MATHEMATICS MIDDLE SCHOOL

What are the solutions of this system of equationThe 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
(5,-24) and (3, -12)
(5,36) and (3, 24)
(-5, -24) and (-3, 12)
(-5, 36) and (-3, 24)
Step
Equation
-x² + 2x -9 = -6x + 6
0 = x2 – 8x+15
0 = (x - 5)(x-3)
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Answers

Answer 1

Answer:

Answer:

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

Step-by-step explanation:

Given the system of two equations

$\left\{\begin{array}{l}y=-x^2+2x-9\n y=-6x+6\end{array}\right.$

Equate right sides of these two equations:

$-x^2+2x-9=-6x+6$

Rewrite this equation:

$-x^2+2x-9+6x-6=0\n \n-x^2+8x-15=0\n \nx^2-8x+15=0\n \nx^2-5x-3x+15=0\n \nx(x-5)-3(x-5)=0\n \n(x-3)(x-5)=0\n \nx_1=3\ \text{or}\ x_2=5$

If x=3, then

$y=-6\cdot 3+6=-18+6=-12$

If x=5, then

$y=-6\cdot 5+6=-30+6=-24$

You get two solutions (3,-12) and (5,-24)

Answer 2

Answer:

The solution set of the system of equation is (5, -24) and (3, -12)

System of equations

The given systen of equation is one linear and one quadratic. Given the syatem of equations

$y = -x^2 +2x - 9\ny = -6x + 6$

Equating bothequations will give

$-x^2 + 2x - 9 = -6x+ 6$

Equate to zero

$-x^2 + 2x - 9 + 6x - 6 =0\nx^2 - 2x +9 - 6x + 6 = 0\nx^2 - 8x + 15 = 0$

Factorize

$x^2 - 3x - 5x + 15 = 0\nx(x-3)-5(x-3) = 0\nx-5 = 0 \ and \ x - 3 = 0\nx = 5 \ and \ 3$

Recall that ;

y = -6x + 6

If x = 5

y = -30 + 6

y = -24

If x = 3

y =-6(3) + 6

y = -12

Hence the solution set of the system of equation is (5, -24) and (3, -12)

Learn more on system of equation here: brainly.com/question/25976025

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