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Chapter 3: Pair of Linear Equations in Two Variables

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Pair of linear equations in two variables CBSE NCERT Solutions

Question:

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

(i) The difference between the 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 is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for traveling a distance of 25 km?

(v) A fraction becomes  , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and denominator it becomes  . Find the fraction.

(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?

Answer:

Solutions:

(i) Let the first number be x and the second number be y.

According to given conditions, we have

x – y = 26 (assuming x > y) … (1)

x = 3y (x > y)… (2)

Putting equation (2) in (1), we get

3y – y = 26

⇒ 2y = 26

⇒ y = 13

Putting the value of y in equation (2), we get

x = 3y = 3  \times 13 = 39

Therefore, two numbers are 13 and 39.

(ii) Let smaller angle = x and let larger angle = y

According to given conditions, we have

y = x + 18 … (1)

Also, x + y = 180 \degree (Sum of supplementary angles) … (2)

Putting (1) in equation (2), we get

x + x + 18 = 180

⇒ 2x = 180 – 18 = 162

⇒ x = 81 \degree

Putting the value of x in equation (1), we get

y = x + 18 = 81 + 18 = 99 \degree

Therefore, the two angles are  81 \degree and 99 \degree .

(iii) Let the cost of each bat = Rs x and let the cost of each ball = Rs y

According to given conditions, we have

7x + 6y = 3800 … (1)

And, 3x + 5y = 1750 … (2)

Using equation (1), we can say that

7x = 3800 − 6y 

 \Rightarrow x = \frac{3800 - 6y}{7}  

Putting this in equation (2), we get

3 \left ( \frac{3800 - 6y}{7} \right ) + 5y = 1750\Rightarrow \left ( \frac{11400 - 18y}{7} \right ) + 5y = 1750

 

\Rightarrow \frac{5y}{1} - \frac{18y}{7} = \frac{1750}{1} - \frac{11400}{7}

⇒ 17y = 850 ⇒ y = 50

Putting the value of y in (2), we get

3x + 250 = 1750

⇒ 3x = 1500

⇒ x = 500

Therefore, the cost of each bat = Rs 500 and the cost of each ball = Rs 50

(iv) Let fixed charge = Rs x and let charge for every km = Rs y

According to given conditions, we have

x + 10y = 105… (1)

x + 15y = 155… (2)

Using equation (1), we can say that

x = 105 − 10y

Putting this in equation (2), we get

105 − 10y + 15y = 155

⇒ 5y = 50 ⇒ y = 10

Putting the value of y in equation (1), we get

x + 10 (10) = 105

⇒ x = 105 – 100 = 5

Therefore, fixed charge = Rs 5 and charge per km = Rs 10

To travel distance of 25 Km, a person will have to pay = Rs (x + 25y)

= Rs (5 + 25 × 10)

= Rs (5 + 250) = Rs 255

(v) Let numerator = x and let denominator = y

According to given conditions, we have

\frac{x + 2}{y + 2} = \frac {9}{11} .... (1)

\frac{x + 3}{y + 3} = \frac {5}{6} .... (2)

Using equation (1), we can say that

11 (x + 2) = 9y + 18

⇒ 11x + 22 = 9y + 18

⇒ 11x = 9y – 4

 \Rightarrow x = \frac{9y - 4}{11}

Putting the value of x in equation (2), we get

6 \left ( \frac{9y - 4}{11} + 3 \right ) = 5 (y + 3)

\Rightarrow \frac{54y}{11} - \frac{24}{11} + 18 = 5y + 15

\Rightarrow \frac{-24}{11} + \frac{3}{1} = \frac{5y}{1} - \frac{54y}{11}

\Rightarrow \frac{-24 + 33}{11} = \frac{55y - 54y}{11}

⇒ y = 9

Putting the value of y in (1), we get

\frac {x + 2}{9 + 2} = \frac {9}{11}

⇒ x + 2 = 9 ⇒ x = 7

Therefore, the fraction is  \frac {x}{y} = \frac {7}{9}  .

(vi) Let present age of Jacob is x years and the present age of Jacob’s son is y years.
According to given conditions, we have
(x + 5) = 3 (y + 5) … (1)
And, (x − 5) = 7 (y − 5) … (2)
From equation (1), we can say that
x + 5 = 3y + 15
⇒ x = 10 + 3y
Putting the value of x = 10 + 3y in equation (2) we get
10 + 3y – 5 = 7y − 35
⇒ −4y = −40
⇒ y = 10 years
Putting the value of y in equation (1), we get
x + 5 = 3 (10 + 5) = 3  15 = 45
⇒ x = 45 – 5 = 40 years
Therefore, the present age of Jacob is 40 years and, the present age of Jacob’s son is 10 years.

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