Solve the system by the method of substitution. $$\left\{ \begin{array} { l }
y = - 4 x \
y = x ^ { 3 } + x ^ { 2 } - x
\end{array} \right.$$
A)(3,4)
B)(0,3)
C)(3,4), (-3,-4)
D)No real solution
E)(0,0)

Question

Solve the system by the method of substitution.​ $$\left\{ \begin{array} { l } 
y = - 4 x \
y = x ^ { 3 } + x ^ { 2 } - x
\end{array} \right.$$ ​
A)(3,4)
B)(0,3)
C)(3,4), (-3,-4)
D)No real solution
E)(0,0)

Quizplus · Accepted Answer

Substitute $ y = -4x $ into $ y = x^3 + x^2 - x $ to get $-4x = x^3 + x^2 - x$. Simplifying gives $x^3 + x^2 + 3x = 0$, which factors to $x(x^2 + x + 3) = 0$. The real solution is $x = 0$, giving $y = -4(0) = 0$, so the solution is (0,0).

Solve the System by the Method of Substitution {y=−4xy=x3+x2−x\left\{ \begin{array} { l } y = - 4 x \\y = x ^ { 3 } + x ^ { 2 } - x\end{array} \right.{y=−4xy=x3+x2−x​

Solve the System by the Method of Substitution $\left\{ \begin{array} { l } y = - 4 x \\y = x ^ { 3 } + x ^ { 2 } - x\end{array} \right.$