Solve the system by the method of substitution. $\left\{ \begin{array} { l }
y = x ^ { 3 } - 6 x ^ { 2 } + 4 \
y = x ^ { 2 } - 10 x + 4
\end{array} \right.$
A)(2, -12), (1, -5)
B)(5, -21), (2, -12), (0, 4)
C)(5, -21), (-1, 15)
D)No real solution
E)(5, -21), (-2, 28), (1, -5)

Question

Solve the system by the method of substitution.  $\left\{ \begin{array} { l } 
y = x ^ { 3 } - 6 x ^ { 2 } + 4 \
y = x ^ { 2 } - 10 x + 4
\end{array} \right.$  
A)(2, -12), (1, -5)
B)(5, -21), (2, -12), (0, 4)
C)(5, -21), (-1, 15)
D)No real solution
E)(5, -21), (-2, 28), (1, -5)

Quizplus · Accepted Answer

By substituting $y = x^2 - 10x + 4$ into $y = x^3 - 6x^2 + 4$, we get $x^3 - 6x^2 + 4 = x^2 - 10x + 4$. Simplifying gives $x^3 - 7x^2 + 10x = 0$, which factors to $x(x-5)(x-2) = 0$. Thus, $x = 0, 2, 5$. Substituting these back into either equation gives the points (0, 4), (2, -12), and (5, -21).

Solve the System by the Method of Substitution {y=x3−6x2+4y=x2−10x+4\left\{ \begin{array} { l } y = x ^ { 3 } - 6 x ^ { 2 } + 4 \\y = x ^ { 2 } - 10 x + 4\end{array} \right.{y=x3−6x2+4y=x2−10x+4​

Solve the System by the Method of Substitution $\left\{ \begin{array} { l } y = x ^ { 3 } - 6 x ^ { 2 } + 4 \\y = x ^ { 2 } - 10 x + 4\end{array} \right.$