Pair of Linear Equations in Two Variables

Form the pair of linear equations in the following problems, and find their solutions graphically.

10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

Answer

Let no. of boys be x and no. of girls be y.

Given,

Total students = 10

⇒ x + y = 10

⇒ y = 10 - x ........(1)

Given,

No. of girls is 4 more than no. of boys.

⇒ y - x = 4

⇒ y = x + 4 ..........(2)

Table of values for equation (1),

Table of values for equation (2),

Steps of construction :

1. Plot the points (5, 5), (6, 4) and (7, 3) and join them to form equation (1).

2. Plot the points (0, 4), (1, 5) and (2, 6) and join them to form equation (2).

3. The lines intersect at K(3, 7), which is the required solution.

Hence, linear equation pair are x + y = 10 and y - x = 4, where x is no. of boys and y is no. of girls and no. of boys = 3 and no. of girls = 7.

Form the pair of linear equations in the following problems, and find their solutions graphically.

5 pencils and 7 pens together cost ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.

Answer

Let cost of one pencil be ₹ x and cost of one pen be ₹ y.

Given,

5 pencils and 7 pens together cost ₹ 50.

⇒ 5x + 7y = 50

⇒ 7y = 50 - 5x

⇒ y = $\dfrac{50 - 5x}{7}$ .........(1)

Also,

7 pencils and 5 pens together cost ₹ 46.

⇒ 7x + 5y = 46

⇒ 5y = 46 - 7x

⇒ y = $\dfrac{46 - 7x}{5}$ .........(2)

Table of values for equation (1),

Table of values for equation (2),

Steps of construction :

1. Plot the points (3, 5) and (10, 0) join them to form equation (1).

2. Plot the points (3, 5) and (8, -2) and join them to form equation (2).

3. The lines intersect at A(3, 5), which is the required solution.

Hence, linear equations pair are 5x + 7y = 50 and 7x + 5y = 46, where x is the cost of one pencil and y is the cost of one pen and cost of one pencil = ₹ 3 and cost of one pen = ₹ 5.

On comparing the ratios $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$ find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident :

5x - 4y + 8 = 0 and 7x + 6y - 9 = 0

Answer

Comparing equations 5x - 4y + 8 = 0 and 7x + 6y - 9 = 0 with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 respectively, we get :

a₁ = 5, b₁ = -4, c₁ = 8, a₂ = 7, b₂ = 6 and c₂ = -9.

$\dfrac{a_1}{a_2} = \dfrac{5}{7}$

$\dfrac{b_1}{b_2} = \dfrac{-4}{6} = -\dfrac{2}{3}$.

Since, $\dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2}$.

Hence, the set of linear equations intersect at a point.

On comparing the ratios $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$ find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident :

9x + 3y + 12 = 0 and 18x + 6y + 24 = 0

Answer

Comparing equations 9x + 3y + 12 = 0 and 18x + 6y + 24 = 0 with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 respectively, we get :

a₁ = 9, b₁ = 3, c₁ = 12, a₂ = 18, b₂ = 6 and c₂ = 24.

$\dfrac{a_1}{a_2} = \dfrac{9}{18} = \dfrac{1}{2}$

$\dfrac{b_1}{b_2} = \dfrac{3}{6} = \dfrac{1}{2}$

$\dfrac{c_1}{c_2} = \dfrac{12}{24} = \dfrac{1}{2}$

Since, $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$.

Hence, the set of linear equations are coincident.

On comparing the ratios $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$ find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident :

6x - 3y + 10 = 0 and 2x - y + 9 = 0

Answer

Comparing equations 6x - 3y + 10 = 0 and 2x - y + 9 = 0 with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 respectively, we get :

a₁ = 6, b₁ = -3, c₁ = 10, a₂ = 2, b₂ = -1 and c₂ = 9.

$\dfrac{a_1}{a_2} = \dfrac{6}{2} = 3$

$\dfrac{b_1}{b_2} = \dfrac{-3}{-1} = 3$

$\dfrac{c_1}{c_2} = \dfrac{10}{9}$

Since, $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}$.

Hence, the set of linear equations are parallel.

On comparing the ratios $\dfrac{a_1}{a_2}, \dfrac{b_1}{b_2}, \dfrac{c_1}{c_2}$ find out whether the following pair of linear equations are consistent, or inconsistent.

3x + 2y = 5 ; 2x - 3y = 7

Answer

Given,

Equations : 3x + 2y = 5 ; 2x - 3y = 7 or,

3x + 2y - 5 = 0 ; 2x - 3y - 7 = 0.

Comparing equations 3x + 2y - 5 = 0 and 2x - 3y - 7 = 0 with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 respectively, we get :

a₁ = 3, b₁ = 2, c₁ = -5, a₂ = 2, b₂ = -3 and c₂ = -7.

$\dfrac{a_1}{a_2} = \dfrac{3}{2}$

$\dfrac{b_1}{b_2} = \dfrac{2}{-3} = -\dfrac{2}{3}$

Since, $\dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2}$.

Hence, the set of linear equations are consistent.

On comparing the ratios $\dfrac{a_1}{a_2}, \dfrac{b_1}{b_2}, \dfrac{c_1}{c_2}$ find out whether the following pair of linear equations are consistent, or inconsistent.

