Answer:
C) (-1, -4) and (4, 6)
Step-by-step explanation:
[tex]\textsf{Equation 1}:y=2x-2[/tex]
[tex]\textsf{Equation 2}:y=x^2-x-6[/tex]
Substitute Equation 1 into Equation 2 and solve for x:
[tex]\implies 2x-2=x^2-x-6[/tex]
[tex]\implies x^2-3x-4=0[/tex]
Find two numbers that multiply to -4 and sum to -3: -4 and 1
Rewrite the middle term as the sum of these two numbers:
[tex]\implies x^2-4x+x-4=0[/tex]
Factorize the first two terms and the last two terms separately:
[tex]\implies x(x-4)+1(x-4)=0[/tex]
Factor out the common term [tex](x-4)[/tex]:
[tex]\implies (x+1)(x-4)=0[/tex]
[tex]\implies (x+1)=0 \implies x=-1[/tex]
[tex]\implies (x-4)=0 \implies x=4[/tex]
Substitute the found values of x into Equation 1 and solve for y:
[tex]x=-1 \implies y=2(-1)-2=-4[/tex]
[tex]x=4 \implies y=2(4)-2=6[/tex]
Therefore, the solution to the system of equations is:
(-1, -4) and (4, 6)
