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Pena Hilton

28/08/2022 · Primary School

Solve the system. Use any method you wish. 

\(x + y + 2 = 0     \)       (1)

\(x^ 2 + y^ 2 + 4y - 3x = - 4\)        (2)

Select the correct choice below and, if necessary, fill in the answer box to complete your choice. 

A. The solution set is \(\{ \} \) .

    (Simplify your answer. Type an ordered pair. Type exact answers, using radicals as needed. Use a comma to separate answer as needed.)

B. There is no solution. 

Answer
expertExpert-Verified Answer

Weber Delgado
Qualified Tutor
4.0 (24votes)

Solve the system of equations: \(x + y + 2 = 0 , x ^ { 2 } + y ^ { 2 } + 4 y - 3 x = - 4\) :\(\left ( \begin{array} { l l } { x = \frac { 3 } { 2 } , } & { y = - \frac { 7 } { 2 } } \\ { x = 0 , } & { y = - 2 } \end{array} \right ) \)

 

Steps 

 

\(\left [ \begin{array} { c } { x + y + 2 = 0 } \\ { x ^ { 2 } + y ^ { 2 } + 4 y - 3 x = - 4 } \end{array} \right ] \) 

Isolate \(y\) for \(x + y + 2 = 0 : y = - x - 2\) 

Plug the solutions \(y = - x - 2\) into \(x ^ { 2 } + y ^ { 2 } + 4 y - 3 x = - 4\) 

For \(x ^ { 2 } + y ^ { 2 } + 4 y - 3 x = - 4\) , subsitute \(y\) with \(- x - 2 : x = \frac { 3 } { 2 } , x = 0\) 

Plug the solutions \(x = \frac { 3 } { 2 } , x = 0\) into \(x + y + 2 = 0\) 

For \(x + y + 2 = 0\) , subsitute \(x\) with \(\frac { 3 } { 2 } : y = - \frac { 7 } { 2 } \) 

For \(x + y + 2 = 0\) , subsitute \(x\) with \(0 : y = - 2\) 

Verify solutions by plugging them into the original equations 

Therefore, the final solutions for \(x + y + 2 = 0 , x ^ { 2 } + y ^ { 2 } + 4 y - 3 x = - 4\) are 

\(\left ( \begin{array} { l } { x = \frac { 3 } { 2 } , y = - \frac { 7 } { 2 } } \\ { x = 0 , y = - 2 } \end{array} \right ) \) 

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