Solve the system by the substitution method.
-$y=3 x^{2}+6 x$
$\mathrm{y}=\mathrm{x}^{2}-4 \mathrm{x}+12$
A) $\{(1,9),(-6,72)\}$
B) $\{(2,0),(-3,9)\}$
C) $\{(-2,0),(-3,9)\}$
D) $\{(2,0),(3,45)\}$

Question

Quizplus · Accepted Answer

To solve the system by substitution, we set the two equations equal to each other since they both equal $y$:$$3x^2 + 6x = x^2 - 4x + 12$$Rearranging the equation to set it to 0:$$2x^2 + 10x - 12 = 0$$Dividing the entire equation by 2 to simplify:$$x^2 + 5x - 6 = 0$$Factoring the quadratic equation:$$(x + 6)(x - 1) = 0$$Thus, the solutions for $x$ are $x = -6$ and $x = 1$.Substituting $x = -6$ into the first equation:$$y = 3(-6)^2 + 6(-6) = 108 - 36 = 72$$Substituting $x = 1$ into the first equation:$$y = 3(1)^2 + 6(1) = 3 + 6 = 9$$Therefore, the solution set is $\{(1,9),(-6,72)\}$.

Solve the System by the Substitution Method y=3x2+6xy=3 x^{2}+6 xy=3x2+6x y=x2−4x+12\mathrm{y}=\mathrm{x}^{2}-4 \mathrm{x}+12y=x2−4x+12

Solve the System by the Substitution Method $y=3 x^{2}+6 x$ $\mathrm{y}=\mathrm{x}^{2}-4 \mathrm{x}+12$