Example 2
Completion requirements
| Example 2 |
Multimedia |
Solve \(\left| 2x + 5\right| = 11\).
Case 1: \(\left|2x + 5\right| = 11\)
\(\begin{align}
2x + 5 &\ge 0 \\
2x &\ge - 5 \\
x &\ge - \frac{5}{2} \\
\end{align}\)
This case occurs when \(x \ge -\frac{5}{2}\). Since \(2x + 5 \ge 0\), the absolute value symbol can be dropped without changing the expression.
\(\begin{align}
\left| {2x + 5} \right| &= 11 \\
2x + 5 &= 11 \\
2x &= 6 \\
x &= 3 \\
\end{align}\)
The solution \(x = 3\) is part of the interval of interest, \(x \ge -\frac{5}{2}\), so it is a solution to the equation.
Case 2: \(2x + 5 \lt 0\)
\(\begin{align}
2x + 5 &< 0 \\
2x &< - 5 \\
x &< -\frac{5}{2} \\
\end{align}\)
This case occurs when \(x \lt -\frac{5}{2}\). Since \(2x + 5 \lt 0\), it becomes negative when the absolute value symbol is dropped.
\(\begin{align}
\left| {2x + 5} \right| &= 11 \\
- \left( {2x + 5} \right) &= 11 \\
- 2x - 5 &= 11 \\
- 2x &= 16 \\
x &= -8 \end{align}\)
The solution \(x = -8\) is part of the internal of interest, \(x \lt -\frac{5}{2}\), so it is a solution to the equation.
The solutions to the equation are \(-8\) and \(3\).
\(\begin{align}
2x + 5 &\ge 0 \\
2x &\ge - 5 \\
x &\ge - \frac{5}{2} \\
\end{align}\)
This case occurs when \(x \ge -\frac{5}{2}\). Since \(2x + 5 \ge 0\), the absolute value symbol can be dropped without changing the expression.
\(\begin{align}
\left| {2x + 5} \right| &= 11 \\
2x + 5 &= 11 \\
2x &= 6 \\
x &= 3 \\
\end{align}\)
The solution \(x = 3\) is part of the interval of interest, \(x \ge -\frac{5}{2}\), so it is a solution to the equation.
Case 2: \(2x + 5 \lt 0\)
\(\begin{align}
2x + 5 &< 0 \\
2x &< - 5 \\
x &< -\frac{5}{2} \\
\end{align}\)
This case occurs when \(x \lt -\frac{5}{2}\). Since \(2x + 5 \lt 0\), it becomes negative when the absolute value symbol is dropped.
\(\begin{align}
\left| {2x + 5} \right| &= 11 \\
- \left( {2x + 5} \right) &= 11 \\
- 2x - 5 &= 11 \\
- 2x &= 16 \\
x &= -8 \end{align}\)
The solution \(x = -8\) is part of the internal of interest, \(x \lt -\frac{5}{2}\), so it is a solution to the equation.
The solutions to the equation are \(-8\) and \(3\).
Verify for \(x = -8\).
| Left Side | Right Side |
|---|---|
| \(\begin{array}{r} \left| {2x + 5} \right| \\ \left| {2\left( { - 8} \right) + 5} \right| \\ \left| { - 11} \right| \\ 11 \\ \end{array}\) |
\(11\) |
| LS = RS | |
Verify for \(x = 3\).
| Left Side | Right Side |
|---|---|
| \(\begin{array}{r} \left| {2x + 5} \right| \\ \left| {2\left( { 3} \right) + 5} \right| \\ \left| { 11} \right| \\ 11 \\ \end{array}\) |
\(11\) |
| LS = RS | |