Example 1
Completion requirements
| Example 1 |
Multimedia |
Solve the given system of equations by substitution. Verify the solution.
\(\left\{ \begin{array}{l}
3x^2 - 42x - y + 159 = 0 \\
3x - y - 3 = 0 \end{array} \right.\)
Although any variable can be isolated, it is a good idea to choose one that will make for an easy substitution.
The \(x\)-values of the solution can be substituted into any equation that contains both an \(x\) and a \(y\). Choose an equation that is simple to work with.
Verification can be completed by substituting the solution(s) back into the original equations or by checking that the graphical solution matches the algebraic solution.
Verify for \((6, 15)\).
Verify for \((9, 24)\).
\(\left\{ \begin{array}{l}
3x^2 - 42x - y + 159 = 0 \\
3x - y - 3 = 0 \end{array} \right.\)
Start by isolating one variable.
\(\begin{align}
3x - y + 3 &= 0 \\
3x - 3 &= y \\
\end{align}\)
Substitute the equivalent expression into the other equation and solve.
\(\begin{align}
3x^2 - 42x - y + 159 &= 0 \\
3x^2 - 42x - \left( {\color{red}{3x - 3}} \right) + 159 &= 0 \\
3x^2 - 45x + 162 &= 0 \\
3\left( {x^2 - 15x + 54} \right) &= 0 \\
3\left( {x - 6} \right)\left( {x - 9} \right) &= 0
\end{align}\)
\(\begin{align}
3x - y + 3 &= 0 \\
3x - 3 &= y \\
\end{align}\)
Substitute the equivalent expression into the other equation and solve.
\(\begin{align}
3x^2 - 42x - y + 159 &= 0 \\
3x^2 - 42x - \left( {\color{red}{3x - 3}} \right) + 159 &= 0 \\
3x^2 - 45x + 162 &= 0 \\
3\left( {x^2 - 15x + 54} \right) &= 0 \\
3\left( {x - 6} \right)\left( {x - 9} \right) &= 0
\end{align}\)
\(\begin{align}
x - 6 &= 0 \\
x &= 6 \end{align}\)
x - 6 &= 0 \\
x &= 6 \end{align}\)
\(\begin{align}
x - 9 &= 0 \\
x &= 9 \end{align}\)
x - 9 &= 0 \\
x &= 9 \end{align}\)
Although any variable can be isolated, it is a good idea to choose one that will make for an easy substitution.
- \(x\) was not isolated in the first equation because \(x\) appears in two terms.
- \(y\) was not isolated in the first equation because the equivalent expression is fairly complex.
- \(x\) was not isolated in the second equation because substituting for \(x\) in the first equation is more complex than substituting for \(y\).
- \(y\) was isolated in the second equation because its equivalent expression is fairly simple and substituting for \(y\) in the first equation is relatively simple.
Now that the \(x\)-values are known, they can be substituted individually into any equation containing both \(x\) and \(y\) to determine the \(y\)-value.
The solutions to the system are \((6, 15)\) and \((9, 24)\).
\(\begin{align}
3x - 3 &= y \\
3\left( 6 \right) - 3 &= y \\
15 &= y \end{align}\)
3x - 3 &= y \\
3\left( 6 \right) - 3 &= y \\
15 &= y \end{align}\)
\(\begin{align}
3x - 3 &= y \\
3\left( 9 \right) - 3 &= y \\
24 &= y \end{align}\)
3x - 3 &= y \\
3\left( 9 \right) - 3 &= y \\
24 &= y \end{align}\)
The \(x\)-values of the solution can be substituted into any equation that contains both an \(x\) and a \(y\). Choose an equation that is simple to work with.
The solutions can be verified by substituting them into the original equations.
Verification can be completed by substituting the solution(s) back into the original equations or by checking that the graphical solution matches the algebraic solution.
Verify for \((6, 15)\).
\(3x^2 - 42x - y + 159 = 0\)
| Left Side | Right Side |
|---|---|
|
\[\begin{array}{r}
3x^2 - 42x - y + 159 \\ 3\left( 6 \right)^2 - 42\left( 6 \right) - \left( {15} \right) + 159 \\ 0 \end{array}\] |
\(0\)
|
|
LS = RS\(\hspace{30pt}\) |
|
\(3x - y - 3 = 0\)
| Left Side | Right Side |
|---|---|
|
\[\begin{array}{r}
3x - y - 3 \\ 3\left( 6 \right) - \left( {15} \right) - 3 \\ 0 \end{array}\] |
\(0\) |
|
LS = RS |
|
Verify for \((9, 24)\).
\(3x^2 - 42x - y + 159 = 0\)
| Left Side | Right Side |
|---|---|
|
\[\begin{array}{r}
3x^2 - 42x - y + 159 \\ 3\left( 9 \right)^2 - 42\left( 9 \right) - \left( {24} \right) + 159 \\ 0 \end{array}\] |
\(0\) |
|
LS = RS |
|
\(3x - y - 3 = 0\)
| Left Side | Right Side |
|---|---|
|
\[\begin{array}{r}
3x - y - 3 \\ 3\left( 9 \right) - \left( {24} \right) - 3 \\ 0 \end{array}\] |
\(0\) |
|
LS = RS |
|