Question

Can someone please help me, thanks

Answer 1

There are a lot of questions here. I'll do problems 7 and 8 to get you started.

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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.

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Answer: There are two solutions and they are (0, -2) and (-2, -8)

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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.

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Answer: There is one solution and it is (-2, 4)