Graph the equations to determine whether the system has any solutions. Find any solutions that exist. { 2 x + y = 12 x 2 + y 2 = 36 \left\{ \begin{array} { r } 2 x + y = 12 \x ^ { 2 } + y ^ { 2 } = 36\end{array} ight. { 2 x + y = 12 x 2 + y 2 = 36 A) ( 6 , 0 ) ( − 18 5 , − 24 5 ) ( 6,0 ) \left( - \frac { 18 } { 5 } , - \frac { 24 } { 5 } ight) ( 6 , 0 ) ( − 5 18 , − 5 24 ) B) ( − 6 , 0 ) ( 18 5 , 24 5 ) ( - 6,0 ) \left( \frac { 18 } { 5 } , \frac { 24 } { 5 } ight) ( − 6 , 0 ) ( 5 18 , 5 24 ) C) ( 6 , 0 ) ( 18 5 , 24 5 ) ( 6,0 ) \left( \frac { 18 } { 5 } , \frac { 24 } { 5 } ight) ( 6 , 0 ) ( 5 18 , 5 24 ) D) ( − 6 , 0 ) ( − 18 5 , − 24 5 ) ( - 6,0 ) \left( - \frac { 18 } { 5 } , - \frac { 24 } { 5 } ight) ( − 6 , 0 ) ( − 5 18 , − 5 24 ) E) no solution exists

Question

Graph the equations to determine whether the system has any solutions. Find any solutions that exist.   { 2 x + y = 12 x 2 + y 2 = 36 \left\{ \begin{array} { r } 2 x + y = 12 \x ^ { 2 } + y ^ { 2 } = 36\end{array} ight. { 2 x + y = 12 x 2 + y 2 = 36 ​ A)    ( 6 , 0 )  ( − 18 5 , − 24 5 )  ( 6,0 )  \left( - \frac { 18 } { 5 } , - \frac { 24 } { 5 } ight)  ( 6 , 0 )  ( − 5 18 ​ , − 5 24 ​ )  B)    ( − 6 , 0 )  ( 18 5 , 24 5 )  ( - 6,0 )  \left( \frac { 18 } { 5 } , \frac { 24 } { 5 } ight)  ( − 6 , 0 )  ( 5 18 ​ , 5 24 ​ )  C)    ( 6 , 0 )  ( 18 5 , 24 5 )  ( 6,0 )  \left( \frac { 18 } { 5 } , \frac { 24 } { 5 } ight)  ( 6 , 0 )  ( 5 18 ​ , 5 24 ​ )  D)    ( − 6 , 0 )  ( − 18 5 , − 24 5 )  ( - 6,0 )  \left( - \frac { 18 } { 5 } , - \frac { 24 } { 5 } ight)  ( − 6 , 0 )  ( − 5 18 ​ , − 5 24 ​ )  E) no solution exists

Dariell Simmons · Accepted Answer

Final Answer : C, Explanation :   We can solve for $y$ in the first equation to get $y=12-2x$ and substitute into the second equation:   x 2 + ( 12 − 2 x ) 2 = 36    ⟹    5 x 2 − 48 x + 72 = 0 x^2+(12-2x)^2=36 \implies 5x^2-48x+72=0 x 2 + ( 12 − 2 x ) 2 = 36 ⟹ 5 x 2 − 48 x + 72 = 0   Solving for $x$ using the quadratic formula we get:   x = 48 ± 4 8 2 − 4 ( 5 ) ( 72 ) 10 = 24 ± 6 21 5 x=\frac{48 \pm \sqrt{48^2-4(5)(72)}}{10}= \frac{24 \pm 6\sqrt{21}}{5} x = 10 48 ± 4 8 2 − 4 ( 5 ) ( 72 ) ​ ​ = 5 24 ± 6 21 ​ ​   Substituting $x$ back into the first equation we get the corresponding solutions for $y$:   x = 24 + 6 21 5    ⟹    y = 24 − 2 ( 24 + 6 21 ) 5 = − 24 5 − 12 21 5 x=\frac{24 + 6\sqrt{21}}{5} \implies y=\frac{24 - 2(24 + 6\sqrt{21})}{5}=-\frac{24}{5} -\frac{12\sqrt{21}}{5} x = 5 24 + 6 21 ​ ​ ⟹ y = 5 24 − 2 ( 24 + 6 21 ​ ) ​ = − 5 24 ​ − 5 12 21 ​ ​     x = 24 − 6 21 5    ⟹    y = 24 − 2 ( 24 − 6 21 ) 5 = − 24 5 + 12 21 5 x=\frac{24 - 6\sqrt{21}}{5} \implies y=\frac{24 - 2(24 - 6\sqrt{21})}{5}=-\frac{24}{5} +\frac{12\sqrt{21}}{5} x = 5 24 − 6 21 ​ ​ ⟹ y = 5 24 − 2 ( 24 − 6 21 ​ ) ​ = − 5 24 ​ + 5 12 21 ​ ​   Therefore, the system has two solutions: $\left(\frac{24 + 6\sqrt{21}}{5},-\frac{24}{5} -\frac{12\sqrt{21}}{5}ight)$ and $\left(\frac{24 - 6\sqrt{21}}{5},-\frac{24}{5} +\frac{12\sqrt{21}}{5}ight)$. The correct choice is $\boxed{	extbf{(C)}}$.

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