$\left\{ \begin{array} { l }
x ^ { 2 } + 4 x + y = 64 \
2 x + y - 16 = 0
\end{array} \right.$
A) (4,8),(0,64)
B) (-6,28)
C) (6,4),(-8,32)
D) (6,4),(-10,36)

Question

$\left\{ \begin{array} { l } 
x ^ { 2 } + 4 x + y = 64 \
2 x + y - 16 = 0
\end{array} \right.$
A)  (4,8),(0,64) 
B)  (-6,28) 
C)  (6,4),(-8,32) 
D)  (6,4),(-10,36)

Quizplus · Accepted Answer

Solving the system of equations by substitution or elimination method, we first express $y$ from the second equation: $y = 16 - 2x$. Substituting this into the first equation gives $x^2 + 4x + 16 - 2x = 64$, simplifying to $x^2 + 2x - 48 = 0$. Factoring or using the quadratic formula, we find $x = 6$ or $x = -8$. Substituting these values back into $y = 16 - 2x$, we get the pairs $(6, 4)$ and $(-8, 32)$.

{x2+4x+y=642x+y−16=0\left\{ \begin{array} { l } x ^ { 2 } + 4 x + y = 64 \\2 x + y - 16 = 0\end{array} \right.{x2+4x+y=642x+y−16=0​

$\left\{ \begin{array} { l } x ^ { 2 } + 4 x + y = 64 \\2 x + y - 16 = 0\end{array} \right.$