What are the solutions of the system?
y=x^2+3x-4
y=2x+2
a) (-3,6) and (2,-4)**
b) (-3,-4) and (2,6)
c) (-3,-4) and (-2,-2)
d) no solutions
2x+2 = x^2 + 3x -4
x^2 + x - 6 = 0
(x-2)(x+3) = 0
x = +2 ---> y = 6 or (2,6)
x = -3 ---> y = -4 or (-3,-4)
so I disagree with you
You got x = -3,2 correct, but your final answer is wrong.
My secon guess was C
guess? No need to guess !!!!
So am I right or wrong
I suck at this
which one is
(-3,-4) and (2,6) ?????
To find the solutions of the system, we need to find the values of x and y that satisfy both equations simultaneously.
Let's start by setting the two equations equal to each other:
x^2 + 3x - 4 = 2x + 2
Next, we can simplify the equation by moving all the terms to one side:
x^2 + 3x - 2x - 4 - 2 = 0
x^2 + x - 6 = 0
Now, we can factor the quadratic equation:
(x + 3)(x - 2) = 0
To find the values of x, we set each factor equal to zero and solve for x:
x + 3 = 0 --> x = -3
x - 2 = 0 --> x = 2
So, we have two possible values for x: -3 and 2.
To find the corresponding y-values, we can substitute these x-values into either of the original equations. Let's use the second equation:
y = 2x + 2
For x = -3:
y = 2(-3) + 2
y = -6 + 2
y = -4
So, one solution is (-3, -4).
For x = 2:
y = 2(2) + 2
y = 4 + 2
y = 6
So, another solution is (2, 6).
Therefore, the correct answer is a) (-3,6) and (2,-4).