big(\dfrac{x}{3} + \dfrac{y}{4}\Big) = 4 \times 12$

⇒ 4x + 3y = 48 ...........(4)

Adding equation (3) and (4), we get :

⇒ 9x - 3y + 4x + 3y = 69 + 48

⇒ 13x = 117

⇒ x = $\dfrac{117}{13}$ = 9.

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

⇒ 3 × 9 - y = 23

⇒ 27 - y = 23

⇒ y = 27 - 23 = 4.

Hence, x = 9 and y = 4.

$\dfrac{5y}{2} - \dfrac{x}{3} = 8$

$\dfrac{y}{2} + \dfrac{5x}{3} = 12$

Answer

Given equations :

$\Rightarrow \dfrac{5y}{2} - \dfrac{x}{3} = 8 ..........(1) \\[1em] \Rightarrow \dfrac{y}{2} + \dfrac{5x}{3} = 12 ......(2)$

Multiplying equation (1) by 30, we get :

$\Rightarrow 30\Big(\dfrac{5y}{2} - \dfrac{x}{3}\Big) = 30 \times 8 \\[1em] \Rightarrow 75y - 10x = 240 ......(3)$

Multiplying equation (2) by 6, we get :

$\Rightarrow 6\Big(\dfrac{y}{2} + \dfrac{5x}{3}\Big) = 12 \times 6 \\[1em] \Rightarrow 3y + 10x = 72 ...........(4)$

Adding equations (3) and (4), we get :

⇒ 75y - 10x + 3y + 10x = 240 + 72

⇒ 78y = 312

⇒ y = $\dfrac{312}{78}$ = 4.

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

$\Rightarrow \dfrac{5 \times 4}{2} - \dfrac{x}{3} = 8 \\[1em] \Rightarrow 10 - \dfrac{x}{3} = 8 \\[1em] \Rightarrow \dfrac{x}{3} = 10 - 8 \\[1em] \Rightarrow x = 3 \times 2 = 6.$

Hence, x = 6 and y = 4.

$\dfrac{1}{5}(x - 2) = \dfrac{1}{4}(1 - y)$

26x + 3y + 4 = 0

Answer

Simplifying first equation :

⇒ $\dfrac{1}{5}(x - 2) = \dfrac{1}{4}(1 - y)$

⇒ 4(x - 2) = 5(1 - y)

⇒ 4x - 8 = 5 - 5y

⇒ 4x + 5y - 8 - 5 = 0

⇒ 4x + 5y - 13 = 0 .......(1)

⇒ 26x + 3y + 4 = 0 .......(2)

Multiplying equation (1) by 3, we get :

⇒ 3(4x + 5y - 13) = 0

⇒ 12x + 15y - 39 = 0 .......(3)

Multiplying equation (2) by 5, we get :

⇒ 5(26x + 3y + 4) = 0

⇒ 130x + 15y + 20 = 0 .......(4)

Subtracting equation (3) from (4), we get :

⇒ 130x + 15y + 20 - (12x + 15y - 39) = 0

⇒ 130x - 12x + 15y - 15y + 20 - (-39) = 0

⇒ 118x + 59 = 0

⇒ 118x = -59

⇒ x = $-\dfrac{59}{118} = -\dfrac{1}{2}$

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

$\Rightarrow 4x + 5y - 13 = 0 \\[1em] \Rightarrow 4 \times -\dfrac{1}{2} + 5y - 13 = 0 \\[1em] \Rightarrow -2 + 5y - 13 = 0 \\[1em] \Rightarrow 5y - 15 = 0 \\[1em] \Rightarrow 5y = 15 \\[1em] \Rightarrow y = \dfrac{15}{5} = 3.$

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

$\dfrac{x - y}{6} = 2(4 - x)$

2x + y = 3(x - 4)

Answer

Simplifying first equation :

⇒ $\dfrac{x - y}{6} = 2(4 - x)$

⇒ x - y = 12(4 - x)

⇒ x - y = 48 - 12x

⇒ x + 12x - y = 48

⇒ 13x - y = 48

⇒ 13x - y - 48 = 0 .......(1)

Simplifying second equation :

⇒ 2x + y = 3(x - 4)

⇒ 2x + y = 3x - 12

⇒ y + 2x - 3x + 12 = 0

⇒ y - x + 12 = 0 .......(2)

Adding equations (1) and (2), we get :

⇒ (13x - y - 48) + (y - x + 12) = 0

⇒ 13x - x - y + y - 48 + 12 = 0

⇒ 12x - 36 = 0

⇒ 12x = 36

⇒ x = $\dfrac{36}{12}$ = 3.

