Solve the system by the method of substitution. { y = x 2 − 2 7 x + 2 y = − 4 \left\{ \begin{aligned}y & = x ^ { 2 } - 2 \7 x + 2 y & = - 4\end{aligned} ight. { y 7 x + 2 y = x 2 − 2 = − 4 A) ( 0 , − 2 ) ( − 7 2 , 41 4 ) ( 0 , - 2 ) \left( - \frac { 7 } { 2 } , \frac { 41 } { 4 } ight) ( 0 , − 2 ) ( − 2 7 , 4 41 ) B) ( 0 , − 2 ) ( 7 2 , 41 4 ) ( 0 , - 2 ) \left( \frac { 7 } { 2 } , \frac { 41 } { 4 } ight) ( 0 , − 2 ) ( 2 7 , 4 41 ) C) ( − 7 2 , 41 4 ) \left( - \frac { 7 } { 2 } , \frac { 41 } { 4 } ight) ( − 2 7 , 4 41 ) D) ( 0 , 2 ) ( − 7 2 , 41 4 ) ( 0,2 ) \left( - \frac { 7 } { 2 } , \frac { 41 } { 4 } ight) ( 0 , 2 ) ( − 2 7 , 4 41 ) E) no solution exists

Question

Solve the system by the method of substitution.   { y = x 2 − 2 7 x + 2 y = − 4 \left\{ \begin{aligned}y  &  = x ^ { 2 } - 2 \7 x + 2 y  &  = - 4\end{aligned} ight. { y 7 x + 2 y ​ = x 2 − 2 = − 4 ​ A)    ( 0 , − 2 )  ( − 7 2 , 41 4 )  ( 0 , - 2 )  \left( - \frac { 7 } { 2 } , \frac { 41 } { 4 } ight)  ( 0 , − 2 )  ( − 2 7 ​ , 4 41 ​ )  B)    ( 0 , − 2 )  ( 7 2 , 41 4 )  ( 0 , - 2 )  \left( \frac { 7 } { 2 } , \frac { 41 } { 4 } ight)  ( 0 , − 2 )  ( 2 7 ​ , 4 41 ​ )  C)    ( − 7 2 , 41 4 )  \left( - \frac { 7 } { 2 } , \frac { 41 } { 4 } ight)  ( − 2 7 ​ , 4 41 ​ )  D)    ( 0 , 2 )  ( − 7 2 , 41 4 )  ( 0,2 )  \left( - \frac { 7 } { 2 } , \frac { 41 } { 4 } ight)  ( 0 , 2 )  ( − 2 7 ​ , 4 41 ​ )  E) no solution exists

Alicia Hosein · Accepted Answer

Final Answer : A, Explanation :   Substitute   y = x 2 − 2 y = x^2 - 2 y = x 2 − 2   into the second equation to get   7 x + 2 ( x 2 − 2 ) = − 4 7x + 2(x^2 - 2) = -4 7 x + 2 ( x 2 − 2 ) = − 4  . Simplify and solve the quadratic equation to find the values of   x x x  , then use these to find corresponding   y y y   values. The solutions are   ( 0 , − 2 ) (0, -2) ( 0 , − 2 )   and   ( − 7 2 , 41 4 ) \left(-\frac{7}{2}, \frac{41}{4}\right) ( − 2 7 ​ , 4 41 ​ )  .

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