- #1
wat2000
- 46
- 0
use the subtitution method to solve.
x^2 + 2y = 9
x - y + 3 = 0
y=x+3 -> (3)
substitute for y with x+3, you get:
x^2+2(x+3)=9 x2+2x−3=0
factorize it to get:
(x+3)(x−1)=0 x=−3,x=1
substitute for each value of x in equation 3
x=−3y=−3+3=0
x=1y=1+3=4
(x,y)=(−3,0),(x,y)=(1,4)
Is this right?
x^2 + 2y = 9
x - y + 3 = 0
y=x+3 -> (3)
substitute for y with x+3, you get:
x^2+2(x+3)=9 x2+2x−3=0
factorize it to get:
(x+3)(x−1)=0 x=−3,x=1
substitute for each value of x in equation 3
x=−3y=−3+3=0
x=1y=1+3=4
(x,y)=(−3,0),(x,y)=(1,4)
Is this right?