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Completion requirements
Solve the given system algebraically.
\(\left\{ \begin{array}{l}
3y = x^2 + 2 \\
2y = - x^2 - 2 \end{array} \right.\)
\(\left\{ \begin{array}{l}
3y = x^2 + 2 \\
2y = - x^2 - 2 \end{array} \right.\)
The two systems are in the same format, so elimination is a good strategy to use for this system. Adding the two equations will eliminate the \(x^2\) term.
\(\begin{align}
3y &= \enspace \thinspace x^2 + 2 \\
+ (2y &= -x^2 - 2) \\
\hline {} \\
5y &= \qquad \enspace 0 \\
y &= \qquad \enspace 0 \end{align}\)
\(\begin{align}
y &= x^2 + 2 \\
0 &= x^2 + 2 \\
- 2 &= x^2 \\
\pm \sqrt { - 2} &= x
\end{align}\)
The square root of a negative number is not a real number, so this system has no real solution.
Graphical verification:
\(\begin{align}
3y &= \enspace \thinspace x^2 + 2 \\
+ (2y &= -x^2 - 2) \\
\hline {} \\
5y &= \qquad \enspace 0 \\
y &= \qquad \enspace 0 \end{align}\)
\(\begin{align}
y &= x^2 + 2 \\
0 &= x^2 + 2 \\
- 2 &= x^2 \\
\pm \sqrt { - 2} &= x
\end{align}\)
The square root of a negative number is not a real number, so this system has no real solution.
Graphical verification: