y=x^2+2x-2
y=3x+4
2007-11-22 12:13:51 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
y = x^2+2x-2 = 3x+4
x^2-x-6 = (x-3)(x+2) = 0
x = 3, y = 3*3+4 = 13
or
x = -2, y = 3(-2)+4 = -2
Therefore, you have two answers: (3, 13) and (-2, -2)
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I have to say that sarkarta got the right answer before me. But I got my current answer before I saw his answer.
2007-11-22 12:25:34 · answer #1 · answered by sahsjing 7 · 0⤊ 0⤋
[04]
y=x^2+2x-2....(1)
y=3x+4.....(2)
As both the equations are equal,we may write
x^2+2x-2=3x+4
x^2+2x-2-3x-4=0
x^2-x-6=0
(x-3)(x+2)=0
x=3 or -2
Plugging the values of x in eqn 2,we get
If x=3,y=3*3+4=13
If x= -2,y=-2*3+4= -2
Ans: x=3 and y=13 or x= -2 and y= -2
2007-11-22 12:30:16 · answer #2 · answered by alpha 7 · 0⤊ 0⤋
x² + 2x - 2 = 3x + 4
x² - x - 6 = 0
(x - 3)(x + 2) = 0
x = 3 , x = - 2
2007-11-23 02:28:17 · answer #3 · answered by Como 7 · 0⤊ 1⤋
3x=y-4
x=(y-4)/3
plug into (1)
x^2+2x-2 becomes
[(y-4)/3]^2+2[(y-3)/4]-2
(y^2-8y+16)/9 +(2y-6)/4-2
36y=4y^2-32y+64 + 18y-54-72
4y^2-50y-62=0
2y^2-25y-31=0
solve this quadratic equation for y
y=13.636 y=-1.136
Use y=3x+4 and solve for x using each value
2007-11-22 12:28:48 · answer #4 · answered by cidyah 7 · 0⤊ 0⤋