Pair of Linear Equations in Two Variables : Exercise 3.3

Q.1      Solve the following pair of linear equations by the substitution method :
(i) x + y = 14                                                         x – y = 4
(ii) s – t = 3                                                           ${s \over 3} + {t \over 2} = 6$
(iii) 3x + y = 3                                                       9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3                                            0.4x + 0.5y = 2.3
(v) $\sqrt 2 x + \sqrt 3 y = 0$                                             $\sqrt 3 x - \sqrt 8 y = 0$
(vi) ${{3x} \over 2} - {{5y} \over 3} = - 2$                                                ${x \over 3} + {y \over 2} = {{13} \over 6}$
Sol.       (i) The given system of equations is
x + y = 14 ...(1)
and, x – y = 4 ...(2)
From (1), y = 14 – x
Substituting y = 14 – x in (2), we get
x – (14 – x) = 14
$\Rightarrow$ x – 14 + x = 4

$\Rightarrow$ 2x = 4 + 14
$\Rightarrow$ 2x = 18

$\Rightarrow$ x = 9
Putting x = 9 in (1), we get
9 + y = 14
$\Rightarrow$ y = 14 – 9
$\Rightarrow$ y = 5

Hence, the solution of the given system of equations is
x = 9, y = 5

(ii) The given system of equations is
s – t = 3 ...(1)
${s \over 3} + {t \over 2} = 6$ ...(2)
From (1), s = 3 + t
Substituting s = 3 + t in (2), we get
${{3 + t} \over 3} + {t \over 2} = 6$
$\Rightarrow 2(3 + t) + 3t = 36$

$\Rightarrow$ 6 + 2t + 3t = 36
$\Rightarrow$ 5t = 30
$\Rightarrow$ t = 6

Putting t = 6 in (1), we get
s – 6 = 3
$\Rightarrow$ s = 3 + 6 = 9

Hence, the solution of the given system of equations is
s = 9, t = 6

(iii) The given system of equations is
3x – y = 3 ...(1)
and, 9x – 3y = 9 ...(2)
From (1), y = 3x – 3
Substituting y = 3x – 3 in (2), we get
9x – 3(3x – 3) = 9
$\Rightarrow$ 9x – 9x + 9 = 9

$\Rightarrow$ 9 = 9
This statement is true for all values of x. However, we do not get a specific value of x as a solution. So, we cannot obtain a specific value of y . This situation has arisen because both the given equations are the same.
Therefore, Equations (1) and (2) have infinitely many solutions.

(iv) The given system of equations is
0.2x + 0.3y = 1.3
$\Rightarrow$ 2x + 3y = 13...(1)

and, 0.4x + 0.5y = 2.3
$\Rightarrow$ 4x + 5y = 23 ...(2)

From (2), 5y = 23 – 4x
$\Rightarrow$ $y = {{23 - 4x} \over 5}$

Substituting $y = {{23 - 4x} \over 5}$ in (1), we get
$2x + 3\left( {{{23 - 4x} \over 5}} \right) = 13$
$\Rightarrow$ 10x + 69 – 12x = 65
$\Rightarrow$ – 2x = – 4
$\Rightarrow$ x = 2

Putting x = 2 in (1), we get
2 × 2 + 3y = 13
$\Rightarrow$ 3y = 13 – 4 = 9

$\Rightarrow$ $y = {9 \over 3} = 3$
Hence, the solution of the given system of equations is
x = 2, y = 3

(v) The given system of equations is
$\sqrt {2x} + \sqrt {3y} = 0$ ...(1)
and, $\sqrt {3x} - \sqrt {8y} = 0$ ...(2)
From (2), $\sqrt {8y} = \sqrt {3x}$
$\Rightarrow$ $y = {{\sqrt {3x} } \over {\sqrt 8 }}$

Substituting $y = {{\sqrt {3x} } \over {\sqrt 8 }}$ in (1), we get
$\sqrt {2x} + \sqrt 3 \left( {{{\sqrt {3x} } \over {\sqrt 8 }}} \right) = 0$
$\Rightarrow$ $\sqrt {2x} + {{3x} \over {\sqrt 8 }} = 0$

