Page 37 - 11-Math-4 Quadratic Equations

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version: 1.1
Example 2: Solve the following equations:

x2 + y2 +=4x 1 and x2 + (y -=1)2 10

Solution: The given system of equations is

x2 + y2 + 4x =1 (i)
 (ii)
 x 2 + y2 - 2y +1 =10

Subtraction gives,

4x + 2 y + 8 =0
⇒ 2x + y + 4 =0

⇒ y =-2x - 4
Putting the value of y in equation (i), (iii)

x2 + ( - 2x - 4)2 + 4x =1⇒ x2 + 4x2 + 16x + 16 + 4x =1

⇒ 5x2 + 20x +15 =0 ⇒ x + 4x + 3 =0
⇒ (x + 3)(x + 1) = 0 ⇒ x = -3 or x = -1

Putting x = -3 in (iii), we get; y =-2( - 3) - 4 =6 - 4 =2

Putting x = -1 in (iii), we get; y =-2( -1) - 4 =2 - 4 =-2

Hence solution set = {( - 3, 2),( -1, - 2)}.

Exercise 4.8

Solve the following systems of equations:

1. 2x - y =4; 2x2 - 4xy - y2 =6 2. x + y =5 ; x2 + 2 y2 =17

3. 3x + 2 y =7; 3x=2 25 + 2 y2 4. x + y= 5; 2 + 3= 2, x ≠0, y ≠0
xy

5. x + y = a + b; a + b = 2 6. 3x + =4 y 25; 3 +=4 2
xy xy

7. (x - 3)2 + y2 =5; 2x= y + 6

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