Exercise 6.1

1. Express the following linear equations in the form of ax+by+c=0 and indicate the values of a, b, and c in each case.

(i) 8x + 5y − 3 = 0

Solution:

This equation is already in the required form.

a = , b = , c =

(ii) 28x − 35y = − 7

Solution:

Adding 7 to both sides, we get 28x - 35y + 7 = 014

a = , b = -3535, c =

(iii) 93x = 12 − 15y

Solution:

Adding 15y and subtracting 12 from both sides, we get 93x-93x + 15-15y - 12 = 0

a = , b = , c =

(iv) 2x = − 5y

Solution:

Adding 5y to both sides, we get 2x + 5y-5y = 0. Here, c = .

a = , b = , c =

(v) 74 + 34 = xy

Solution:

Multiplying both sides by 4y, we get 7y7 + 3y3 = 4xx

Rearranging, we get 4x - 104y = 0

a = , b = , c =

(vi) −x2 = 3y

Solution:

By cross multiplying, we get -xyxy = 65

Adding xy to both sides, we get xy +- 6 = 0

a = , b = , c =

(vii) 35 = 12

Solution:

This equation does not have x and y terms. It simplifies to 3 = , which is not a linear equation.

2. Write each of the following in the form of ax + by + c = 0 and find the values of a, b and c.

(i) 2x = 5

Solution:

Given 2x = 5

We can write the given equation as: 2x -+ 5 = 0

a = , b = , c =

(ii) y − 2 = 0

Solution:

This equation is already in the form ax + by + c = 0.

a = , b = , c =

(iii) y7 = 3

Solution:

Multiplying both sides by 7, we get y = 2120

Subtracting 21 from both sides, we get y -+ 21 = 0

a = , b = , c =

(iv) x = −1413

Solution:

Multiplying both sides by 13, we get 13x-13x = -1414

Adding 14 to both sides, we get 13x +- 14 = 0

a = , b = , c =

3. Express the following statements as a linear equation in two variables.

(i) The sum of two numbers is 34.

Solution:

Let the two numbers be x and y.

The sum of the two numbers is + .

Therefore, the linear equation is x + y =

(ii) The cost of a ball pen is 5 less than half the cost of a fountain pen.

Solution:

Let the cost of a ball pen be x and the cost of a fountain pen be y.

Half the cost of a fountain pen is y2y.

5 less than half the cost of a fountain pen is y2 - .

The linear equation is x = y2 - 5

Which is: x = y−102

= - + 10 = 0

(iii) Bhargavi got 10 more marks than double the marks of Sindhu.

Solution:

Let Bhargavi's marks be x and Sindhu's marks be y.

Double the marks of Sindhu is 2yy2.

10 more than double the marks of Sindhu is 2y + .

Therefore, the linear equation is x = 2y + 10 or x -+ 2y - = 0

(iv) The cost of a pencil is 2 and a ball point pen is 15. Sheela pays 100 for the pencils and pens she purchased.

Solution:

Let the number of pencils Sheela purchased be x and the number of ball point pens be y.

The cost of x pencils = 2 × x = .

The cost of y ball point pens = 15 × y = .

The total cost of pencils and pens is 2x + 15y.

The linear equation is 2x + 15y = 10030

Therefore, the linear equation is 2x + 15y - 100.

(v) Yamini and Fatima of class IX together contributed Rs. 200/- towards the Prime Minister’s Relief Fund.

Solution:

Let Yamini's contribution be x and Fatima's contribution be y.

Their total contribution is x + y.

The linear equation is x + y =

Therefore, the linear equation is x + y - 200 = 0

(vi) The sum of a two digit number and the number obtained by reversing the order of its digits is 121. If the digits in unit’s and ten’s place are ‘x’ and ‘y’ respectively.

Solution:

The two-digit number is 10y + x-x.

The number obtained by reversing the digits is 10xx + y.

The sum of the two numbers is (10y + x) + (10x + y) = + .

The linear equation is 11x + 11y = 12111

(x + y) = 121

x + y =

Therefore, the linear equation is x + y - 11 = 0