There are a lot of questions here. I'll do problems 7 and 8 to get you started.
====================================================
Problem 7
The given equations are
y = x^2 + 5x - 2
y = 3x - 2
Since both right hand sides are equal to y, we can equate them, get everything to one side, and solve
x^2 + 5x - 2 = 3x - 2
x^2 + 5x - 2 - 3x + 2 = 0
x^2 + 2x = 0
x(x + 2) = 0
x = 0 or x + 2 = 0 ........ zero product property
x = 0 or x = -2
Now plug each x value back into one of the original equations given
If x = 0, then
y = 3x-2
y = 3*0-2
y = -2
Making (0,-2) as one solution. Repeat for the x value x = -2
y = 3x-2
y = 3(-2)-2
y = -8
The point (-2, -8) is the other solution to the system.
-------------------------------
Answer: There are two solutions and they are (0, -2) and (-2, -8)
====================================================
Problem 8
y = -x^2 - 3x + 2
y = x + 6
We'll use the same idea as before
y = -x^2 - 3x + 2
x + 6 = -x^2 - 3x + 2
x + 6 + x^2 + 3x - 2 = 0
x^2 + 4x + 4 = 0
Now factor the left side to solve for x
(x+2)(x+2) = 0
(x+2)^2 = 0
x+2 = 0
x = -2
Use this x value to find y
y = x+6
y = -2+6
y = 4
Or we can cay
y = -x^2 - 3x + 2
y = -(-2)^2 - 3(-2) + 2
y = 4
We get the same y value either way.
-------------------------------
Answer: There is one solution and it is (-2, 4)
