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

Answer:

$(5,-24)$ and  $(3,-12)$

Step-by-step 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

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

$x=\frac{-8(+/-)\sqrt{64-60}} {-2}$

$x=\frac{-8(+/-)2}{-2}$

$x=\frac{-8+2}{-2}=3$

$x=\frac{-8-2}{-2}=5$

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 $(5,-24)$

Answer 2

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