2x - 3y = 8 ; 4x - 6y = 9

Answer

Given,

Equations : 2x - 3y = 8 ; 4x - 6y = 9 or,

2x - 3y - 8 = 0 ; 4x - 6y - 9 = 0.

Comparing equations 2x - 3y - 8 = 0 and 4x - 6y - 9 = 0 with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 respectively, we get :

a₁ = 2, b₁ = -3, c₁ = -8, a₂ = 4, b₂ = -6 and c₂ = -9.

$\dfrac{a_1}{a_2} = \dfrac{2}{4} = \dfrac{1}{2}$

$\dfrac{b_1}{b_2} = \dfrac{-3}{-6} = \dfrac{1}{2}$

$\dfrac{c_1}{c_2} = \dfrac{-8}{-9} = \dfrac{8}{9}$

Since, $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}$.

Hence, the set of linear equations are inconsistent.

On comparing the ratios $\dfrac{a_1}{a_2}, \dfrac{b_1}{b_2}, \dfrac{c_1}{c_2}$ find out whether the following pair of linear equations are consistent, or inconsistent.

$\dfrac{3}{2}x + \dfrac{5}{3}y = 7$ ; 9x - 10y = 14

Answer

Given,

Equations : $\dfrac{3}{2}x + \dfrac{5}{3}y = 7$ ; 9x - 10y = 14 or,

$\Rightarrow \dfrac{9x + 10y}{6} = 7$; 9x - 10y = 14 or,

⇒ 9x + 10y = 42; 9x - 10y = 14 or,

⇒ 9x + 10y - 42 = 0; 9x - 10y - 14 = 0.

Comparing equations 9x + 10y - 42 = 0 and 9x - 10y - 14 = 0 with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 respectively, we get :

a₁ = 9, b₁ = 10, c₁ = -42, a₂ = 9, b₂ = -10 and c₂ = -14.

$\dfrac{a_1}{a_2} = \dfrac{9}{9} = 1$

$\dfrac{b_1}{b_2} = \dfrac{10}{-10} = -1$

Since, $\dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2}$.

Hence, the set of linear equations are consistent.

On comparing the ratios $\dfrac{a_1}{a_2}, \dfrac{b_1}{b_2}, \dfrac{c_1}{c_2}$ find out whether the following pair of linear equations are consistent, or inconsistent.

5x - 3y = 11 ; -10x + 6y = -22

Answer

Given,

Equations : 5x - 3y = 11 ; -10x + 6y = -22 or,

5x - 3y - 11 = 0 ; -10x + 6y + 22 = 0

Comparing equations 5x - 3y - 11 = 0 and -10x + 6y + 22 = 0 with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 respectively, we get :

a₁ = 5, b₁ = -3, c₁ = -11, a₂ = -10, b₂ = 6 and c₂ = 22.

$\dfrac{a_1}{a_2} = \dfrac{5}{-10} = -\dfrac{1}{2}$

$\dfrac{b_1}{b_2} = \dfrac{-3}{6} = -\dfrac{1}{2}$

$\dfrac{c_1}{c_2} = \dfrac{-11}{22} = -\dfrac{1}{2}$

Since, $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$.

Hence, the set of linear equations are consistent.

On comparing the ratios $\dfrac{a_1}{a_2}, \dfrac{b_1}{b_2}, \dfrac{c_1}{c_2}$ find out whether the following pair of linear equations are consistent, or inconsistent.

$\dfrac{4}{3}x + 2y = 8; 2x + 3y = 12$

Answer

Given,

Equations : $\dfrac{4}{3}x + 2y = 8; 2x + 3y = 12$ or,

⇒ 4x + 6y = 24; 2x + 3y = 12

⇒ 4x + 6y - 24 = 0 ; 2x + 3y - 12 = 0.

Comparing equations 4x + 6y - 24 = 0 and 2x + 3y - 12 = 0 with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 respectively, we get :

a₁ = 4, b₁ = 6, c₁ = -24, a₂ = 2, b₂ = 3 and c₂ = -12.

$\dfrac{a_1}{a_2} = \dfrac{4}{2} = 2$

$\dfrac{b_1}{b_2} = \dfrac{6}{3} = 2$

$\dfrac{c_1}{c_2} = \dfrac{-24}{-12} = 2$

Since, $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$.

Hence, the set of linear equations are consistent.

Which of the following pairs of linear equations are consistent/inconsistent ? If consistent, obtain the solution graphically :

x + y = 5, 2x + 2y = 10

Answer

Given,

Equations :

⇒ x + y = 5, 2x + 2y = 10

⇒ x + y - 5 = 0, 2x + 2y - 10 = 0

Comparing equations x + y - 5 = 0 and 2x + 2y - 10 = 0 with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 respectively, we get :

a₁ = 1, b₁ = 1, c₁ = -5, a₂ = 2, b₂ = 2 and c₂ = -10.

$\dfrac{a_1}{a_2} = \dfrac{1}{2}$

$\dfrac{b_1}{b_2} = \dfrac{1}{2}$

$\dfrac{c_1}{c_2} = \dfrac{-5}{-10} = \dfrac{1}{2}$

Since, $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$.

Hence, pair of lines are consistent.

