Solve system of equations by the method of substitution.
$$\left\{ \begin{aligned}
12 x ^ { 3 } - 9 y & = 0 \
12 x - y & = 0
\end{aligned} \right.$$
A) $( 0,0 ) , ( - 3 , - 36 ) , ( 3,36 )$
B) $( 0,0 ) , ( - 3,36 ) , ( 3,36 )$
C) $( 0,0 ) , ( 3 , - 36 ) , ( 3,36 )$
D) $( 0,0 ) , ( - 3,37 ) , ( 3,37 )$
E) $( 0,0 ) , ( - 3 , - 35 ) , ( 3,35 )$

Question

Quizplus · Accepted Answer

From the second equation, $y = 12x$. Substituting $y$ in the first equation gives $12x^3 - 9(12x) = 0$, which simplifies to $x^3 - 9x = 0$, or $x(x^2 - 9) = 0$, leading to $x = 0$, $x = 3$, and $x = -3$. Substituting these $x$ values back into $y = 12x$ gives the pairs $(0,0)$, $(3,36)$, and $(-3,-36)$.

Solve System of Equations by the Method of Substitution {12x3−9y=012x−y=0\left\{ \begin{aligned}12 x ^ { 3 } - 9 y & = 0 \\12 x - y & = 0\end{aligned} \right.{12x3−9y12x−y​=0=0​

Solve System of Equations by the Method of Substitution $\left\{ \begin{aligned}12 x ^ { 3 } - 9 y & = 0 \\12 x - y & = 0\end{aligned} \right.$