Question

the first three steps in determining the solution set of the system of equations algebraically are shown in the table. y = −x2 +2x − 9 y = −6x + 6 What are the solutions of this system of equations? (5, −24) and (3, −12) (5, 36) and (3, 24) (−5, −24) and (−3, 12) (−5, 36) and (−3, 24)

Answer 1

it is A! I know because i just took the test!

Answer 2

`(5,-24)` and
`(3,-12)`

Explanation:

we have

`y=-x^(2) +2x-9` -----> equation A

`y=-6x+6` -----> equation B

equate the equation A and equation B

`-x^(2) +2x-9=-6x+6`

`-x^(2) +2x-9+6x-6=0`

`-x^(2) +8x-15=0`

The formula to solve a quadratic equation of the form
`ax^(2) +bx+c=0` is equal to

`x=\frac{-b(+/-)\sqrt{b^(2)-4ac}} {2a}`

in this problem we have

`-x^(2) +8x-15=0`

so

`a=-1\\b=8\\c=-15`

substitute in the formula

`x=\frac{-8(+/-)\sqrt{8^(2)-4(-1)(-15)}} {2(-1)}`

`x=\frac{-8(+/-)√(64-60)} {-2}`

`x=(-8(+/-)2)/(-2)`

`x=(-8+2)/(-2)=3`

`x=(-8-2)/(-2)=5`

Find the values of y

For
`x=3`

`y=-6x+6` ------>
`y=-6(3)+6=-12`

For
`x=5`

`y=-6x+6` ------>
`y=-6(5)+6=-24`

The solutions are the points
`(3,-12)` and
`(5,-24)`