1st equation : x + y - 5 = 0

⇒ y = 5 - x .......(1)

2nd equation : 2x + 2y - 10 = 0

⇒ 2y = 10 - 2x

⇒ y = $\dfrac{10 - 2x}{2}$ .......(2)

Table for equation (1) :

Table for equation (2) :

Steps of construction :

1. Plot the points (1, 4), (2, 3) and (3, 2) and join them to form equation (1).

2. Plot the points (1, 4), (2, 3) and (3, 2) and join them to form equation (2).

From graph,

Both the lines are coincident.

Hence, all the points on the co-incident line are solution i.e. infinitely many solutions.

Which of the following pairs of linear equations are consistent/inconsistent ? If consistent, obtain the solution graphically :

x - y = 8, 3x - 3y = 16

Answer

Given,

Equations :

⇒ x - y = 8, 3x - 3y = 16

⇒ x - y - 8 = 0, 3x - 3y - 16 = 0.

Comparing equations x - y - 8 = 0 and 3x - 3y - 16 = 0 with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 respectively, we get :

a₁ = 1, b₁ = -1, c₁ = -8, a₂ = 3, b₂ = -3 and c₂ = -16.

$\dfrac{a_1}{a_2} = \dfrac{1}{3}$

$\dfrac{b_1}{b_2} = \dfrac{-1}{-3} = \dfrac{1}{3}$

$\dfrac{c_1}{c_2} = \dfrac{-8}{-16} = \dfrac{1}{2}$

Since, $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}$.

Hence, pair of lines are inconsistent.

Which of the following pairs of linear equations are consistent/inconsistent ? If consistent, obtain the solution graphically :

2x + y - 6 = 0, 4x - 2y - 4 = 0

Answer

Given,

Equations :

⇒ 2x + y - 6 = 0, 4x - 2y - 4 = 0

Comparing equations 2x + y - 6 = 0 and 4x - 2y - 4 = 0 with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 respectively, we get :

a₁ = 2, b₁ = 1, c₁ = -6, a₂ = 4, b₂ = -2 and c₂ = -4.

$\dfrac{a_1}{a_2} = \dfrac{2}{4} = \dfrac{1}{2}$

$\dfrac{b_1}{b_2} = \dfrac{1}{-2} = -\dfrac{1}{2}$

Since, $\dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2}$.

Hence, pair of lines are consistent.

1st equation : 2x + y - 6 = 0

⇒ y = 6 - 2x ..........(1)

2nd equation : 4x - 2y - 4 = 0

⇒ 2y = 4x - 4

⇒ 2y = 2(2x - 2)

⇒ y = 2x - 2 .........(2)

Table of values of equation (1),

Table of values of equation (2),

Steps of construction :

1. Plot the points (2, 2), (3, 0) and (4, -2) and join them to form equation (1).

2. Plot the points (0, -2), (1, 0) and (2, 2) and join them to form equation (2).

3. The lines intersect at A(2, 2), which is the required solution.

Hence, above pair of equations has a unique solution i.e. x = 2 and y = 2.

Which of the following pairs of linear equations are consistent/inconsistent ? If consistent, obtain the solution graphically :

2x - 2y - 2 = 0, 4x - 4y - 5 = 0

Answer

Given,

Equations :

⇒ 2x - 2y - 2 = 0, 4x - 4y - 5 = 0

Comparing equations 2x - 2y - 2 = 0 and 4x - 4y - 5 = 0 with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 respectively, we get :

a₁ = 2, b₁ = -2, c₁ = -2, a₂ = 4, b₂ = -4 and c₂ = -5.

$\dfrac{a_1}{a_2} = \dfrac{2}{4} = \dfrac{1}{2}$

$\dfrac{b_1}{b_2} = \dfrac{-2}{-4} = \dfrac{1}{2}$

$\dfrac{c_1}{c_2} = \dfrac{-2}{-5} = \dfrac{2}{5}$

Since, $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}$.

Hence, pair of lines are inconsistent.

Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Answer

Let width of rectangular garden be x meters.

Length = (x + 4) meters.

Given,

Half of perimeter = 36 m

By formula,

⇒ Perimeter = 2(Length + Width)

⇒ Half of perimeter = (Length + Width)

⇒ 36 = [x + (x + 4)]

⇒ 2x + 4 = 36

⇒ 2x = 36 - 4

⇒ 2x = 32

⇒ x = $\dfrac{32}{2}$ = 16 meters.

⇒ x + 4 = 16 + 4 = 20 meters.

Hence, length = 20 meters and width = 16 meters.

Given the linear equation 2x + 3y - 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is :

(i) intersecting lines

(ii) parallel lines

(iii) coincident lines

Answer

For any pair of linear equations,

a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

(i) For intersecting lines

Condition : $\dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2}$.

For,

2x + 3y - 8 = 0

a₁ = 2 and b₁ = 3

So, considering a₂ = 3 and b₂ = 2 will satisfy the condition for intersecting lines. c₂ can be any value.

Let c₂ be -7.

So,

⇒ a₂x + b₂y + c₂ = 0

⇒ 3x + 2y - 7 = 0.

Hence, equation of required line is 3x + 2y - 7 = 0.

(ii) For parallel lines

Condition : $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}$.

For,

2x + 3y - 8 = 0

a₁ = 2, b₁ = 3 and c₁ = -8.

