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Completion requirements
Solve the given system of equations by elimination. Verify the solution.
The equations are in different formats. Begin by rearranging them into the same format.
Subtract the two equations to eliminate the \(p\)-variable, then solve for \(q\).
\(\begin{align}
&2p + 3q - 12 = 0 \\
-( &2p - 3q - \thinspace \thinspace 6 = 0) \\
\hline {} \\
&0p + 6q - \thinspace \thinspace 6 = 0 \\
\end{align}\)
\(\begin{align}
6q - 6 &= 0 \\
6q &= 6 \\
q &= 1 \\
\end{align}\)
Substitute this known value into either of the original equations to determine the \(p\)-value.
The solution is \(p = \frac{9}{2}\) and \(q = 1\).
Verify the solution using the original equations.
Try to solve the given system of equations by elimination before continuing with the lesson.
\(\left\{ \begin{array}{l}
y = x^2 - x - 3 \\
y = x - 2 \end{array} \right.\)
\[\left\{ \begin{array}{l}
2 - \frac{p}{3} = \frac{q}{2} \\
\frac{5}{3}\left( {2p - 3q} \right) = 10 \\
\end{array} \right.\]
2 - \frac{p}{3} = \frac{q}{2} \\
\frac{5}{3}\left( {2p - 3q} \right) = 10 \\
\end{array} \right.\]
The equations are in different formats. Begin by rearranging them into the same format.
\[\begin{align}
2 - \frac{p}{3} &= \frac{q}{2} \\
2 - \frac{p}{3} - \frac{q}{2} &= 0 \\
- \frac{1}{3}p - \frac{1}{2}q + 2 &= 0 \\
- 6\left( { - \frac{1}{3}p - \frac{1}{2}q + 2} \right) &= - 6\left( 0 \right) \\
2p + 3q - 12 &= 0 \\
\end{align}\]
\[\begin{align}
\frac{5}{3}\left( {2p - 3q} \right) &= 10 \\
2p - 3q &= 6 \\
2p - 3q - 6 &= 0 \\
\end{align}\]
2 - \frac{p}{3} &= \frac{q}{2} \\
2 - \frac{p}{3} - \frac{q}{2} &= 0 \\
- \frac{1}{3}p - \frac{1}{2}q + 2 &= 0 \\
- 6\left( { - \frac{1}{3}p - \frac{1}{2}q + 2} \right) &= - 6\left( 0 \right) \\
2p + 3q - 12 &= 0 \\
\end{align}\]
\[\begin{align}
\frac{5}{3}\left( {2p - 3q} \right) &= 10 \\
2p - 3q &= 6 \\
2p - 3q - 6 &= 0 \\
\end{align}\]
\(\begin{align}
&2p + 3q - 12 = 0 \\
-( &2p - 3q - \thinspace \thinspace 6 = 0) \\
\hline {} \\
&0p + 6q - \thinspace \thinspace 6 = 0 \\
\end{align}\)
\(\begin{align}
6q - 6 &= 0 \\
6q &= 6 \\
q &= 1 \\
\end{align}\)
Substitute this known value into either of the original equations to determine the \(p\)-value.
\[\begin{align}
2p + 3q - 12 &= 0 \\
2p + 3\left( 1 \right) - 12 &= 0 \\
2p &= 9 \\
p &= \frac{9}{2}
\end{align}\]
2p + 3q - 12 &= 0 \\
2p + 3\left( 1 \right) - 12 &= 0 \\
2p &= 9 \\
p &= \frac{9}{2}
\end{align}\]
The solution is \(p = \frac{9}{2}\) and \(q = 1\).
Verify the solution using the original equations.
\(2 - \frac{p}{3} = \frac{q}{2}\)
| Left Side | Right Side |
|---|---|
| \[\begin{array}{r} 2 - \frac{p}{3} \\ 2 - \frac{{\left( {\frac{9}{2}} \right)}}{3} \\ 2 - \frac{3}{2} \\ \frac{1}{2} \end{array}\] |
\[\begin{array}{l} \frac{q}{2} \\ \frac{1}{2} \\ \end{array}\] |
| LS = RS | |
\(\frac{5}{3}\left( {2p - 3q} \right) = 10\)
| Left Side | Right Side |
|---|---|
| \[\begin{array}{r} \frac{5}{3}\left( {2p - 3q} \right) \\ \frac{5}{3}\left( {2\left( {\frac{9}{2}} \right) - 3\left( 1 \right)} \right) \\ 10 \end{array}\] |
\(10\) |
| LS = RS | |
Try to solve the given system of equations by elimination before continuing with the lesson.
\(\left\{ \begin{array}{l}
y = x^2 - x - 3 \\
y = x - 2 \end{array} \right.\)