Solve the Inequality for c |2c-1|<=7

$| 2 c - 1 | \leq 7$

Step 1

Write $| 2 c - 1 | \leq 7$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$2 c - 1 \geq 0$

Solve the inequality.

Add $1$ to both sides of the inequality.

$2 c \geq 1$

Divide each term in $2 c \geq 1$ by $2$ and simplify.

Divide each term in $2 c \geq 1$ by $2$ .

$\frac{2 c}{2} \geq \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 c}{2} \geq \frac{1}{2}$

Divide $c$ by $1$ .

$c \geq \frac{1}{2}$

In the piece where $2 c - 1$ is non-negative, remove the absolute value.

$2 c - 1 \leq 7$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$2 c - 1 < 0$

Solve the inequality.

Add $1$ to both sides of the inequality.

$2 c < 1$

Divide each term in $2 c < 1$ by $2$ and simplify.

Divide each term in $2 c < 1$ by $2$ .

$\frac{2 c}{2} < \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 c}{2} < \frac{1}{2}$

Divide $c$ by $1$ .

$c < \frac{1}{2}$

In the piece where $2 c - 1$ is negative, remove the absolute value and multiply by $- 1$ .

$- (2 c - 1) \leq 7$

Write as a piecewise.

${\begin{cases} 2 c - 1 \leq 7 & c \geq \frac{1}{2} \\ - (2 c - 1) \leq 7 & c < \frac{1}{2} \end{cases}$

Simplify $- (2 c - 1) \leq 7$ .

Apply the distributive property.

${\begin{cases} 2 c - 1 \leq 7 & c \geq \frac{1}{2} \\ - (2 c) - - 1 \leq 7 & c < \frac{1}{2} \end{cases}$

Multiply $2$ by $- 1$ .

${\begin{cases} 2 c - 1 \leq 7 & c \geq \frac{1}{2} \\ - 2 c - - 1 \leq 7 & c < \frac{1}{2} \end{cases}$

Multiply $- 1$ by $- 1$ .

${\begin{cases} 2 c - 1 \leq 7 & c \geq \frac{1}{2} \\ - 2 c + 1 \leq 7 & c < \frac{1}{2} \end{cases}$

Step 2

Solve $2 c - 1 \leq 7$ when $c \geq \frac{1}{2}$ .

Solve $2 c - 1 \leq 7$ for $c$ .

Move all terms not containing $c$ to the right side of the inequality.

Add $1$ to both sides of the inequality.

$2 c \leq 7 + 1$

Add $7$ and $1$ .

$2 c \leq 8$

Divide each term in $2 c \leq 8$ by $2$ and simplify.

Divide each term in $2 c \leq 8$ by $2$ .

$\frac{2 c}{2} \leq \frac{8}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 c}{2} \leq \frac{8}{2}$

Divide $c$ by $1$ .

$c \leq \frac{8}{2}$

Simplify the right side.

Divide $8$ by $2$ .

$c \leq 4$

Find the intersection of $c \leq 4$ and $c \geq \frac{1}{2}$ .

$\frac{1}{2} \leq c \leq 4$

Step 3

Solve $- 2 c + 1 \leq 7$ when $c < \frac{1}{2}$ .

Solve $- 2 c + 1 \leq 7$ for $c$ .

Move all terms not containing $c$ to the right side of the inequality.

Subtract $1$ from both sides of the inequality.

$- 2 c \leq 7 - 1$

Subtract $1$ from $7$ .

$- 2 c \leq 6$

Divide each term in $- 2 c \leq 6$ by $- 2$ and simplify.

Divide each term in $- 2 c \leq 6$ by $- 2$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 2 c}{- 2} \geq \frac{6}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 c}{- 2} \geq \frac{6}{- 2}$

Divide $c$ by $1$ .

$c \geq \frac{6}{- 2}$

Simplify the right side.

Divide $6$ by $- 2$ .

$c \geq - 3$

Find the intersection of $c \geq - 3$ and $c < \frac{1}{2}$ .

$- 3 \leq c < \frac{1}{2}$

Step 4

Find the union of the solutions.

$- 3 \leq c \leq 4$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$- 3 \leq c \leq 4$

Interval Notation:

$[- 3, 4]$

Step 6

Solve the Inequality for c |2c-1|<=7

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