$\Rightarrow$ $\sqrt {16x} + 3x = 0$
$\Rightarrow$ 4x + 3x = 0
$\Rightarrow$ 7x = 0
$\Rightarrow$ x = 0

Putting x = 0 in (1), we get
0 + $\sqrt 3 y$ = 0
$\Rightarrow$ y = 0

Hence, the solution of the given system of equations is
x = 0, y = 0

(vi) The given system of equations is
${{3x} \over 2} - {{5y} \over 3} = - 2$
$\Rightarrow$ 9x – 10y = – 12 ...(1)

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

From (1), 10y = 9x + 12
$\Rightarrow$ $y = {{9x + 12} \over {10}}$

Substituting $y = {{9x + 12} \over {10}}$ in (2), we get
$2x + 3\left( {{{9x + 12} \over {10}}} \right) = 13$
$\Rightarrow$ 20x + 27x + 36 = 130
$\Rightarrow$ 47x = 130 – 36
$\Rightarrow$ 47x = 94

$\Rightarrow$ $x = {{94} \over {47}} = 2$
Putting x = 2 in (1), we get
9 × 2 – 10y = – 12
$\Rightarrow$ –10y = –12 – 18
$\Rightarrow$ – 10y = – 30
$\Rightarrow$ $y = {{ - 30} \over { - 10}} = 3$
Hence, the solution of the given system of equations is
x = 2, y = 3

Q.2          Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of 'm' for which y = mx + 3.
Sol.          The given system of equations is
2x + 3y = 11
and, 2x – 4y = – 24
From(1), 3y = 11 – 2x
$\Rightarrow y = {{11 - 2x} \over 3}$
Substituting $y = {{11 - 2x} \over 3}$ in (2), we get
$2x - 4\left( {{{11 - 2x} \over 3}} \right) = - 24$
$\Rightarrow$ 6x – 44 + 8x = – 72
$\Rightarrow$ 14x = –72 + 44
$\Rightarrow$ 14x = –28
$\Rightarrow$ $x = {{ - 28} \over {14}} = - 2$
Putting x = –2 in (1), we get
2(– 2) + 3y = 11
$\Rightarrow$ – 4 + 3y = 11
$\Rightarrow$ 3y = 11 + 4
$\Rightarrow$ 3y = 15
$\Rightarrow$ $y = {{15} \over 3} = 5$
Putting x = – 2, y = 5 in y = mx + 3, we get
5 = m(–2) + 3
$\Rightarrow$ –2m = 5 – 3
$\Rightarrow$ – 2m = 2
$\Rightarrow$ m = – 1

Q.3      Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.
(iv) 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 Rs. 105 and for a journey of 15 km, the charge paid Rs. 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 ?
(v) A fraction becomes ${9 \over {11}}$, if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it become ${5 \over 6}$. Find the fraction.
(vi) Five years hence, the age of Jacob will three times that of his son. Five years ago, Jacob age was seven times that of his son. What are the present ages?
Sol.      (i) Let the numbers be x and y (x > y). Then,
x – y = 26
and, x = 3y
From (2), $y = {x \over 3}$ in (1), we get
$x - {x \over 3} = 26$
$\Rightarrow 3x - x = 78$
$\Rightarrow$ 2x = 78
$\Rightarrow$ x = 39
Putting x = 39 in (2), we get
$\Rightarrow$ 39 = 3y
$\Rightarrow$ $y = {{39} \over 3} = 13$
Hence, the required numbers are 39 and 13.

(ii) Let the angles be x° and y°(x > y). Then,
x + y = 180
and, x = y + 18
Substituting x = y + 18 in (1), we get
y + 18 + y = 180
$\Rightarrow$ 2y = 180 – 18

$\Rightarrow$ 2y = 162
$\Rightarrow$ $y = {{162} \over 2} = 81$

Putting y = 81 in (2), we get
x = 81 + 18 = 99
Thus, the angles are 99° and 81°.

