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

Question

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

Zybrea Knight · Accepted Answer

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

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