Solve the system by the method of substitution. $$\left\{ \begin{aligned}
x & = \sqrt { y + 10 } \
3 x + 2 y & = - 20
\end{aligned} \right.$$
A) $( 0 , - 10 ) \left( - \frac { 3 } { 2 } , - \frac { 31 } { 4 } \right)$
B) $\left( \frac { 3 } { 2 } , - \frac { 31 } { 4 } \right)$
C) $( 0 , - 10 )$
D) $( 0,10 ) \left( \frac { 3 } { 2 } , - \frac { 31 } { 4 } \right)$
E)no solution exist

Question

Quizplus · Accepted Answer

Substitute $x = \sqrt{y + 10}$ into the second equation to get $3\sqrt{y + 10} + 2y = -20$. Squaring both sides and solving for $y$ yields two solutions: $y = -10$ and $y = -\frac{31}{4}$. Substituting these values back into the first equation gives the corresponding $x$ values, resulting in the solutions $(0, -10)$ and $\left(-\frac{3}{2}, -\frac{31}{4}\right)$.

Solve the System by the Method of Substitution {x=y+103x+2y=−20\left\{ \begin{aligned}x & = \sqrt { y + 10 } \\3 x + 2 y & = - 20\end{aligned} \right.{x3x+2y​=y+10​=−20​

Solve the System by the Method of Substitution $\left\{ \begin{aligned}x & = \sqrt { y + 10 } \\3 x + 2 y & = - 20\end{aligned} \right.$