3.2  SOLUTION OF A LINEAR EQUATION IN TWO VARIABLES

The solution of a linear equation in two variables is a pair of values, one for x and the other for y, which makes the two sides of the equation equal.

Eg:  If   2x + y = 4, then ( 0, 4 ) is one of its solutions as it satisfies the equation. A linear equation in two variables has infinitely many solutions.

A pair of values of variables ‘x‘ and ‘y’ which satisfy both the equations in the given system of equations is said to be a solution of the simultaneous pair of linear equation.

EXERCISE 3.1

Aftab tells his daughter, ‘Seven years ago, I was seven times as old as u were then. Also three years from now, I shall be three times as old as you will be.’ (Isn’t this interesting?) Represent this situation algebraically and graphically.

Solution : At present: Let Aftab’s age = x years

We plot both the equations, we obtain a straight lines.

2. The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 3 more balls of the same kind for Rs 1300. Represent this situation algebraically and graphically.

Solution. Let cost of 1 cricket bat   =   Rs x    and let cost of 1 cricket ball  =   Rs y

According to given conditions, we have

3 x   +   6 y   =   3900        ⇒    x   +   2 y   =   1300     ….… (1)

And  x  +  3 y   =   1300                    ….… (2)

To represent them graphically, we will find 3 sets of points which lie on the lines.

For equation   x   +  2 y   =   1300, we have following points which lie on the line

We plot both the equations, and the  equations intersect at (1300, 0).

Q3. The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically.

Solution: Let cost of 1 kg of apples   =   Rs x         and let cost of 1 kg of grapes  =   Rs y

According to given conditions, we have

2 x  +  y   =   160               ….… (1)

4x  +  2 y   =   300

⇒ 2 x  +  y   =   150         ….… (2)

So, we have equations (1) and (2),    2 x  +  y   =   160    and 2 x  +  y   =   150 which represent given situation algebraically.

For equation   2 x  +  y   =   160, we have following points which lie on the line.