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.

1. Two air crafts are represented  by  the  equations x  +  2 y  –  4   =   0 and  2 x  +  4 y  –  12   =   0. Represent this situation geometrically.

2. Rehana went to a stationery shop and purchased 12 pencils and 13 erasers for Rs 280. Her friend Sheetal saw the new variety of pencils and erasers with Rehana, and she also bought 4 pencils and 6 erasers of the same kind for Rs 300. Represent this situation algebraically and graphically.

3. Ankita goes to a fair with Rs 200 and wants to have rides on the Giant Wheel and play Hoopla. Represent this situation algebraically and graphically (geometrically).

4. Ashish tells his son, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” Represent this situation algebraically and graphically.

5. The cost of 5 kg of rice and 3 kg of wheat on a day was found to be Rs 245. After a month, the cost of 4 kg of rice and 4 kg of is wheat Rs 350. Represent the situation algebraically and geometrically. We plot both linear equations and get two parallel straight lines.