So, considering a₂ = 2, b₂ = 3 and c₂ = -12 will satisfy the condition for parallel lines.

$\dfrac{a_1}{a_2} = \dfrac{2}{2} = 1$

$\dfrac{b_1}{b_2} = \dfrac{3}{3} = 1$

$\dfrac{c_1}{c_2} = \dfrac{-8}{-12} = \dfrac{2}{3}$

Since, $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}$.

Substituting values we get :,

⇒ a₂x + b₂y + c₂ = 0

⇒ 2x + 3y - 12 = 0.

Hence, equation of required line is 2x + 3y - 12 = 0.

(iii) For coincident lines

Condition : $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$.

For,

2x + 3y - 8 = 0

a₁ = 2, b₁ = 3 and c₁ = -8.

So, considering a₂ = 4, b₂ = 6 and c₂ = -16 will satisfy the condition for coincident lines.

$\dfrac{a_1}{a_2} = \dfrac{2}{4} = \dfrac{1}{2}$

$\dfrac{b_1}{b_2} = \dfrac{3}{6} = \dfrac{1}{2}$

$\dfrac{c_1}{c_2} = \dfrac{-8}{-16} = \dfrac{1}{2}$

Since, $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$.

Substituting values we get :,

⇒ a₂x + b₂y + c₂ = 0

⇒ 4x + 6y - 16 = 0.

Hence, equation of required line is 4x + 6y - 16 = 0.

Draw the graphs of the equations x - y + 1 = 0 and 3x + 2y - 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Answer

Given,

1st equation : x - y + 1 = 0

⇒ y = x + 1 .........(1)

2nd equation : 3x + 2y - 12 = 0

⇒ 2y = 12 - 3x

⇒ y = $\dfrac{12 - 3x}{2}$ .........(2)

Table for equation (1),

Table for equation (2),

Steps of construction :

1. Plot the points (0, 1) and (4, 5) join them to form equation (1).

2. Plot the points (0, 6) and (2, 3) and join them to form equation (2).

3. The lines intersect at K(2, 3).

From graph,

x - y + 1 = 0 meets x-axis at point (-1, 0) and 3x + 2y - 12 = 0 meets x-axis at point (4, 0).

Hence, coordinates of vertices of triangle are (2, 3), (-1, 0) and (4, 0).

Solve the following pair of linear equations by the substitution method.

x + y = 14 and x - y = 4

Answer

Given,

x + y = 14 .......(1)

x - y = 4 ........(2)

Solving equation (1),

⇒ x + y = 14

⇒ y = 14 - x ...........(3)

Substituting above value of y in x - y = 4, we get :

⇒ x - (14 - x) = 4

⇒ x - 14 + x = 4

⇒ 2x = 4 + 14

⇒ 2x = 18

⇒ x = 9.

Substituting value of x in equation (3), we get :

⇒ y = 14 - 9 = 5.

Hence, x = 9 and y = 5.

Solve the following pair of linear equations by the substitution method.

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

Answer

Given,

s - t = 3 .........(1)

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

Solving equation (1),

⇒ s - t = 3

⇒ s = t + 3 ...........(3)

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

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

Substituting value of t in equation (3), we get :

⇒ s = 6 + 3 = 9.

Hence, s = 9 and t = 6.

Solve the following pair of linear equations by the substitution method.

3x - y = 3 and 9x - 3y = 9

Answer

Given,

3x - y = 3 .........(1)

9x - 3y = 9 ........(2)

Solving equation (1),

⇒ 3x - y = 3

⇒ y = 3x - 3

Solving equation (2),

⇒ 9x - 3y = 9

⇒ 3(3x - y) = 9

⇒ 3x - y = 3

⇒ y = 3x - 3

We get, y = 3x - 3 by solving both equation (1) and (2).

Hence, y = 3x - 3, where x can take any value, i.e., infinitely many solutions.

Solve the following pair of linear equations by the substitution method.

0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3

Answer

Given,

0.2x + 0.3y = 1.3 ............(1)

0.4x + 0.5y = 2.3 ............(2)

Solving equation (1),

⇒ 0.2x + 0.3y = 1.3

⇒ 0.2x = 1.3 - 0.3y

⇒ x = $\dfrac{1.3 - 0.3y}{0.2}$ ............(3)

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

$\Rightarrow 0.4 \times \Big(\dfrac{1.3 - 0.3y}{0.2}\Big) + 0.5y = 2.3 \\[1em] \Rightarrow 2(1.3 - 0.3y) + 0.5y = 2.3 \\[1em] \Rightarrow 2.6 - 0.6y + 0.5y = 2.3 \\[1em] \Rightarrow 2.6 - 0.1y = 2.3 \\[1em] \Rightarrow 0.1y = 2.6 - 2.3 \\[1em] \Rightarrow 0.1y = 0.3 \\[1em] \Rightarrow y = \dfrac{0.3}{0.1} = 3.$

Substituting value of y in equation (3), we get :

⇒ x = $\dfrac{1.3 - 0.3 \times 3}{0.2}$

⇒ x = $\dfrac{1.3 - 0.9}{0.2} = \dfrac{0.4}{0.2}$ = 2.

Hence, x = 2 and y = 3.

Solve the following pair of linear equations by the substitution method.

