Page 29 - 10-Math-2 THEORY OF QUADRATIC EQUATIONS

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Example 1: Solve the system of equations Version 1.1
3x + y = 4 and 3x2 + y2 = 52.

Solution: The given equations are
3x + y = 4 (i)
and 3x2 + y2 = 52 (ii)
From eq. (i) y = 4 - 3x (iii)

In an ordered pair (x, y), x always occupies first place and y
second place.

Put value of y in eq. (ii)
3x2 + (4 - 3x)2 = 52
3x2 + 16 - 24x + 9x2 - 52 = 0
12x2 - 24x - 36 = 0 or x2 - 2x - 3 = 0
By factorization
x2 - 3x + x - 3 = 0
x (x - 3) + 1 (x - 3) = 0
⇒ (x - 3) (x + 1) = 0

Either x - 3 = 0 or x + 1 = 0, that is,
x = 3 or x = -1
Put the values of x in eq. (iii)
When x = 3 When x = -1
y = 4 - 3x y = 4 - 3x
y = 4 - 3 (3) = 4 - 9 y = 4 - 3(-1) = 4 + 3
y = -5 y = 7
∴ The ordered pairs are (3, -5) and (-1,7)
Thus, the solution set is {(3, -5), (-1, 7)}

(b) when both the equations are quadratic.
The method to solve the equations is illustrated through the
following examples.

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