Long Beck
02/27/2024 · High School
Solve for \( x \) and \( y \) simultaneously: \( y+x=2 \) and \( x^{2}+3 x y+8=0 \)
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Tutor-Verified Answer
Step-by-step Solution
Solve the system of equations \( y+x=2;x^2+3xy+8=0 \).
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
\(\left\{ \begin{array}{l}y+x=2\\x^{2}+3xy+8=0\end{array}\right.\)
- step1: Solve the equation:
\(\left\{ \begin{array}{l}x=2-y\\x^{2}+3xy+8=0\end{array}\right.\)
- step2: Substitute the value of \(x:\)
\(\left(2-y\right)^{2}+3\left(2-y\right)y+8=0\)
- step3: Simplify:
\(12+2y-2y^{2}=0\)
- step4: Factor the expression:
\(2\left(3-y\right)\left(2+y\right)=0\)
- step5: Divide the terms:
\(\left(3-y\right)\left(2+y\right)=0\)
- step6: Separate into possible cases:
\(\begin{align}&3-y=0\\&2+y=0\end{align}\)
- step7: Solve the equation:
\(\begin{align}&y=3\\&y=-2\end{align}\)
- step8: Calculate:
\(y=3\cup y=-2\)
- step9: Rearrange the terms:
\(\left\{ \begin{array}{l}x=2-y\\y=3\end{array}\right.\cup \left\{ \begin{array}{l}x=2-y\\y=-2\end{array}\right.\)
- step10: Calculate:
\(\left\{ \begin{array}{l}x=-1\\y=3\end{array}\right.\cup \left\{ \begin{array}{l}x=4\\y=-2\end{array}\right.\)
- step11: Check the solution:
\(\left\{ \begin{array}{l}x=-1\\y=3\end{array}\right.\cup \left\{ \begin{array}{l}x=4\\y=-2\end{array}\right.\)
- step12: Rewrite:
\(\left(x,y\right) = \left(-1,3\right)\cup \left(x,y\right) = \left(4,-2\right)\)
The solutions for the system of equations \( y+x=2 \) and \( x^{2}+3xy+8=0 \) are \( (x,y) = (-1,3) \) and \( (x,y) = (4,-2) \).
Quick Answer
The solutions are \( (x,y) = (-1,3) \) and \( (x,y) = (4,-2) \).
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