$\sqrt{2}x + \sqrt{3}y = 0 \text{ and } \sqrt{3}x - \sqrt{8}y = 0$

Answer

Given,

$\sqrt{2}x + \sqrt{3}y = 0$ .........(1)

$\sqrt{3}x - \sqrt{8}y = 0$ ..........(2)

Solving equation (1),

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

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

$\Rightarrow \sqrt{3} \times -\dfrac{\sqrt{3}y}{\sqrt{2}} - \sqrt{8}y = 0 \\[1em] \Rightarrow -\dfrac{3y}{\sqrt{2}} - \sqrt{8}y = 0 \\[1em] \Rightarrow \dfrac{-3y - \sqrt{2}\sqrt{8}y}{\sqrt{2}} = 0 \\[1em] \Rightarrow \dfrac{-3y - \sqrt{16}y}{\sqrt{2}} = 0 \\[1em] \Rightarrow -3y - \sqrt{16}y = 0 \\[1em] \Rightarrow -3y - 4y = 0 \\[1em] \Rightarrow -7y = 0 \\[1em] \Rightarrow y = 0.$

Substituting value of y in equation (3), we get :

$\Rightarrow x = -\dfrac{\sqrt{3}}{\sqrt{2}} \times 0 \\[1em] \Rightarrow x = 0.$

Hence, x = 0 and y = 0.

Solve the following pair of linear equations by the substitution method.

$\dfrac{3x}{2} - \dfrac{5y}{3} = -2 \text{ and } \dfrac{x}{3} + \dfrac{y}{2} = \dfrac{13}{6}$

Answer

Given,

1st equation :

$\Rightarrow \dfrac{3x}{2} - \dfrac{5y}{3} = -2 \\[1em] \Rightarrow \dfrac{9x - 10y}{6} = -2 \\[1em] \Rightarrow 9x - 10y = -12 \\[1em] \Rightarrow 9x = 10y - 12 \\[1em] \Rightarrow x = \dfrac{10y - 12}{9}\text{ .........(1)}$

2nd equation :

$\Rightarrow \dfrac{x}{3} + \dfrac{y}{2} = \dfrac{13}{6}$ .........(2)

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

$\Rightarrow \dfrac{\dfrac{10y - 12}{9}}{3} + \dfrac{y}{2} = \dfrac{13}{6} \\[1em] \Rightarrow \dfrac{10y - 12}{27} + \dfrac{y}{2} = \dfrac{13}{6} \\[1em] \Rightarrow \dfrac{2(10y - 12) + 27y}{54} = \dfrac{13}{6} \\[1em] \Rightarrow \dfrac{20y - 24 + 27y}{54} = \dfrac{13}{6} \\[1em] \Rightarrow 47y - 24 = \dfrac{13}{6} \times 54 \\[1em] \Rightarrow 47y - 24 = 13 \times 9 \\[1em] \Rightarrow 47y - 24 = 117 \\[1em] \Rightarrow 47y = 117 + 24 \\[1em] \Rightarrow 47y = 141 \\[1em] \Rightarrow y = \dfrac{141}{47} = 3.$

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

$\Rightarrow x = \dfrac{10y - 12}{9} \\[1em] \Rightarrow x = \dfrac{10 \times 3 - 12}{9} \\[1em] \Rightarrow x = \dfrac{30 - 12}{9} \\[1em] \Rightarrow x = \dfrac{18}{9} \\[1em] \Rightarrow x = 2.$

Hence, x = 2 and y = 3.

Solve 2x + 3y = 11 and 2x - 4y = -24 and hence find the value of ‘m’ for which y = mx + 3.

Answer

Given,

2x + 3y = 11 ..........(1)

2x - 4y = -24 .........(2)

Solving equation (1), we get :

⇒ 2x + 3y = 11

⇒ 2x = 11 - 3y

⇒ x = $\dfrac{11 - 3y}{2}$ ...........(3)

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

$\Rightarrow 2 \times \Big(\dfrac{11 - 3y}{2}\Big) - 4y = -24 \\[1em] \Rightarrow 11 - 3y - 4y = -24 \\[1em] \Rightarrow 11 - 7y = -24 \\[1em] \Rightarrow 7y = 11 + 24 \\[1em] \Rightarrow 7y = 35 \\[1em] \Rightarrow y = \dfrac{35}{7} = 5.$

Substituting value of y in equation (3), we get :

$\Rightarrow x = \dfrac{11 - 3 \times 5}{2} \\[1em] \Rightarrow x = \dfrac{11 - 15}{2} \\[1em] \Rightarrow x = \dfrac{-4}{2} = -2.$

Substituting value of x and y in y = mx + 3, we get :

⇒ 5 = -2m + 3

⇒ -2m = 5 - 3

⇒ -2m = 2

⇒ m = $\dfrac{2}{-2}$ = -1.

Hence, x = -2, y = 5 and m = -1.

Form the pair of linear equations for the following problem and find their solution by substitution method.

The difference between two numbers is 26 and one number is three times the other. Find them.

Answer

Let first number be x, second number be y and x > y.

Given,

Difference between two numbers is 26.

∴ x - y = 26 ........(1)

Also,

One number is three times the other.

∴ x = 3y ............(2)

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

⇒ 3y - y = 26

⇒ 2y = 26

⇒ y = $\dfrac{26}{2}$ = 13.

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

⇒ x = 3 × 13 = 39.