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

⇒ y - 3 + 12 = 0

⇒ y + 9 = 0

⇒ y = -9.

Hence, x = 3 and y = -9.

2x - 3y - 3 = 0

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

Answer

Given, equations :

⇒ 2x - 3y - 3 = 0 .............(1)

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

Simplifying second equation :

$\Rightarrow \dfrac{2x}{3} + 4y + \dfrac{1}{2} = 0 \\[1em] \Rightarrow \dfrac{4x + 24y + 3}{6} = 0 \\[1em] \Rightarrow 4x + 24y + 3 = 0 \text{ .......(3)}$

Multiplying equation (1) by 2, we get :

⇒ 2(2x - 3y - 3) = 2 × 0

⇒ 4x - 6y - 6 = 0 .........(4)

Subtracting equation (4) from (3), we get :

⇒ 4x + 24y + 3 - (4x - 6y - 6) = 0

⇒ 4x - 4x + 24y + 6y + 3 + 6 = 0

⇒ 30y + 9 = 0

⇒ 30y = -9

⇒ y = $-\dfrac{9}{30} = -\dfrac{3}{10}$.

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

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

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

13x + 11y = 70

11x + 13y = 74

Answer

Given, equations :

⇒ 13x + 11y = 70 ..........(1)

⇒ 11x + 13y = 74 ..........(2)

Multiplying equation (1) by 11, we get :

⇒ 11(13x + 11y) = 11 × 70

⇒ 143x + 121y = 770 ........(3)

Multiplying equation (2) by 13, we get :

⇒ 13(11x + 13y) = 13 × 74

⇒ 143x + 169y = 962 ........(4)

Subtracting equation (3) from (4), we get :

⇒ 143x + 169y - (143x + 121y) = 962 - 770

⇒ 143x - 143x + 169y - 121y = 192

⇒ 48y = 192

⇒ y = $\dfrac{192}{48}$ = 4

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

⇒ 13x + 11(4) = 70

⇒ 13x + 44 = 70

⇒ 13x = 70 - 44

⇒ 13x = 26

⇒ x = $\dfrac{26}{13}$ = 2.

Hence, x = 2 and y = 4.

41x + 53y = 135

53x + 41y = 147

Answer

Given, equations :

⇒ 41x + 53y = 135 .........(1)

⇒ 53x + 41y = 147 .........(2)

Multiplying equation (1) by 53, we get :

⇒ 53(41x + 53y) = 53 × 135

⇒ 2173x + 2809y = 7155 .......(3)

Multiplying equation (2) by 41, we get :

⇒ 41(53x + 41y) = 41 × 147

⇒ 2173x + 1681y = 6027 .......(4)

Subtracting equation (4) from (3), we get :

⇒ 2173x + 2809y - (2173x + 1681y) = 7155 - 6027

⇒ 2173x - 2173x + 2809y - 1681y = 1128

⇒ 1128y = 1128

⇒ y = $\dfrac{1128}{1128}$ = 1.

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

⇒ 41x + 53y = 135

⇒ 41x + 53(1) = 135

⇒ 41x + 53 = 135

⇒ 41x = 135 - 53

⇒ 41x = 82

⇒ x = $\dfrac{82}{41}$ = 2.

Hence, x = 2 and y = 1.

If 2x + y = 23 and 4x - y = 19; find the values of x - 3y and 5y - 2x.

Answer

Given,

Equations : 2x + y = 23 and 4x - y = 19

⇒ 2x + y = 23

⇒ y = 23 - 2x .......(1)

Substituting value of y from equation (1) in 4x - y = 19, we get :

⇒ 4x - (23 - 2x) = 19

⇒ 4x - 23 + 2x = 19

⇒ 6x = 19 + 23

⇒ 6x = 42

⇒ x = $\dfrac{42}{6}$ = 7.

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

⇒ y = 23 - 2(7) = 23 - 14 = 9.

⇒ x - 3y = 7 - 3 × 9 = 7 - 27 = -20

⇒ 5y - 2x = 5 × 9 - 2 × 7 = 45 - 14 = 31.

Hence, x - 3y = -20 and 5y - 2x = 31.

If 10y = 7x - 4 and 12x + 18y = 1; find the values of 4x + 6y and 8y - x.

