Example 2
Completion requirements
| Example 2 |
Solve the given system of equations by elimination. Verify the solution.
\(\left\{ \begin{array}{l}
2x = 5y - 38 \\
6y = 8 - 7x \end{array} \right.\)
\(\left\{ \begin{array}{l}
2x = 5y - 38 \\
6y = 8 - 7x \end{array} \right.\)
Begin by rearranging the equations into the same format.
\(\begin{align}
2x &= 5y - 38 \\
2x - 5y &= - 38 \end{align}\)
\(\begin{align}
6y &= 8 - 7x \\
6y + 7x &= 8 \\
7x + 6y &= 8 \\
\end{align}\)
The equations are now in the same format, but neither variable has the same coefficient in each equation. Multiply one or both equations by a constant to give one variable the same coefficient in each equation. Determining a Least Common Multiple (LCM) may help decide which constants to multiply by. In this case, the coefficients of \(x\) have an LCM of \(14\). Multiplying the first equation by \(7\) and the second equation by \(2\) will result in the required common coefficient.
\(\begin{align}
7\left( {2x - 5y} \right) &= 7\left( { - 38} \right) \\
14x - 35y &= - 266 \\
\end{align}\)
\(\begin{align}
2\left( {7x + 6y} \right) &= 2\left( 8 \right) \\
14x + 12y &= 16 \\
\end{align}\)
Now the two \(x\)-variables have the same coefficient. Subtract the two equations to eliminate the \(x\)-variable, then solve for \(y\).
\(\begin{align}
14x -35y &= -266 \\
-(14x + 12y &= \thinspace 16) \\
\hline {} \\
0x - 47y &= - 282 \\
- 47y &= - 282 \\
y &= \thinspace \thinspace 6 \\
\end{align}\)
Substitute this known value into either of the original equations to determine the \(x\)-value.
\(\begin{align}
2x &= 5y - 38 \\
2x &= 5\left( 6 \right) - 38 \\
2x &= - 8 \\
x &= - 4 \\
\end{align}\)
The solution is \((–4, 6)\).
Verify the solution using the original equations.
\(\begin{align}
2x &= 5y - 38 \\
2x - 5y &= - 38 \end{align}\)
\(\begin{align}
6y &= 8 - 7x \\
6y + 7x &= 8 \\
7x + 6y &= 8 \\
\end{align}\)
The equations are now in the same format, but neither variable has the same coefficient in each equation. Multiply one or both equations by a constant to give one variable the same coefficient in each equation. Determining a Least Common Multiple (LCM) may help decide which constants to multiply by. In this case, the coefficients of \(x\) have an LCM of \(14\). Multiplying the first equation by \(7\) and the second equation by \(2\) will result in the required common coefficient.
\(\begin{align}
7\left( {2x - 5y} \right) &= 7\left( { - 38} \right) \\
14x - 35y &= - 266 \\
\end{align}\)
\(\begin{align}
2\left( {7x + 6y} \right) &= 2\left( 8 \right) \\
14x + 12y &= 16 \\
\end{align}\)
Now the two \(x\)-variables have the same coefficient. Subtract the two equations to eliminate the \(x\)-variable, then solve for \(y\).
\(\begin{align}
14x -35y &= -266 \\
-(14x + 12y &= \thinspace 16) \\
\hline {} \\
0x - 47y &= - 282 \\
- 47y &= - 282 \\
y &= \thinspace \thinspace 6 \\
\end{align}\)
Substitute this known value into either of the original equations to determine the \(x\)-value.
\(\begin{align}
2x &= 5y - 38 \\
2x &= 5\left( 6 \right) - 38 \\
2x &= - 8 \\
x &= - 4 \\
\end{align}\)
The solution is \((–4, 6)\).
Verify the solution using the original equations.
Verify for \(2x = 5y - 38\).
| Left Side | Right Side |
|---|---|
| \(\begin{array}{r} 2x \\ 2\left( { - 4} \right) \\ - 8 \end{array}\) |
\(\begin{array}{l} 5y - 38 \\ 5\left( 6 \right) - 38 \\ - 8 \end{array}\) |
| LS = RS | |
Verify for \(6y = 8 - 7x\).
| Left Side | Right Side |
|---|---|
| \(\begin{array}{r} 6y \\ 6\left( 6 \right) \\ 36 \end{array}\) |
\(\begin{array}{l} 8 - 7x \\ 8 - 7\left( { - 4} \right) \\ 36 \end{array}\) |
| LS = RS | |