Hence, pair of linear equations are x - y = 26 and x = 3y, where x and y are two numbers such that x > y; x = 39 and y = 13.

Form the pair of linear equations for the following problem and find their solution by substitution method.

The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Answer

Let x be the larger angle and y be the smaller angle.

Given,

Angles are supplementary.

∴ x + y = 180° ...........(1)

Given,

Larger of two supplementary angles exceeds the smaller by 18 degrees.

∴ x - y = 18° .............(2)

Solving equation (2),

⇒ x - y = 18°

⇒ x = 18° + y ...........(3)

Substituting value of x from equation (3) in equation (1), we get :

⇒ 18° + y + y = 180°

⇒ 2y = 180° - 18°

⇒ 2y = 162°

⇒ y = $\dfrac{162°}{2}$ = 81°.

Substituting above value of y in equation (3), we get :

⇒ x = 18° + 81° = 99°.

Hence, pair of linear equations are x + y = 180° and x - y = 18°, where x and y are two angles such that x > y; x = 99° and y = 81°.

Form the pair of linear equations for the following problem and find their solution by substitution method.

The coach of a cricket team buys 7 bats and 6 balls for ₹3800. Later, she buys 3 bats and 5 balls for ₹1750. Find the cost of each bat and each ball.

Answer

Let cost of each bat be ₹ x and each ball be ₹ y. Given, cost of 7 bats and 6 balls ₹3800.

⇒ 7x + 6y = 3800 ............(1)

Given,

Cost of 3 bats and 5 balls for ₹1750.

⇒ 3x + 5y = 1750 ............(2)

Solving equation (1), we get :

⇒ 7x + 6y = 3800

⇒ 6y = 3800 - 7x

⇒ y = $\dfrac{3800 - 7x}{6}$ ............(3)

Substituting value of y from equation (3) in equation (2), we get :

$\Rightarrow 3x + 5 \times \dfrac{3800 - 7x}{6} = 1750 \\[1em] \Rightarrow 3x + \dfrac{19000 - 35x}{6} = 1750 \\[1em] \Rightarrow \dfrac{18x + 19000 - 35x}{6} = 1750 \\[1em] \Rightarrow \dfrac{19000 - 17x}{6} = 1750 \\[1em] \Rightarrow 19000 - 17x = 10500 \\[1em] \Rightarrow 17x = 19000 - 10500 \\[1em] \Rightarrow 17x = 8500 \\[1em] \Rightarrow x = \dfrac{8500}{17} = ₹500.$

Substituting value of x in equation (3), we get :

$\Rightarrow y = \dfrac{3800 - 7x}{6} \\[1em] \Rightarrow y = \dfrac{3800 - 7 \times 500}{6} \\[1em] \Rightarrow y = \dfrac{3800 - 3500}{6} \\[1em] \Rightarrow y = \dfrac{300}{6} \\[1em] \Rightarrow y = ₹50.$

Hence, pair of linear equations are 7x + 6y = 3800, 3x + 5y = 1750, where x and y are the costs (in ₹) of one bat and one ball respectively; x = 500, y = 50.

Form the pair of linear equations for the following problem and find their solution by substitution method.

The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹105 and for a journey of 15 km, the charge paid is ₹155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

Answer

Let ₹x be the fixed charge and ₹y be the charge for every distance covered.

Given,

For a distance of 10 km, the charge paid is ₹105.

x + 10y = 105 ......(1)

For a distance of 15 km, the charge paid is ₹ 155.

x + 15y = 155 .......(2)

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

⇒ x + 15y - (x + 10y) = 155 - 105

⇒ x - x + 15y - 10y = 50

⇒ 5y = 50

⇒ y = $\dfrac{50}{5}$ = 10.

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

⇒ x + 10 × 10 = 105

⇒ x + 100 = 105

⇒ x = 5.

For 25 km charge, will be :

⇒ x + 25y = 5 + 25 × 10 = 5 + 250 = ₹255.

Hence, pair of linear equations are x + 10y = 105 and x + 15y = 155, where x is the fixed charge (in ₹) and y is the charge (in ₹ per km); x = 5, y = 10; and charge for 25 km = ₹255.

Form the pair of linear equations for the following problems and find their solution by substitution method.

A fraction becomes $\dfrac{9}{11}$, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes $\dfrac{5}{6}$. Find the fraction.

Answer

Let numerator be x and denominator be y.

Fraction = $\dfrac{x}{y}$

Given,

If 2 is added to both numerator and denominator, fraction becomes $\dfrac{9}{11}$.

$\Rightarrow \dfrac{x + 2}{y + 2} = \dfrac{9}{11} \\[1em] \Rightarrow 11(x + 2) = 9(y + 2) \\[1em] \Rightarrow 11x + 22 = 9y + 18 \\[1em] \Rightarrow 11x - 9y + 22 - 18 = 0 \\[1em] \Rightarrow 11x - 9y + 4 = 0 \text{ ..........(1)}$

Given,

If 3 is added to both numerator and denominator, fraction becomes $\dfrac{5}{6}$.