(iii) Let the cost of one bat and one ball be Rs. x and Rs. y respectively. Then,
7x + 6y = 3800 .....(1)
and, 3x + 5y = 1750 ...(2)
From (2), 5y = 1750 – 3x
$\Rightarrow$ $y = {{1750 - 3x} \over 5}$

Substituting $y = {{1750 - 3x} \over 5}$ in (1), we get
$7x + 6\left( {{{1750 - 3x} \over 5}} \right) = 3800$
$\Rightarrow$ 35x + 10500 – 18x = 19000
$\Rightarrow$ 17x = 19000 – 10500
$\Rightarrow$ 17x = 8500
$\Rightarrow$ $x = {{8500} \over {17}} = 500$
Putting x = 500 in (2), we get
3(500) + 5y = 1750
$\Rightarrow$ 5y = 1750 – 1500
$\Rightarrow$ 5y = 250
$\Rightarrow$ $y = {{250} \over 5} = 50$
Hence, the cost of one bat is Rs. 500 and the cost of one ball is Rs 50.

(iv) Let the fixed charges of taxi be Rs. x per km and the running charges be Rs y km/hr.
According to the given condition, we have
x + 10y = 105 ...(1)
x + 15y = 155 ...(2)
From (1), x = 105 – 10y
Substituting x = 105 – 10y in (2), we get
105 – 10y + 15y = 155
$\Rightarrow$ 105 + 5y = 155

$\Rightarrow$ 5y = 155 – 105
$\Rightarrow$ 5y = 50

$\Rightarrow$ y = 10
Putting y = 10 in (1), we get
x + 10 × 10 = 105
$\Rightarrow$ x = 105 – 100

$\Rightarrow$ x = 5
Total charges for travelling a distance of 25 km
= x + 25y = Rs (5 + 25 × 10)
= Rs. 255
Hence, the fixed charge is Rs 5, the charge per km is Rs. 10 and the total charge for travelling a distance of 25 km is Rs 255.

(v) Let the fraction be ${x \over y}$.
Then, according to the given conditions, we have
${{x + 2} \over {y + 2}} = {9 \over {11}}$ and ${{x + 3} \over {y + 3}} = {5 \over 6}$
$\Rightarrow$ 11x + 22 = 9y + 18 and 6x + 18 = 5y + 15
$\Rightarrow$ 11x – 9y = – 4 ...(1)
and, 6x – 5y = – 3....(2)
From (2), 5y = 6x + 3
$\Rightarrow$ $y = {{6x + 3} \over 5}$
Substituting $y = {{6x + 3} \over 5}$ in (1), we get
$11x - 9\left( {{{6x + 3} \over 5}} \right) = - 4$
$\Rightarrow$ 55x – 54x – 27 = –20

$\Rightarrow$ x = – 20 + 27
$\Rightarrow$ x = 7

Putting x = 7 in (1), we get
11(7) – 9y = – 4
$\Rightarrow$ – 9y = – 4 – 77

$\Rightarrow$ –9y = – 81
$\Rightarrow$ $y = {{ - 81} \over { - 9}} = 9$

Hence, the given fraction is ${7 \over 9}$.

(vi) Let the present age of Jacob be x years and the present age of his son be y years.
Five years hence, Jacob's age = (x + 5) years
Son's age = (y + 5) years
Five years ago, Jacob's age = (y – 5) years
Son's age = (y – 5) years
As per question, we get
(x + 5) = 3(y + 5)
$\Rightarrow$ x + 5 = 3y + 15
$\Rightarrow$ x – 3y =10
$\Rightarrow$ x – 3y = 15 – 5
and, (x – 5) = 7(y – 5)
$\Rightarrow$ x – 5 = 7y – 35
$\Rightarrow$ x – 7y = – 30
$\Rightarrow$ x – 7y = – 35 + 5
From (1), x = 3y + 10
Substituting x = 3y +10 in (2), we get
3y + 10 – 7y = – 30

$\Rightarrow$ – 4y = – 30 – 10
$\Rightarrow$ – 4y = – 40
$\Rightarrow$ y = 10
Putting y = 10 in (1),we get
x – 3 × 10 = 10
$\Rightarrow$ x = 10 + 30 = 40
Hence, present age of Jacob is 40 years and that his son is 10 years.