Answer:
(3,-12) and (5,-24)
Step-by-step explanation:
Given the system of two equations
[tex]\left\{\begin{array}{l}y=-x^2+2x-9\\ y=-6x+6\end{array}\right.[/tex]
Equate right sides of these two equations:
[tex]-x^2+2x-9=-6x+6[/tex]
Rewrite this equation:
[tex]-x^2+2x-9+6x-6=0\\ \\-x^2+8x-15=0\\ \\x^2-8x+15=0\\ \\x^2-5x-3x+15=0\\ \\x(x-5)-3(x-5)=0\\ \\(x-3)(x-5)=0\\ \\x_1=3\ \text{or}\ x_2=5[/tex]
If x=3, then
[tex]y=-6\cdot 3+6=-18+6=-12[/tex]
If x=5, then
[tex]y=-6\cdot 5+6=-30+6=-24[/tex]
You get two solutions (3,-12) and (5,-24)
The solution set of the system of equation is (5, -24) and (3, -12)
The given systen of equation is one linear and one quadratic. Given the syatem of equations
[tex]y = -x^2 +2x - 9\\y = -6x + 6[/tex]
Equating both equations will give
[tex]-x^2 + 2x - 9 = -6x+ 6[/tex]
Equate to zero
[tex]-x^2 + 2x - 9 + 6x - 6 =0\\x^2 - 2x +9 - 6x + 6 = 0\\x^2 - 8x + 15 = 0[/tex]
Factorize
[tex]x^2 - 3x - 5x + 15 = 0\\x(x-3)-5(x-3) = 0\\x-5 = 0 \ and \ x - 3 = 0\\x = 5 \ and \ 3[/tex]
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)
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