Pair of Linear Equations in Two Variables - Solved Examples

Exercise 3.3

Exercise 3.1

For equation $ x + y = 10$

For equation $ -x + y = 4$

We will plot both the equations on the graph.

From the above graph, it is evident that the 2 lines intersect at point (3, 7).
Hence this is the solution of the given pair of equations.
So the number of boys are 3 and the number of girls are 7


(ii)
Let the the number of pencils be $x$ and the number of pens as $y$.
5 pencils and 7 pens together cost ₹ 50. This can be algebraically represented as
$5x + 7y = 50$

7 pencils and 5 pens together cost ₹ 46. This can be algebraically represented as
$ 7x + 5y = 46$

To obtain the equivalent graphical representation, let us find a few points on the line representing each equation.

For equation $ x + y = 10$

For equation $ -x + y = 4$

We will plot both the equations on the graph.

From the above graph, it is evident that the 2 lines intersect at point (3, 5).
Hence this is the solution of the given pair of equations.
So the cost of pencil is 3 and the cost of 1 pen is 5.

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2

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

(i)
$5x - 4y + 8 = 0$
$7x + 6y - 9 = 0$

(ii)
$9x - 3y + 12 = 0$
$18x + 6y + 24 = 0$

(iii)
$6x - 3y + 10 = 0$
$2x - y + 9 = 0$

Solution

We know that if the lines represented by a pair of linear equations
$a_1x + b_1y + c_1 = 0$ and
$a_2x + b_2y + c_2 = 0$
are
• intersecting , then $\dfrac{a_1}{a_2}$ ≠ $\dfrac{b_1}{b_2}$

• coincident, then $\dfrac{a_1}{a_2} $ = $\dfrac{b_1}{b_2} $ = $\dfrac{c_1}{c_2}$

• parallel, then $\dfrac{a_1}{a_2}$ = $\dfrac{b_1}{b_2}$ ≠ $\dfrac{c_1}{c_2}$

(i)
$5x - 4y + 8 = 0$
$7x + 6y - 9 = 0$

$a_1$ = 5, $b_1$ = -4, $c_1$ = 8,
$a_2$ = 7, $b_2$ = 6, $c_2$ = -9,

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

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

As $\dfrac{a_1}{a_2}$ ≠ $\dfrac{b_1}{b_2}$, these lines intersect at a point.

(ii)
$9x - 3y + 12 = 0$
$18x + 6y + 24 = 0$

$a_9$ = 9, $b_1$ = 3, $c_1$ = 12,
$a_2$ = 18, $b_2$ = 6, $c_2$ = 24,

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

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

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

As $\dfrac{a_1}{a_2}$ = $\dfrac{b_1}{b_2}$ = $\dfrac{c_1}{c_2}$ = $\dfrac{1}{2}$, these lines are coincident.

(iii)
$6x - 3y + 10 = 0$
$2x - y + 9 = 0$

$a_9$ = 6, $b_1$ = -3, $c_1$ = 10,
$a_2$ = 2, $b_2$ = -1, $c_2$ = 9,

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

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

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

As $\dfrac{a_1}{a_2}$ = $\dfrac{b_1}{b_2}$ ≠ $\dfrac{c_1}{c_2}$ = $\dfrac{1}{2}$, these lines are parallel.

3

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

(i) $3x + 2y = 5$ ; $2x - 3y = 7$
(ii) $2x - 3y = 8$ ; $4x - 6y = 9$
(iii) $\dfrac{3}{2}x + \dfrac{5}{3}y = 7$ ; $9x - 10y = 14$
(iv) $5x - 3y = 11$ ; $-10x + 6y = -22$
(v) $\frac{4}{3}x + 2y = 8$ ; $2x + 3y = 12$

Solution

We know that if the lines represented by a pair of linear equations
$a_1x + b_1y + c_1 = 0$ and
$a_2x + b_2y + c_2 = 0$
are
• consistent if they have one solution or infinitely many solutions i.e.
⇒ intersecting lines ⇒ $\dfrac{a_1}{a_2}$ ≠ $\dfrac{b_1}{b_2}$ and
⇒ coincident lines ⇒ $\dfrac{a_1}{a_2} $ = $\dfrac{b_1}{b_2} $ = $\dfrac{c_1}{c_2}$

• inconsistent ⇒ parallel lines ⇒ $\dfrac{a_1}{a_2}$ = $\dfrac{b_1}{b_2}$ ≠ $\dfrac{c_1}{c_2}$

(i) $3x + 2y = 5$ ; $2x - 3y = 7$

$a_1$ = 3, $b_1$ = 2, $c_1$ = -5,
$a_2$ = 2, $b_2$ = -3, $c_2$ = -7,

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

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

As $\dfrac{a_1}{a_2}$ ≠ $\dfrac{b_1}{b_2}$, these lines intersect at a point.
Hence, the pair of equations are consistent.

