EXERCISE 3.2

1. Solve the following pairs of linear equations by the substitution method

(i) Equations:

(1) $x+y=14$
(2) $x-y=4$

From equation (2), we get:

$x=4+y$ --- (3)

Substitute equation (3) into equation (1):

$(4+y)+y=14$

$4+2y=14\Rightarrow 2y=10\Rightarrow y=5$

Substitute y = 5 into equation (3):

$x=4+5=9$

Final Answer: x = 9, y = 5

(ii) Equations:

(1) $s-t=3$
(2) $\frac{s}{3}+\frac{t}{2}=6$

Step 1: From equation (1), express s in terms of t:

$s=3+t$ --- (3)

Step 2: Substitute equation (3) into equation (2):

$\frac{3+t}{3}+\frac{t}{2}=6$

Multiply the entire equation by 6 (LCM) to clear fractions:

$2(3+t)+3t=36$

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

Step 3: Substitute t = 6 into equation (3):

$s=3+6=9$

Final Answer: s = 9, t = 6

(iii) Equations:

(1) $3x-y=3$
(2) $9x-3y=9$

Step 1: From equation (1), express y in terms of x:

$y=3x-3$ --- (3)

Step 2: Substitute equation (3) into equation (2):

$9x-3(3x-3)=9$

$9x-9x+9=9\Rightarrow 9=9$

Hence Since this statement is always true, the system has infinitely many solutions.

Final Answer: Infinitely many solutions, where y = 3x - 3

(iv) Equations:

(1) $0.2x+0.3y=1.3$
(2) $0.4x+0.5y=2.3$

Multiply both equations by 10 to clear decimals:

(1a) $2x+3y=13$
(2a) $4x+5y=23$

Step 1: From equation (1a), express x in terms of y:

$2x=13-3y\Rightarrow x=\frac{13-3y}{2}$ --- (3)

Step 2: Substitute equation (3) into equation (2a):

$4\left(\frac{13-3y}{2}\right)+5y=23$

$26-6y+5y=23\Rightarrow 26-y=23\Rightarrow y=3$

Step 3: Substitute y = 3 into equation (3):

$x=\frac{13-3\left(3\right)}{2}=\frac{13-9}{2}=2$

Final Answer: x = 2, y = 3

(v) Equations:

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

Step 1: From equation (1), express x in terms of y:

$\sqrt{2}x=-\sqrt{3}y\Rightarrow x=-\frac{\sqrt{3}}{\sqrt{2}}y$ --- (3)

Step 2: Substitute equation (3) into equation (2):

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

$-\frac{3}{\sqrt{2}}y-\sqrt{8}y=0$

Factor out y:

$y(-\frac{3}{\sqrt{2}}-\sqrt{8})=0\Rightarrow y=0$

Step 3: Substitute y = 0 into equation (3):

$x=-\frac{\sqrt{3}}{\sqrt{2}}\left(0\right)=0$

Final Answer: x = 0, y = 0

(vi) Equations:

(1) $\frac{3x}{2}-\frac{5y}{3}=-2$
(2) $\frac{x}{3}+\frac{y}{2}=\frac{13}{6}$

Multiply equation (1) by 6 and equation (2) by 6 to clear fractions:

(1a) $9x-10y=-12$
(2a) $2x+3y=13$

Step 1: From equation (2a), express x in terms of y:

$2x=13-3y\Rightarrow x=\frac{13-3y}{2}$ --- (3)

Step 2: Substitute equation (3) into equation (1a):

$9\left(\frac{13-3y}{2}\right)-10y=-12$

Multiply the entire equation by 2 to clear the fraction:

$9(13-3y)-20y=-24$

$117-27y-20y=-24$

$117-47y=-24\Rightarrow -47y=-141\Rightarrow y=3$

Step 3: Substitute y = 3 into equation (3):

$x=\frac{13-3\left(3\right)}{2}=\frac{13-9}{2}=2$

Final Answer: x = 2, y = 3

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

2x + 3y = 11

2x − 4y = −24

Subtracting the second equation from the first:

(2x + 3y) − (2x − 4y) = 11 − (−24)

7y = 35

y = 5

Substituting y = 5 in 2x + 3y = 11:

2x + 15 = 11

2x = −4

x = −2

Given:

y = mx + 3

Substituting x = −2 and y = 5:

5 = m(−2) + 3

2 = −2m

m = −1

Answer: x = −2, y = 5 and m = −1.

Question 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.

Let the numbers be x and y.

x − y = 26

x = 3y

Substituting x = 3y:

3y − y = 26

2y = 26

y = 13

x = 39

Answer: The numbers are 39 and 13.

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

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

x + y = 180°

x − y = 18°

From x − y = 18:

x = y + 18

Substituting:

y + 18 + y = 180

2y = 162

y = 81°

x = 99°

Answer: The angles are 99° and 81°.

(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.

Let the cost of one bat be ₹x and the cost of one ball be ₹y.

7x + 6y = 3800

3x + 5y = 1750

From:

3x + 5y = 1750

3x = 1750 − 5y

x = (1750 − 5y)/3

Substituting in 7x + 6y = 3800:

7(1750 − 5y)/3 + 6y = 3800

12250 − 35y + 18y = 11400

12250 − 17y = 11400

17y = 850

y = 50

Substituting in 3x + 5y = 1750:

3x + 250 = 1750

3x = 1500

x = 500

Answer:

• Cost of one bat = ₹500
• Cost of one ball = ₹50

(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 ₹ 105 and for a journey of 15 km, the charge paid is ₹155. What are the fixed charges and the charge p

Let the fixed charge be ₹x and the charge per kilometre be ₹y.

x + 10y = 105

x + 15y = 155

From the first equation:

x = 105 − 10y

Substituting:

105 − 10y + 15y = 155

5y = 50

y = 10

x = 105 − 100

x = 5

For 25 km:

Fare = 5 + (25 × 10)

= 255

Answer:

• Fixed charge = ₹5
• Charge per km = ₹10
• Fare for 25 km = ₹255

(v) A fraction becomes 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 5/6 . Find the fraction.

Let the fraction be x/y.

Given:

(x + 2)/(y + 2) = 9/11

11(x + 2) = 9(y + 2)

11x − 9y = −4     (1)

Also:

(x + 3)/(y + 3) = 5/6

6(x + 3) = 5(y + 3)

6x − 5y = −3     (2)

From (2):

x = (5y − 3)/6

Substituting in (1):

11(5y − 3)/6 − 9y = −4

55y − 33 − 54y = −24

y = 9

x = 7

Answer: The fraction is ⁷⁄₉.

(vi) 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?

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

Five years hence:

x + 5 = 3(y + 5)

x − 3y = 10     (1)

Five years ago:

x − 5 = 7(y − 5)

x − 7y = −30     (2)

From (1):

x = 3y + 10

Substituting in (2):

3y + 10 − 7y = −30

−4y = −40

y = 10

x = 40

Answer:

• Jacob's present age = 40 years
• Son's present age = 10 years