Solve for x |(2x)/39-1|<=5

$| \frac{2 x}{39} - 1 | \leq 5$

Step 1

Write $| \frac{2 x}{39} - 1 | \leq 5$ as a piecewise.

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

$\frac{2 x}{39} - 1 \geq 0$

Solve the inequality.

Add $1$ to both sides of the inequality.

$\frac{2 x}{39} \geq 1$

Multiply both sides by $39$ .

$\frac{2 x}{39} \cdot 39 \geq 1 \cdot 39$

Simplify.

Simplify the left side.

Cancel the common factor of $39$ .

Cancel the common factor.

$\frac{2 x}{39} \cdot 39 \geq 1 \cdot 39$

Rewrite the expression.

$2 x \geq 1 \cdot 39$

Simplify the right side.

Multiply $39$ by $1$ .

$2 x \geq 39$

Divide each term in $2 x \geq 39$ by $2$ and simplify.

Divide each term in $2 x \geq 39$ by $2$ .

$\frac{2 x}{2} \geq \frac{39}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} \geq \frac{39}{2}$

Divide $x$ by $1$ .

$x \geq \frac{39}{2}$

In the piece where $\frac{2 x}{39} - 1$ is non-negative, remove the absolute value.

$\frac{2 x}{39} - 1 \leq 5$

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

$\frac{2 x}{39} - 1 < 0$

Solve the inequality.

Add $1$ to both sides of the inequality.

$\frac{2 x}{39} < 1$

Multiply both sides by $39$ .

$\frac{2 x}{39} \cdot 39 < 1 \cdot 39$

Simplify.

Simplify the left side.

Cancel the common factor of $39$ .

Cancel the common factor.

$\frac{2 x}{39} \cdot 39 < 1 \cdot 39$

Rewrite the expression.

$2 x < 1 \cdot 39$

Simplify the right side.

Multiply $39$ by $1$ .

$2 x < 39$

Divide each term in $2 x < 39$ by $2$ and simplify.

Divide each term in $2 x < 39$ by $2$ .

$\frac{2 x}{2} < \frac{39}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} < \frac{39}{2}$

Divide $x$ by $1$ .

$x < \frac{39}{2}$

In the piece where $\frac{2 x}{39} - 1$ is negative, remove the absolute value and multiply by $- 1$ .

$- (\frac{2 x}{39} - 1) \leq 5$

Write as a piecewise.

${\begin{cases} \frac{2 x}{39} - 1 \leq 5 & x \geq \frac{39}{2} \\ - (\frac{2 x}{39} - 1) \leq 5 & x < \frac{39}{2} \end{cases}$

Simplify $- (\frac{2 x}{39} - 1) \leq 5$ .

Apply the distributive property.

${\begin{cases} \frac{2 x}{39} - 1 \leq 5 & x \geq \frac{39}{2} \\ - \frac{2 x}{39} - - 1 \leq 5 & x < \frac{39}{2} \end{cases}$

Multiply $- 1$ by $- 1$ .

${\begin{cases} \frac{2 x}{39} - 1 \leq 5 & x \geq \frac{39}{2} \\ - \frac{2 x}{39} + 1 \leq 5 & x < \frac{39}{2} \end{cases}$

Step 2

Solve $\frac{2 x}{39} - 1 \leq 5$ when $x \geq \frac{39}{2}$ .

Solve $\frac{2 x}{39} - 1 \leq 5$ for $x$ .

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

Add $1$ to both sides of the inequality.

$\frac{2 x}{39} \leq 5 + 1$

Add $5$ and $1$ .

$\frac{2 x}{39} \leq 6$

Multiply both sides by $39$ .

$\frac{2 x}{39} \cdot 39 \leq 6 \cdot 39$

Simplify.

Simplify the left side.

Cancel the common factor of $39$ .

Cancel the common factor.

$\frac{2 x}{39} \cdot 39 \leq 6 \cdot 39$

Rewrite the expression.

$2 x \leq 6 \cdot 39$

Simplify the right side.

Multiply $6$ by $39$ .

$2 x \leq 234$

Divide each term in $2 x \leq 234$ by $2$ and simplify.

Divide each term in $2 x \leq 234$ by $2$ .

$\frac{2 x}{2} \leq \frac{234}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} \leq \frac{234}{2}$

Divide $x$ by $1$ .

$x \leq \frac{234}{2}$

Simplify the right side.

Divide $234$ by $2$ .

$x \leq 117$

Find the intersection of $x \leq 117$ and $x \geq \frac{39}{2}$ .

$\frac{39}{2} \leq x \leq 117$

Step 3

Solve $- \frac{2 x}{39} + 1 \leq 5$ when $x < \frac{39}{2}$ .

Solve $- \frac{2 x}{39} + 1 \leq 5$ for $x$ .

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

Subtract $1$ from both sides of the inequality.

$- \frac{2 x}{39} \leq 5 - 1$

Subtract $1$ from $5$ .

$- \frac{2 x}{39} \leq 4$

Divide each term in $- \frac{2 x}{39} \leq 4$ by $- 1$ and simplify.

Divide each term in $- \frac{2 x}{39} \leq 4$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- \frac{2 x}{39}}{- 1} \geq \frac{4}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{\frac{2 x}{39}}{1} \geq \frac{4}{- 1}$

Divide $\frac{2 x}{39}$ by $1$ .

$\frac{2 x}{39} \geq \frac{4}{- 1}$

Simplify the right side.

Divide $4$ by $- 1$ .

$\frac{2 x}{39} \geq - 4$

Multiply both sides by $39$ .

$\frac{2 x}{39} \cdot 39 \geq - 4 \cdot 39$

Simplify.

Simplify the left side.

Cancel the common factor of $39$ .

Cancel the common factor.

$\frac{2 x}{39} \cdot 39 \geq - 4 \cdot 39$

Rewrite the expression.

$2 x \geq - 4 \cdot 39$

Simplify the right side.

Multiply $- 4$ by $39$ .

$2 x \geq - 156$

Divide each term in $2 x \geq - 156$ by $2$ and simplify.

Divide each term in $2 x \geq - 156$ by $2$ .

$\frac{2 x}{2} \geq \frac{- 156}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} \geq \frac{- 156}{2}$

Divide $x$ by $1$ .

$x \geq \frac{- 156}{2}$

Simplify the right side.

Divide $- 156$ by $2$ .

$x \geq - 78$

Find the intersection of $x \geq - 78$ and $x < \frac{39}{2}$ .

$- 78 \leq x < \frac{39}{2}$

Step 4

Find the union of the solutions.

$- 78 \leq x \leq 117$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$- 78 \leq x \leq 117$

Interval Notation:

$[- 78, 117]$

Step 6

Solve for x |(2x)/39-1|<=5

Related Questions

Support LCMGCF.com by disabling your adblocker.