Answer

Given,

Equations : 10y = 7x - 4 and 12x + 18y = 1

⇒ 10y = 7x - 4

⇒ y = $\dfrac{7x - 4}{10}$ .........(1)

Substituting value of y from equation (1) in 12x + 18y = 1, we get :

$\Rightarrow 12x + 18 \times \Big(\dfrac{7x - 4}{10}\Big) = 1 \\[1em] \Rightarrow 12x + \dfrac{126x - 72}{10} = 1 \\[1em] \Rightarrow \dfrac{120x + 126x - 72}{10} = 1\\[1em] \Rightarrow 246x - 72 = 10 \\[1em] \Rightarrow 246x = 10 + 72 \\[1em] \Rightarrow 246x = 82 \\[1em] \Rightarrow x = \dfrac{82}{246} = \dfrac{1}{3}.$

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

$\Rightarrow y = \dfrac{7 \times \dfrac{1}{3} - 4}{10}\\[1em] = \dfrac{\dfrac{7}{3} - 4}{10} \\[1em] = \dfrac{\dfrac{7 - 12}{3}}{10} \\[1em] = \dfrac{-5}{3 \times 10} \\[1em] = \dfrac{-5}{30} \\[1em] = -\dfrac{1}{6}.$

Substituting value of x and y in 4x + 6y and 8y - x, we get :

$\Rightarrow 4x + 6y = 4 \times \dfrac{1}{3} + 6 \times -\dfrac{1}{6} \\[1em] = \dfrac{4}{3} + (-1) \\[1em] = \dfrac{4 - 3}{3} \\[1em] = \dfrac{1}{3}. \\[1em] \Rightarrow 8y - x = 8 \times -\dfrac{1}{6} - \dfrac{1}{3} \\[1em] = -\dfrac{4}{3} - \dfrac{1}{3} \\[1em] = \dfrac{-4 - 1}{3} \\[1em] = -\dfrac{5}{3}.$

Hence, $4x + 6y = \dfrac{1}{3} \text{ and } 8y - x = -\dfrac{5}{3}$.

Solve for x and y :

$\dfrac{y + 7}{5} = \dfrac{2y - x}{4} + 3x - 5$

$\dfrac{7 - 5x}{2} + \dfrac{3 - 4y}{6} = 5y - 18$

Answer

Simplifying first equation :

$\Rightarrow \dfrac{y + 7}{5} = \dfrac{2y - x}{4} + 3x - 5 \\[1em] \Rightarrow \dfrac{y + 7}{5} = \dfrac{2y - x + 12x - 20}{4} \\[1em] \Rightarrow 4(y + 7) = 5(2y + 11x - 20) \\[1em] \Rightarrow 4y + 28 = 10y + 55x - 100 \\[1em] \Rightarrow 55x + 10y - 4y = 100 + 28 \\[1em] \Rightarrow 55x + 6y = 128 \\[1em] \Rightarrow 55x = 128 - 6y \\[1em] \Rightarrow x = \dfrac{128 - 6y}{55}\text{ .......(1)}$

Simplifying second equation :

$\Rightarrow \dfrac{7 - 5x}{2} + \dfrac{3 - 4y}{6} = 5y - 18 \\[1em] \Rightarrow \dfrac{3(7 - 5x) + 3 - 4y}{6} = 5y - 18 \\[1em] \Rightarrow \dfrac{21 - 15x + 3 - 4y}{6} = 5y - 18 \\[1em] \Rightarrow 24 - 15x - 4y = 6(5y - 18) \\[1em] \Rightarrow 24 - 15x - 4y = 30y - 108 \\[1em] \Rightarrow 15x + 30y + 4y = 108 + 24 \\[1em] \Rightarrow 15x + 34y = 132 \text{ ......(2)}.$

Substituting value of x from equation (1) in (2), we get :

$\Rightarrow 15 \times \dfrac{128 - 6y}{55} + 34y = 132 \\[1em] \Rightarrow \dfrac{3}{11} \times (128 - 6y) + 34y = 132 \\[1em] \Rightarrow \dfrac{384 - 18y + 374y}{11} = 132 \\[1em] \Rightarrow 384 - 18y + 374y = 1452 \\[1em] \Rightarrow 384 + 356y = 1452 \\[1em] \Rightarrow 356y = 1452 - 384 \\[1em] \Rightarrow 356y = 1068 \\[1em] \Rightarrow y = \dfrac{1068}{356} = 3.$

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

$\Rightarrow x = \dfrac{128 - 6 \times 3}{55} \\[1em] = \dfrac{128 - 18}{55}\\[1em] = \dfrac{110}{55} \\[1em] = 2.$

Hence, x = 2 and y = 3.

Solve for x and y :

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

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

Answer

Simplifying first equation :

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

Simplifying second equation :

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

From equation (1) and (2), we get :

⇒ 33x - 136 = 8 - 3x

⇒ 33x + 3x = 136 + 8

⇒ 36x = 144

⇒ x = $\dfrac{144}{36}$ = 4.

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

⇒ y = 8 - 3(4) = 8 - 12 = -4.

Hence, x = 4 and y = -4.