(ii) $2x - 3y = 8$ ; $4x - 6y = 9$
$a_1$ = 2, $b_1$ = -3, $c_1$ = -8,
$a_2$ = 4, $b_2$ = -6, $c_2$ = -9,

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

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

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

As $\dfrac{a_1}{a_2}$ = $\dfrac{b_1}{b_2}$ ≠ $\dfrac{c_1}{c_2}$, these lines are parallel and have no solution.
Hence, the pair of equations are inconsistent.

(iii) $\dfrac{3}{2}x + \dfrac{5}{3}y = 7$ ; $9x - 10y = 14$
$a_1 = \dfrac{3}{2}$, $b_1 = \dfrac{5}{3}$, $c_1$ = -7,
$a_2$ = 9, $b_2$ = -10, $c_2$ = -14,

$\dfrac{a_1}{a_2}$ = $\dfrac{\dfrac{3}{2}}{9}$ = $\dfrac{3}{18}$ = $\dfrac{1}{6}$

$\dfrac{b_1}{b_2}$ = $\dfrac{\dfrac{5}{3}}{-10}$ = $\dfrac{5}{-30}$ = $\dfrac{-1}{6}$

$\dfrac{c_1}{c_2}$ = $\dfrac{-7}{-14}$ = $\dfrac{1}{2}$

As $\dfrac{a_1}{a_2}$ ≠ $\dfrac{b_1}{b_2}$, these lines are intersecting and have one solution.
Hence, they are consistent.

(iv) $5x - 3y = 11$ ; $-10x + 6y = -22$
$a_1 = 5$, $b_1 = -3$, $c_1 = -11$,
$a_2 = -10$, $b_2 = 6$, $c_2 = 22$

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

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

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

As $\dfrac{a_1}{a_2}$ = $\dfrac{b_1}{b_2}$ = $\dfrac{c_1}{c_2}$, these lines are coincident and have infinitely many solutions.
Hence, they are consistent.

(v) $\frac{4}{3}x + 2y = 8$ ; $2x + 3y = 12$
$a_1 = \dfrac{4}{3}$, $b_1 = 2$, $c_1 = -8$,
$a_2 = 2$, $b_2 = 3$, $c_2 = -12$

$\dfrac{a_1}{a_2}$ = $\dfrac{\dfrac{4}{3}}{2}$ = $\dfrac{2}{3}$

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

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

As $\dfrac{a_1}{a_2}$ = $\dfrac{b_1}{b_2}$ = $\dfrac{c_1}{c_2}$, these lines are coincident and have infinitely many solutions.
Hence, they are consistent.

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4

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

(i) $x + y = 5$ ; $2x + 2y = 10$
(ii) $x - y = 8$ ; $3x - 3y = 16$
(iii) $2x + y - 6 = 0$ ; $4x - 2y - 4 = 0$
(iv) $2x - 2y - 2 = 0$ ; $4x - 4y - 5 = 0$

Solution

We know that if the lines represented by a pair of linear equations
$a_1x + b_1y + c_1 = 0$ and
$a_2x + b_2y + c_2 = 0$
are
• consistent if they have one solution or infinitely many solutions i.e.
⇒ intersecting lines ⇒ $\dfrac{a_1}{a_2}$ ≠ $\dfrac{b_1}{b_2}$ and
⇒ coincident lines ⇒ $\dfrac{a_1}{a_2} $ = $\dfrac{b_1}{b_2} $ = $\dfrac{c_1}{c_2}$

• inconsistent ⇒ parallel lines ⇒ $\dfrac{a_1}{a_2}$ = $\dfrac{b_1}{b_2}$ ≠ $\dfrac{c_1}{c_2}$

(i) $x + y = 5$ ; $2x + 2y = 10$
The two equations are
$x + y - 5 = 0$ and
$2x + 2y - 10 = 0$

$a_1 = 1$, $b_1 = 1$, $c_1 = -5$,
$a_2 = 2$, $b_2 = 2$, $c_2 = -10$

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

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

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

As $\dfrac{a_1}{a_2}$ = $\dfrac{b_1}{b_2}$ = $\dfrac{c_1}{c_2}$, these lines are coincident and have infinitely many solutions.
Hence, they are consistent. To obtain the equivalent graphical representation, let us find 2 points on each of these lines.

