Example 1
Completion requirements
| Example 1 |
Solve the given system of equations. Verify the solution.
- Equation 1: \(3x + 5y + 24 = 0\)
- Equation 2: \(2x + 2y = -8\)
To solve a system of equations by substitution, you can isolate either variable. However, it is often more efficient to choose the variable that is easier to isolate and whose equivalent expression will be simpler to work with.
Begin by isolating a variable.
\(\begin{align}
2x + 2y &= - 8 \\
2x &= - 8 - 2y \\
x &= - 4 - y \end{align}\)
The variable \(x\) was isolated using Equation 2, so substitute \(-4 - y\) for \(x\) in Equation 1, and solve for \(y\).
\(\begin{align}
3x + 5y + 24 &= 0 \\
3\left( { - 4 - y} \right) + 5y + 24 &= 0 \\
- 12 - 3y + 5y + 24 &= 0 \\
2y + 12 &= 0 \\
2y &= - 12 \\
y &= - 6 \end{align}\)
The \(y\)-value is \(–6\). Substitute this value into either of the original equations to determine the \(x\)-value. Recall that the equation \(x = - 4 - y\) is simply a rearranged version of the second original equation, \(2x + 2y = -8\).
\(\begin{align}
x &= - 4 - y \\
x &= - 4 - \left( { - 6} \right) \\
x &= 2 \end{align}\)
The solution is \((2, –6)\).
Verify the solution using the original equations.
\(\begin{align}
2x + 2y &= - 8 \\
2x &= - 8 - 2y \\
x &= - 4 - y \end{align}\)
The variable \(x\) was isolated using Equation 2, so substitute \(-4 - y\) for \(x\) in Equation 1, and solve for \(y\).
\(\begin{align}
3x + 5y + 24 &= 0 \\
3\left( { - 4 - y} \right) + 5y + 24 &= 0 \\
- 12 - 3y + 5y + 24 &= 0 \\
2y + 12 &= 0 \\
2y &= - 12 \\
y &= - 6 \end{align}\)
The \(y\)-value is \(–6\). Substitute this value into either of the original equations to determine the \(x\)-value. Recall that the equation \(x = - 4 - y\) is simply a rearranged version of the second original equation, \(2x + 2y = -8\).
\(\begin{align}
x &= - 4 - y \\
x &= - 4 - \left( { - 6} \right) \\
x &= 2 \end{align}\)
The solution is \((2, –6)\).
Verify the solution using the original equations.
\(3x + 5y + 24 = 0\)
| Left Side | Right Side |
|---|---|
|
\[\begin{array}{r} 3x + 5y + 24 \\ 3\left( 2 \right) + 5\left( { - 6} \right) + 24 \\ 0 \\ \end{array}\] |
\(0\) |
|
LS = RS |
|
\(2x + 2y = - 8\)
| Left Side | Right Side |
|---|---|
|
\[\begin{array}{r} 2x + 2y \\ 2\left( 2 \right) + 2\left( { - 6} \right) \\ - 8 \\ \end{array}\] |
\(-8\) |
|
LS = RS |
|