Question

Solve the system of equations:

Answer 1

Answer:

C)  (-1, -4) and (4, 6)

Step-by-step explanation:

$\textsf{Equation 1}:y=2x-2$

$\textsf{Equation 2}:y=x^2-x-6$

Substitute Equation 1 into Equation 2 and solve for x:

$\implies 2x-2=x^2-x-6$

$\implies x^2-3x-4=0$

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:

$\implies x^2-4x+x-4=0$

Factorize the first two terms and the last two terms separately:

$\implies x(x-4)+1(x-4)=0$

Factor out the common term $(x-4)$:

$\implies (x+1)(x-4)=0$

$\implies (x+1)=0 \implies x=-1$

$\implies (x-4)=0 \implies x=4$

Substitute the found values of x into Equation 1 and solve for y:

$x=-1 \implies y=2(-1)-2=-4$

$x=4 \implies y=2(4)-2=6$

Therefore, the solution to the system of equations is:

(-1, -4) and (4, 6)