Solve the system algebraically.
-$$\begin{array} { l }
x ^ { 2 } + y ^ { 2 } = 12 \
3 x - y = - 2
\end{array}$$
A) $\left( \frac { - 3 + \sqrt { 29 } } { 5 } , \frac { 1 + 3 \sqrt { 29 } } { 5 } \right)$ and $\left( \frac { - 3 - \sqrt { 29 } } { 5 } , \frac { 1 - 3 \sqrt { 29 } } { 5 } \right)$
B) $\left\{ \frac { - 3 + 2 \sqrt { 29 } } { 5 } , \frac { 1 + 3 \sqrt { 29 } } { 5 } \right)$ and $\left( \frac { - 3 - 6 \sqrt { 29 } } { 5 } , \frac { 1 - 6 \sqrt { 29 } } { 5 } \right)$
C) $\left\{ \frac { - 3 + \sqrt { 29 } } { 5 } , \frac { - 7 + 3 \sqrt { 29 } } { 5 } \right)$ and $\left( \frac { - 3 - \sqrt { 29 } } { 5 } , \frac { - 7 - 3 \sqrt { 29 } } { 5 } \right)$
D) $\left( \frac { - 3 + \sqrt { 29 } } { 5 } , \frac { 1 - 3 \sqrt { 29 } } { 5 } \right)$ and $\left( \frac { - 3 - \sqrt { 29 } } { 5 } , \frac { 1 + 3 \sqrt { 29 } } { 5 } \right)$

Question

Solve the system algebraically.
-$$\begin{array} { l } 
x ^ { 2 } + y ^ { 2 } = 12 \
3 x - y = - 2
\end{array}$$
A) $\left( \frac { - 3 + \sqrt { 29 } } { 5 } , \frac { 1 + 3 \sqrt { 29 } } { 5 } \right)$ and $\left( \frac { - 3 - \sqrt { 29 } } { 5 } , \frac { 1 - 3 \sqrt { 29 } } { 5 } \right)$
B) $\left\{ \frac { - 3 + 2 \sqrt { 29 } } { 5 } , \frac { 1 + 3 \sqrt { 29 } } { 5 } \right)$ and $\left( \frac { - 3 - 6 \sqrt { 29 } } { 5 } , \frac { 1 - 6 \sqrt { 29 } } { 5 } \right)$
C) $\left\{ \frac { - 3 + \sqrt { 29 } } { 5 } , \frac { - 7 + 3 \sqrt { 29 } } { 5 } \right)$ and $\left( \frac { - 3 - \sqrt { 29 } } { 5 } , \frac { - 7 - 3 \sqrt { 29 } } { 5 } \right)$
D) $\left( \frac { - 3 + \sqrt { 29 } } { 5 } , \frac { 1 - 3 \sqrt { 29 } } { 5 } \right)$ and $\left( \frac { - 3 - \sqrt { 29 } } { 5 } , \frac { 1 + 3 \sqrt { 29 } } { 5 } \right)$

Quizplus · Accepted Answer

Solving the system algebraically involves expressing one variable in terms of the other using one equation and substituting it into the other equation. From the second equation, $y = 3x + 2$. Substituting this into the first equation gives a quadratic equation in terms of $x$, which can be solved to find the values of $x$. Substituting these $x$ values back into $y = 3x + 2$ gives the corresponding $y$ values. The correct solutions are those that satisfy both original equations, which are given in option A.

Solve the System Algebraically x2+y2=123x−y=−2\begin{array} { l } x ^ { 2 } + y ^ { 2 } = 12 \\3 x - y = - 2\end{array}x2+y2=123x−y=−2​

Solve the System Algebraically $\begin{array} { l } x ^ { 2 } + y ^ { 2 } = 12 \\3 x - y = - 2\end{array}$