Question

Find 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 solution

Answer 1

b. (-3, -4) and (2, 6)

Explanation:

By the transitive property of equality, if y equals thing 1 and y also equals thing 2, then thing1 and thing 2 are also equal. So we will set them equal to each other and factor to solve for the 2 values of x:

`2x+2=x^2+3x-4`

Get everything on one side of the equals sign, set the whole mess equal to 0, and combine like terms to get:

`0=x^2+x-6`

Because this is a second degree polynomial, a quadratic to be precise, it has 2 solutions. We need to find those 2 values of x and then use them in either one of the original equations to solve for the y values that go with each x.

Factoring that polynomial above gives you the x values of x = -3 and 2. Sub in -3 first:

y = 2(-3) + 2 and

y = -6 + 2 so

y = -4

Therefore, the coordinate is (-3, -4).

Onto the next x value of 2:

y = 2(2) + 2 and

y = 4 + 2 so

y = 6

Therefore, the coordinate is (2, 6)