Example 1
Completion requirements
| Example 1 |
Solve the rational equation \(\frac{2}{3} = \frac{{4x + 1}}{{5x}}\). Identify any non-permissible values, and verify the solution(s).
Step 1: Identify any non-permissible values.
\(x \ne 0\)
Step 2: Determine the lowest common denominator, and multiply both sides by this expression.
The LCD is \(15x\).
Step 3: Solve for \(x\).
\(\begin{align}
10x &= 12x + 3 \\
-3 &= 2x \\
-\frac{3}{2} &= x \\
\end{align}\)
Step 4: Verify for \(x = -\frac{3}{2}\).
Because the left side equals the right side, and \(x = -\frac{3}{2}\) is not an NPV, \(x = -\frac{3}{2}\) is the solution to the equation.
\(x \ne 0\)
Step 2: Determine the lowest common denominator, and multiply both sides by this expression.
The LCD is \(15x\).
\[\begin{align}
\left( {\frac{2}{3}} \right)\left( {15x} \right) &= \left( {\frac{{4x + 1}}{{5x}}} \right)\left( {15x} \right) \\
\left( 2 \right)\left( {5x} \right) &= \left( {4x + 1} \right)\left( 3 \right) \\
10x &= 12x + 3 \\
\end{align}\]
\left( {\frac{2}{3}} \right)\left( {15x} \right) &= \left( {\frac{{4x + 1}}{{5x}}} \right)\left( {15x} \right) \\
\left( 2 \right)\left( {5x} \right) &= \left( {4x + 1} \right)\left( 3 \right) \\
10x &= 12x + 3 \\
\end{align}\]
Step 3: Solve for \(x\).
\(\begin{align}
10x &= 12x + 3 \\
-3 &= 2x \\
-\frac{3}{2} &= x \\
\end{align}\)
Step 4: Verify for \(x = -\frac{3}{2}\).
| Left Side | Right Side |
|---|---|
|
\[\frac{2}{3}\] |
\[\begin{array}{l} \frac{{4x + 1}}{{5x}} \\ \frac{{4\left( { - \frac{3}{2}} \right) + 1}}{{5\left( { - \frac{3}{2}} \right)}} \\ \frac{{ - 6 + 1}}{{ - \frac{{15}}{2}}} \\ \frac{{ - 5}}{{ - \frac{{15}}{2}}} \\ 5 \times \frac{2}{{15}} \\ \frac{2}{3} \\ \end{array}\] |
| \(\hspace {25pt}\) LS = RS | |
Because the left side equals the right side, and \(x = -\frac{3}{2}\) is not an NPV, \(x = -\frac{3}{2}\) is the solution to the equation.