$\Rightarrow \dfrac{x + 3}{y + 3} = \dfrac{5}{6} \\[1em] \Rightarrow 6(x + 3) = 5(y + 3) \\[1em] \Rightarrow 6x + 18 = 5y + 15 \\[1em] \Rightarrow 6x - 5y + 18 - 15 = 0 \\[1em] \Rightarrow 6x - 5y + 3 = 0 \text{ ..........(2)}$

Multiplying equation (1) by 5, we get :

⇒ 55x - 45y + 20 = 0 .............(3)

Multiplying equation (2) by 9, we get :

⇒ 54x - 45y + 27 = 0 ..........(4)

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

⇒ 55x - 45y + 20 - (54x - 45y + 27) = 0

⇒ 55x - 54x - 45y + 45y + 20 - 27 = 0

⇒ x - 7 = 0

⇒ x = 7.

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

⇒ 6 × 7 - 5y + 3 = 0

⇒ 42 - 5y + 3 = 0

⇒ 45 - 5y = 0

⇒ 5y = 45

⇒ y = 9.

Fraction = $\dfrac{x}{y} = \dfrac{7}{9}$.

Hence, pair of linear equations are 11x - 9y + 4 = 0, 6x - 5y + 3 = 0, where x and y are numerator and denominator and fraction = $\dfrac{7}{9}$.

Form the pair of linear equations for the following problems and find their solution by substitution method.

Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Answer

Let present age of Jacob be x years and his son's age be y years.

Given,

Five years later, the age of Jacob will be three times that of his son.

⇒ (x + 5) = 3(y + 5)

⇒ x + 5 = 3y + 15

⇒ x - 3y + 5 - 15 = 0

⇒ x - 3y - 10 = 0 ........(1)

Given,

Five years ago, Jacob's age was seven times that of his son.

⇒ (x - 5) = 7(y - 5)

⇒ x - 5 = 7y - 35

⇒ x - 7y - 5 + 35 = 0

⇒ x - 7y + 30 = 0 ..........(2)

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

⇒ x - 3y - 10 - (x - 7y + 30) = 0

⇒ x - x - 3y + 7y - 10 - 30 = 0

⇒ 4y - 40 = 0

⇒ 4y = 40

⇒ y = $\dfrac{40}{4}$ = 10.

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

⇒ x - 3 × 10 - 10 = 0

⇒ x - 30 - 10 = 0

⇒ x - 40 = 0

⇒ x = 40.

Hence, pair of linear equations are x - 3y - 10 = 0 and x - 7y + 30 = 0, where x and y are the ages in years of Jacob and his son; x = 40, y = 10.

Solve the following pair of linear equations by the elimination method and the substitution method :

x + y = 5 and 2x - 3y = 4

Answer

Given,

x + y = 5 .........(1)

2x - 3y = 4 ........(2)

Multiplying equation (1) by 3, we get :

3x + 3y = 15 ........(3)

Adding equation (2) and (3), we get :

⇒ 2x - 3y + 3x + 3y = 4 + 15

⇒ 5x = 19

⇒ x = $\dfrac{19}{5}$

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

$\Rightarrow \dfrac{19}{5} + y = 5 \\[1em] \Rightarrow y = 5 - \dfrac{19}{5} \\[1em] \Rightarrow y = \dfrac{25 - 19}{5} \\[1em] \Rightarrow y = \dfrac{6}{5}.$

Hence, x = $\dfrac{19}{5} \text{ and y} = \dfrac{6}{5}$.

Solve the following pair of linear equations by the elimination method and the substitution method :

3x + 4y = 10 and 2x - 2y = 2

Answer

Given,

3x + 4y = 10 ........(1)

2x - 2y = 2 .........(2)

Multiplying equation (2) by 2, we get :

4x - 4y = 4 ...........(3)

Adding equation (1) and equation (3), we get :

⇒ 3x + 4y + 4x - 4y = 10 + 4

⇒ 7x = 14

⇒ x = $\dfrac{14}{7}$

⇒ x = 2.

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

⇒ 3(2) + 4y = 10

⇒ 6 + 4y = 10

⇒ 4y = 10 - 6

⇒ 4y = 4

⇒ y = $\dfrac{4}{4}$

⇒ y = 1.

Hence, x = 2 and y = 1.

Solve the following pair of linear equations by the elimination method and the substitution method :

3x - 5y - 4 = 0 and 9x = 2y + 7

Answer

Given,

⇒ 3x - 5y - 4 = 0

⇒ 3x - 5y = 4 ..........(1)

⇒ 9x = 2y + 7

⇒ 9x - 2y = 7 ..........(2)

Multiplying equation (1) by 3, we get :

9x - 15y = 12 ..........(3)

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

⇒ 9x - 2y - (9x - 15y) = 7 - 12

⇒ 9x - 9x - 2y + 15y = -5

⇒ 13y = -5

⇒ y = $-\dfrac{5}{13}$.

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

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

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

Solve the following pair of linear equations by the elimination method and the substitution method :

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

Answer

Given,

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

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

Multiplying equation (2) by 2, we get :

$2x - \dfrac{2y}{3} = 6$ ............(3)

Adding equation (1) and (3), we get :

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

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

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

Hence, x = 2 and y = -3.

Form the pair of linear equations in the following problem, and find their solutions (if they exist) by the elimination method :

If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes $\dfrac{1}{2}$ if we only add 1 to the denominator. What is the fraction ?

Answer

Let x be the numerator and y be the denominator.

