Question

Find the solutions of the system. y=x^2+3x-4 y=2x + 2, have to show work

Answer 1

This one is a real easy one.

I'll use the substitution method. Since the second equation is a "y =" statement, I can just do:

2x + 2 = x² + 3x - 4

Now we just subtract by 2x and 2 on both sides then combine like terms. This will give you:

x² + x - 6 = 0

Now factor to get (x + 3)(x - 2) = 0 and set each binomial equal to 0. This gives you x = - 3 and x = 2.

Now plug in - 3 and 2 into either equation. I'd pick the second equation, y = 2x + 2, because it's easy.

For x = - 3:

y = 2(- 3) + 2
y = - 6 + 2
y = - 4

This ordered pair solution would be (- 3, - 4)

For x = 2:

y = 2(2) + 2
y = 4 + 2
y = 6

This ordered pair solution would be: (2, 6)

Answer 2

Hint: There are several ways to proceed in this kind of problems, but usually the simplest and the one that will yield the most success on average is substituting.

Look at y variable and the expressions to which it is equal; substitute the easier-looking one in the more complicated.
Applying this, we get: 2x+2=x^2+3x-4
x^2+x-6=0.

Solving this equation of 2nd degree, we have a=1, b=1, c=-6 and we get Δ=25 and the solutions are -3 and 2. Hence, we get that the only possible values for x are -3 and 2.
To get the solutions, we have to also find the corresponding y-values:
For x=-3, y=2x+2=-4. For x=2, y=6.

Hence, the 2 pairs of solutions are: (-3,-4) and (2,6)