**Exercise 4.1 Page: 68**

**1. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.**

**(Take the cost of a notebook to be Rs x and that of a pen to be Rs y)**

Solution:

Let the cost of a notebook to be = Rs x

Let the cost of a pen to be = Rs y

According to the question,

The cost of a notebook is twice the cost of a pen.

i.e., Cost of a notebook = 2Ã—Cost of a pen

x = 2Ã—y

x = 2y

x-2y = 0

x-2y = 0 is the linear equation in two variables to represent the statement â€˜The cost of a notebook is twice the cost of a penâ€™.

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

**(ii) x â€“(y/5)â€“10 = 0**

Solution:

The equation x â€“(y/5)-10 = 0 can be written as,

1x+(-1/5)y +(â€“10) = 0

Now comparing x+(-1/5)y+(â€“10) = 0 with ax+by+c = 0

We get,

a = 1

b = -(1/5)

c = -10

**(iii) â€“2x+3y = 6**

Solution:

â€“2x+3y = 6

Re-arranging the equation, we get,

â€“2x+3yâ€“6 = 0

The equation â€“2x+3yâ€“6 = 0 can be written as,

(â€“2)x+3y+(â€“ 6) = 0

Now comparing (â€“2)x+3y+(â€“6) = 0 with ax+by+c = 0

We get, a = â€“2

b = 3

c =6

**(iv) x = 3y**

Solution:

x = 3y

Re-arranging the equation, we get,

x-3y = 0

The equation x-3y=0 can be written as,

1x+(-3)y+(0)c = 0

Now comparing 1x+(-3)y+(0)c = 0 with ax+by+c = 0

We get, a = 1

b = -3

c =0

**(v) 2x = â€“5y**

Solution:

2x = â€“5y

Re-arranging the equation, we get,

2x+5y = 0

The equation 2x+5y = 0 can be written as,

2x+5y+0 = 0

Now comparing 2x+5y+0= 0 with ax+by+c = 0

We get, a = 2

b = 5

c = 0

**(vi) 3x+2 = 0**

Solution:

3x+2 = 0

The equation 3x+2 = 0 can be written as,

3x+0y+2 = 0

Now comparing 3x+0+2= 0 with ax+by+c = 0

We get, a = 3

b = 0

c = 2

**(vii) yâ€“2 = 0**

Solution:

yâ€“2 = 0

The equation yâ€“2 = 0 can be written as,

0x+1y+(â€“2) = 0

Now comparing 0x+1y+(â€“2) = 0with ax+by+c = 0

We get, a = 0

b = 1

c = â€“2

**(viii) 5 = 2x**

Solution:

5 = 2x

Re-arranging the equation, we get,

2x = 5

i.e., 2xâ€“5 = 0

The equation 2xâ€“5 = 0 can be written as,

2x+0yâ€“5 = 0

Now comparing 2x+0yâ€“5 = 0 with ax+by+c = 0

We get, a = 2

b = 0

c = -5