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

Question

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

Quizplus · Accepted Answer

To solve the system by substitution, set the two equations equal to each other since they both equal $y$: $$3x^2 + 6x = x^2 - 4x + 12.$$ Rearrange the equation to set it to 0: $$2x^2 + 10x - 12 = 0.$$ Simplify by dividing the entire equation by 2: $$x^2 + 5x - 6 = 0.$$ Factor the quadratic equation: $$(x + 6)(x - 1) = 0.$$ Solve for $x$: $x = -6$ and $x = 1.$ Substitute $x = -6$ and $x = 1$ back into one of the original equations to find $y$: For $x = 1$, $y = 3(1)^2 + 6(1) = 9.$ For $x = -6$, $y = 3(-6)^2 + 6(-6) = 72.$ Therefore, the solutions are $(1, 9)$ and $(-6, 72)$.

Solve the System by the Substitution Method y=3x2+6xy=x2−4x+12\begin{array} { l } y = 3 x ^ { 2 } + 6 x \\y = x ^ { 2 } - 4 x + 12\end{array}y=3x2+6xy=x2−4x+12​

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