For equation $ x + y - 5 = 0$

For equation $ 2x + 2y - 10 = 0$

We will plot both the equations on the graph.



All the points on coincident lines are solutions for the given pair of equations.

(ii) $x - y = 8$ ; $3x - 3y = 16$
The two equations are
$x - y - 8 = 0$ and
$3x - 3y - 16 = 0$

$a_1 = 1$, $b_1 = -1$, $c_1 = -8$,
$a_2 = 3$, $b_2 = -3$, $c_2 = -16$

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

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

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

As $\dfrac{a_1}{a_2}$ = $\dfrac{b_1}{b_2}$ ≠ $\dfrac{c_1}{c_2}$, these lines are parallel and have no solution.
Hence, they are inconsistent.

(iii) $2x + y - 6 = 0$ ; $4x - 2y - 4 = 0$
The two equations are
$2x + y - 6 = 0$ and
$4x - 2y - 4 = 0$

$a_1 = 2$, $b_1 = 1$, $c_1 = -6$,
$a_2 = 4$, $b_2 = -2$, $c_2 = -4$

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

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

$\dfrac{c_1}{c_2}$ = $\dfrac{-6}{-4}$ = $\dfrac{3}{2}$

As $\dfrac{a_1}{a_2}$ ≠ $\dfrac{b_1}{b_2}$, these lines are intersecting and have one solution.
Hence, they are consistent.

For equation $2x + y - 6 = 0$

For equation $4x - 2y - 4 = 0$

We will plot both the equations on the graph.



x = 2 and y = 2 is the solution for the given pair of equations.

(iv) $2x - 2y - 2 = 0$ ; $4x - 4y - 5 = 0$
The two equations are
$2x - 2y - 2 = 0$ and
$4x - 4y - 5 = 0$

$a_1 = 2$, $b_1 = -2$, $c_1 = -2$,
$a_2 = 4$, $b_2 = -4$, $c_2 = -5$

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

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

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

As $\dfrac{a_1}{a_2}$ = $\dfrac{b_1}{b_2}$ ≠ $\dfrac{c_1}{c_2}$, these lines are parallel and have no solution.
Hence, they are inconsistent.

5

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

Solution

Let the length of a rectangular garden be $x$ and it's width be $y$.
∴ Perimeter of this rectangular garden = $2(x + y)$
Half of the perimeter = $(x + y) = 36$

It is also given that $x = y + 4$ $⇒ x – y = 4$

Adding the two equations, we get 2x = 40 ⇒ x = 20
Subtracting the two equations, we get 2y = 32 ⇒ y = 16

Hence, length = 20 m and width = 16 m

6

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

Solution

Let the second line be equal to $a_2x + b_2y + c_2 = 0$

Comparing given line $2x + 3y – 8 = 0$ with $a_1x + b_1y + c_1 = 0$
We get $a_1 = 2$, $b_1 = 3$ and $c_1 = -8$

(i) Two lines intersect with each other if $\dfrac{a_1}{a_2}$ ≠ $\dfrac{b_1}{b_2}$

So, second equation can be x + 2y = 3 because $\dfrac{2}{3}$ ≠ $\dfrac{1}{2}$

(ii) Lines are parallel when $\dfrac{a_1}{a_2}$ = $\dfrac{b_1}{b_2}$ ≠ $\dfrac{c_1}{c_2}$

So the second equation can be 2x + 3y – 6 = 0 as in that case, we satisfy the above criteria.

$\dfrac{2}{2}$ = $\dfrac{3}{3}$ ≠ $\dfrac{-8}{-6}$

(iii) Two lines are coincident if $\dfrac{a_1}{a_2} $ = $\dfrac{b_1}{b_2} $ = $\dfrac{c_1}{c_2}$

So, second equation can be 4x + 6y – 16 = 0 because $\dfrac{2}{4} $ = $\dfrac{3}{6} $ = $\dfrac{-8}{-16}$

7

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.

Solution

Let us find 2 points each which lie on these lines.

For equation $x – y + 1 = 0$


For equation $3x + 2y - 12 = 0$

We will plot both the equations on the graph.



We can see from the graphs that vertices of the triangle formed by these lines and the x-axis are (–1, 0), (2, 3) and (4, 0).