Solve system of equations by the method of substitution.
$$\left\{ \begin{array} { r }
4 x ^ { 3 } - 25 y = 0 \
4 x - y = 0
\end{array} \right.$$
A) $\quad ( 0,0 ) , ( - 5 , - 20 ) , ( 5,20 )$
B) $\quad ( 0,0 ) , ( - 5,20 ) , ( 5,20 )$
C) $\quad ( 0,0 ) , ( 5 , - 20 ) , ( 5,20 )$
D) $\quad ( 0,0 ) , ( - 5,21 ) , ( 5,21 )$
E) $\quad ( 0,0 ) , ( - 5 , - 19 ) , ( 5,19 )$

Question

Solve system of equations by the method of substitution.
$$\left\{ \begin{array} { r } 
4 x ^ { 3 } - 25 y = 0 \
4 x - y = 0
\end{array} \right.$$
A) $\quad ( 0,0 ) , ( - 5 , - 20 ) , ( 5,20 )$
B) $\quad ( 0,0 ) , ( - 5,20 ) , ( 5,20 )$
C) $\quad ( 0,0 ) , ( 5 , - 20 ) , ( 5,20 )$
D) $\quad ( 0,0 ) , ( - 5,21 ) , ( 5,21 )$
E) $\quad ( 0,0 ) , ( - 5 , - 19 ) , ( 5,19 )$

Quizplus · Accepted Answer

From the second equation, $y = 4x$. Substituting $y$ in the first equation gives $4x^3 - 25(4x) = 0$, which simplifies to $x^3 - 25x = x(x^2 - 25) = x(x-5)(x+5) = 0$. Thus, $x = 0, 5, -5$. Substituting these values back into $y = 4x$ gives the pairs $(0,0)$, $(5,20)$, and $(-5,-20)$.

Solve System of Equations by the Method of Substitution {4x3−25y=04x−y=0\left\{ \begin{array} { r } 4 x ^ { 3 } - 25 y = 0 \\4 x - y = 0\end{array} \right.{4x3−25y=04x−y=0​

Solve System of Equations by the Method of Substitution $\left\{ \begin{array} { r } 4 x ^ { 3 } - 25 y = 0 \\4 x - y = 0\end{array} \right.$