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Completion requirements
Solve \(-\frac{3}{2}x - 1 = \left|\frac{1}{2}x - 1\right|\). Verify the solutions by substitution.
Case 1: \(\frac{1}{2}x - 1 \ge 0\)
This case occurs when \(x \ge 2\). Since \(\frac{1}{2}x - 1 \ge 0\), the absolute value symbol can be dropped without changing the expression.
Extraneous solutions can be found by comparing the solution to the interval of interest or by verifying the solution by substitution.
Case 2: \(\frac{1}{2}x - 1 \lt 0\)
This case occurs when \(x \lt 2\). Since \(\frac{1}{2}x - 1 \lt 0\), it becomes negative when the absolute value symbol is dropped.
The solution \(x = -2\) is part of the internal of interest, \(x \lt 2\), so it is a solution to the equation.
The solution to the equation is \(x = -2\).
\[\begin{align}
\frac{1}{2}x - 1 &\ge 0 \\
\frac{1}{2}x &\ge 1 \\
x &\ge 2 \\
\end{align}\]
\frac{1}{2}x - 1 &\ge 0 \\
\frac{1}{2}x &\ge 1 \\
x &\ge 2 \\
\end{align}\]
This case occurs when \(x \ge 2\). Since \(\frac{1}{2}x - 1 \ge 0\), the absolute value symbol can be dropped without changing the expression.
\[\begin{align}
- \frac{3}{2}x - 1 &= \left| {\frac{1}{2}x - 1} \right| \\
- \frac{3}{2}x - 1 &= \frac{1}{2}x - 1 \\
0 &= 2x \\
0 &= x \\
\end{align}\]
- \frac{3}{2}x - 1 &= \left| {\frac{1}{2}x - 1} \right| \\
- \frac{3}{2}x - 1 &= \frac{1}{2}x - 1 \\
0 &= 2x \\
0 &= x \\
\end{align}\]
The solution \(x = 0\) is not part of the interval of interest,
\(x \ge 2\), so it is an extraneous solution, and is not a solution to the equation.
Extraneous solutions can be found by comparing the solution to the interval of interest or by verifying the solution by substitution.
Case 2: \(\frac{1}{2}x - 1 \lt 0\)
\[\begin{align}
\frac{1}{2}x - 1 &< 0 \\
\frac{1}{2}x &< 1 \\
x &< 2 \\
\end{align}\]
\frac{1}{2}x - 1 &< 0 \\
\frac{1}{2}x &< 1 \\
x &< 2 \\
\end{align}\]
This case occurs when \(x \lt 2\). Since \(\frac{1}{2}x - 1 \lt 0\), it becomes negative when the absolute value symbol is dropped.
\[\begin{align}
- \frac{3}{2}x - 1 &= \left| {\frac{1}{2}x - 1} \right| \\
- \frac{3}{2}x - 1 &= - \left( {\frac{1}{2}x - 1} \right) \\
- \frac{3}{2}x - 1 &= - \frac{1}{2}x + 1 \\
- 2 &= x \\
\end{align}\]
- \frac{3}{2}x - 1 &= \left| {\frac{1}{2}x - 1} \right| \\
- \frac{3}{2}x - 1 &= - \left( {\frac{1}{2}x - 1} \right) \\
- \frac{3}{2}x - 1 &= - \frac{1}{2}x + 1 \\
- 2 &= x \\
\end{align}\]
The solution \(x = -2\) is part of the internal of interest, \(x \lt 2\), so it is a solution to the equation.
The solution to the equation is \(x = -2\).
Verify for \(x = 0\).
This confirms \(x = 0\) is not a solution.
| Left Side | Right Side |
|---|---|
|
\[\begin{array}{r}
- \frac{3}{2}x - 1 \\ - \frac{3}{2}\left( 0 \right) - 1 \\ - 1 \end{array}\] |
\[\begin{array}{l}
\left| {\frac{1}{2}x - 1} \right| \\ \left| {\frac{1}{2}\left( 0 \right) - 1} \right| \\ \left| { - 1} \right| \\ 1 \end{array}\] |
| LS \(\ne\) RS | |
This confirms \(x = 0\) is not a solution.
Verify for \(x = -2\).
| Left Side | Right Side |
|---|---|
|
\[\begin{array}{r}
- \frac{3}{2}x - 1 \\ - \frac{3}{2}\left( { - 2} \right) - 1 \\ 2 \end{array}\] |
\[\begin{array}{l}
\left| {\frac{1}{2}x - 1} \right| \\ \left| {\frac{1}{2}\left( { - 2} \right) - 1} \right| \\ \left| { - 2} \right| \\ 2 \end{array}\] |
| LS = RS | |