Original fraction = $\dfrac{x}{y}$

Given,

If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1.

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

Given,

Fraction becomes $\dfrac{1}{2}$ if we only add 1 to the denominator.

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

Multiplying equation (1) by 2, we get :

2x - 2y + 4 = 0 ........(3)

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

⇒ 2x - y - 1 - (2x - 2y + 4) = 0

⇒ 2x - 2x - y + 2y - 1 - 4 = 0

⇒ y - 5 = 0

⇒ y = 5.

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

⇒ x - 5 + 2 = 0

⇒ x - 3 = 0

⇒ x = 3.

Fraction = $\dfrac{x}{y} = \dfrac{3}{5}$.

Hence, pair of linear equations are x - y + 2 = 0, 2x - y - 1 = 0, where x and y are the numerator and denominator of the fraction and fraction = $\dfrac{3}{5}$.

Form the pair of linear equations in the following problem, and find their solutions (if they exist) by the elimination method :

Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

Answer

Let present age of Nuri be x years and age of Sonu be y years.

Given,

Five years ago, Nuri was thrice as old as Sonu.

⇒ (x - 5) = 3(y - 5)

⇒ x - 5 = 3y - 15

⇒ x - 3y - 5 + 15 = 0

⇒ x - 3y + 10 = 0 ........(1)

Given,

Ten years later, Nuri will be twice as old as Sonu.

⇒ (x + 10) = 2(y + 10)

⇒ x + 10 = 2y + 20

⇒ x - 2y + 10 - 20 = 0

⇒ x - 2y - 10 = 0 ........(2)

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

⇒ x - 2y - 10 - (x - 3y + 10) = 0

⇒ x - 2y - 10 - x + 3y - 10 = 0

⇒ y - 20 = 0

⇒ y = 20.

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

⇒ x - 3(20) + 10 = 0

⇒ x - 60 + 10 = 0

⇒ x - 50 = 0

⇒ x = 50.

Hence, pair of linear equations are x - 3y + 10 = 0, x - 2y - 10 = 0, where x and y are the ages (in years) of Nuri and Sonu respectively; Age of Nuri (x) = 50, Age of Sonu (y) = 20.

Form the pair of linear equations in the following problem, and find their solutions (if they exist) by the elimination method :

The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Answer

Let x be the digit at ten's place and y be the digit at units place.

Number = 10x + y

Given,

Sum of two digits is 9.

∴ x + y = 9 .......(1)

Given,

Nine times this number is twice the number obtained by reversing the order of the digits.

∴ 9(10x + y) = 2(10y + x)

⇒ 90x + 9y = 20y + 2x

⇒ 90x - 2x + 9y - 20y = 0

⇒ 88x - 11y = 0

⇒ 11(8x - y) = 0

⇒ 8x - y = 0 ........(2)

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

⇒ x + y + (8x - y) = 9

⇒ x + 8x + y - y = 9

⇒ 9x = 9

⇒ x = $\dfrac{9}{9}$

⇒ x = 1.

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

⇒ 8(1) - y = 0

⇒ 8 - y = 0

⇒ y = 8.

Number = (10x + y) = 10 × 1 + 8 = 10 + 8 = 18.

Hence, pair of linear equations are x + y = 9, 8x - y = 0, where x and y are respectively the tens and units digits of the number and number = 18.

Form the pair of linear equations in the following problem, and find their solutions (if they exist) by the elimination method :

Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.

Answer

Let no. of ₹ 50 notes be x and no. of ₹ 100 notes be y.

Given,

Meena got 25 notes in all.

x + y = 25 ........(1)

Given,

Total money = ₹ 2000

⇒ 50x + 100y = 2000

⇒ 50(x + 2y) = 2000

⇒ x + 2y = $\dfrac{2000}{50}$

⇒ x + 2y = 40 .........(2)

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

⇒ x + 2y - (x + y) = 40 - 25

⇒ x - x + 2y - y = 15

⇒ y = 15.

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

⇒ x + 15 = 25

⇒ x = 25 - 15 = 10.

Hence, pair of linear equations are x + 2y = 40, x + y = 25, where x and y are respectively the number of ₹50 and ₹100 notes; x = 10, y = 15.

Form the pair of linear equations in the following problem, and find their solutions (if they exist) by the elimination method :

A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Answer

Let fixed charge be ₹ x and ₹ y be the additional charge per day.

Given,

Saritha paid ₹27 for a book kept for seven days.

For the first 3 days, she was charged the fix charge of ₹ x. For the remaining 4 days she was charged ₹ y per day.

∴ x + 4y = 27 ........(1)

Susy paid ₹ 21 for a book for five days.

For the first 3 days, Susy was charged the fix charge of ₹ x. For the remaining 2 days she was charged ₹ y per day.

∴ x + 2y = 21 .........(2)

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

⇒ x + 4y - (x + 2y) = 27 - 21

⇒ x - x + 4y - 2y = 6

⇒ 2y = 6

⇒ y = $\dfrac{6}{2}$ = 3.

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

⇒ x + 2(3) = 21

⇒ x + 6 = 21

⇒ x = 21 - 6

⇒ x = 15.

Hence, pair of linear equations are x + 4y = 27, x + 2y = 21, where x is the fixed charge (in ₹) and y is the additional charge (in ₹) per day; x = 15, y = 3.