Question

The first three steps in determining the solution set of the system of equations algebraically are shown.

y = x2 − x − 3
y = −3x + 5



What are the solutions of this system of equations?

(−2, −1) and (4, 17)
(−2, 11) and (4, −7)
(2, −1) and (−4, 17)
(2, 11) and (−4, −7)

Answer 1

Your answer is (2,-1), (-4,17).

Answer 2

In order to solve the system of equations algebraically you have to follow those steps

we have

`y=x^(2)-x-3` ------> equation A

`y=-3x+5` -------> equation B

Step 1

Equate the equation A and equation B

`x^(2)-x-3=-3x+5`

`x^(2)-x-3+3x-5=0`

`x^(2)+2x-8=0`

Step 2

Convert the quadratic equation in factored form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

`x^(2)+2x=8`

Complete the square. Remember to balance the equation by adding the same constants to each side.

`x^(2)+2x+1=8+1`

`x^(2)+2x+1=9`

Rewrite as perfect squares

`(x+1)^(2)=9`

Square root both sides

`x+1=(+/-)3`

`x=-1(+/-)3`

`x1=-1+3=2`

`x2=-1-3=-4`

Step 3

Find the values of y

Substitute the value of x in the equation B

For
`x=2`

`y=-3*2+5=-1`

For
`x=-4`

`y=-3*(-4)+5=17`

therefore

the answer is

`(2,-1)` and
`(-4,17)`