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

Question

Quizplus · Accepted Answer

From the second equation, $y = 4x$. Substituting $y$ in the first equation gives $4x^3 - 25(4x) = 0$, which simplifies to $4x^3 - 100x = 0$. Factoring out $4x$, we get $4x(x^2 - 25) = 0$, leading to $x = 0$, $x = 5$, and $x = -5$. Substituting these $x$ values back into $y = 4x$ gives the solutions $(0,0)$, $(5, 20)$, and $(-5, -20)$.

Solve System of Equations by the Method of Substitution {4x3−25y=04x−y=0\left\{ \begin{aligned}4 x ^ { 3 } - 25 y & = 0 \\4 x - y & = 0\end{aligned} \right.{4x3−25y4x−y​=0=0​

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