Originally Posted by

**spirit79** I can't find the right answer to this question... Is it just me, or are the solutions given wrong?

x^2 + 3x + 5y = 20

x + 3y = 1

Given solutions are x = 3 or x = -9/2

Where did I go wrong?

X = 1 – 3y ---- (1)

X^2 + 3x + 5y = 20 --- (2)

Sub (1) into (2), (1 – 3y)^2 + 3(1 – 3y) + 5y = 20

1 – 6y + 9y^2 + 3 – 9y + 5y = 20

9y^2 – 10y – 16 = 0

(9y + 8 ) (y – 2) = 0

y = -8/9 or y = 2

Sub y = -8/9 in (1),

x = 1 - 3(-8/9)

= 11/3

Sub y = 2 in (1),

x = 1 - 3(2)

= -5

Are the given solutions correct? And are what I worked the right answers? If not, please tell me where I made a mistake. I'm not too sure if I substituted the y values correctly...