Solve the Inequality for x 2-|2/5x+3|>3/5

$2 - | \frac{2}{5} x + 3 | > \frac{3}{5}$

Write $2 - | \frac{2}{5} x + 3 | > \frac{3}{5}$ as a piecewise.

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

$\frac{2}{5} x + 3 \geq 0$

Solve the inequality.

Combine $\frac{2}{5}$ and $x$ .

$\frac{2 x}{5} + 3 \geq 0$

Subtract $3$ from both sides of the inequality.

$\frac{2 x}{5} \geq - 3$

Multiply both sides by $5$ .

$\frac{2 x}{5} \cdot 5 \geq - 3 \cdot 5$

Simplify.

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{2 x}{5} \cdot 5 \geq - 3 \cdot 5$

Rewrite the expression.

$2 x \geq - 3 \cdot 5$

Simplify the right side.

Multiply $- 3$ by $5$ .

$2 x \geq - 15$

Divide each term in $2 x \geq - 15$ by $2$ and simplify.

Divide each term in $2 x \geq - 15$ by $2$ .

$\frac{2 x}{2} \geq \frac{- 15}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} \geq \frac{- 15}{2}$

Divide $x$ by $1$ .

$x \geq \frac{- 15}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x \geq - \frac{15}{2}$

In the piece where $\frac{2}{5} x + 3$ is non-negative, remove the absolute value.

$2 - (\frac{2}{5} x + 3) > \frac{3}{5}$

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

$\frac{2}{5} x + 3 < 0$

Solve the inequality.

Combine $\frac{2}{5}$ and $x$ .

$\frac{2 x}{5} + 3 < 0$

Subtract $3$ from both sides of the inequality.

$\frac{2 x}{5} < - 3$

Multiply both sides by $5$ .

$\frac{2 x}{5} \cdot 5 < - 3 \cdot 5$

Simplify.

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{2 x}{5} \cdot 5 < - 3 \cdot 5$

Rewrite the expression.

$2 x < - 3 \cdot 5$

Simplify the right side.

Multiply $- 3$ by $5$ .

$2 x < - 15$

Divide each term in $2 x < - 15$ by $2$ and simplify.

Divide each term in $2 x < - 15$ by $2$ .

$\frac{2 x}{2} < \frac{- 15}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} < \frac{- 15}{2}$

Divide $x$ by $1$ .

$x < \frac{- 15}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x < - \frac{15}{2}$

In the piece where $\frac{2}{5} x + 3$ is negative, remove the absolute value and multiply by $- 1$ .

$2 - - (\frac{2}{5} x + 3) > \frac{3}{5}$

Write as a piecewise.

${\begin{cases} 2 - (\frac{2}{5} x + 3) > \frac{3}{5} & x \geq - \frac{15}{2} \\ 2 - - (\frac{2}{5} x + 3) > \frac{3}{5} & x < - \frac{15}{2} \end{cases}$

Simplify $2 - (\frac{2}{5} x + 3) > \frac{3}{5}$ .

Simplify each term.

Combine $\frac{2}{5}$ and $x$ .

${\begin{cases} 2 - (\frac{2 x}{5} + 3) > \frac{3}{5} & x \geq - \frac{15}{2} \\ 2 - - (\frac{2}{5} x + 3) > \frac{3}{5} & x < - \frac{15}{2} \end{cases}$

Apply the distributive property.

${\begin{cases} 2 - \frac{2 x}{5} - 1 \cdot 3 > \frac{3}{5} & x \geq - \frac{15}{2} \\ 2 - - (\frac{2}{5} x + 3) > \frac{3}{5} & x < - \frac{15}{2} \end{cases}$

Multiply $- 1$ by $3$ .

${\begin{cases} 2 - \frac{2 x}{5} - 3 > \frac{3}{5} & x \geq - \frac{15}{2} \\ 2 - - (\frac{2}{5} x + 3) > \frac{3}{5} & x < - \frac{15}{2} \end{cases}$

Subtract $3$ from $2$ .

${\begin{cases} - \frac{2 x}{5} - 1 > \frac{3}{5} & x \geq - \frac{15}{2} \\ 2 - - (\frac{2}{5} x + 3) > \frac{3}{5} & x < - \frac{15}{2} \end{cases}$

Simplify $2 - - (\frac{2}{5} x + 3) > \frac{3}{5}$ .

Simplify each term.

Combine $\frac{2}{5}$ and $x$ .

${\begin{cases} - \frac{2 x}{5} - 1 > \frac{3}{5} & x \geq - \frac{15}{2} \\ 2 - - (\frac{2 x}{5} + 3) > \frac{3}{5} & x < - \frac{15}{2} \end{cases}$

Apply the distributive property.

${\begin{cases} - \frac{2 x}{5} - 1 > \frac{3}{5} & x \geq - \frac{15}{2} \\ 2 - (- \frac{2 x}{5} - 1 \cdot 3) > \frac{3}{5} & x < - \frac{15}{2} \end{cases}$

Multiply $- 1$ by $3$ .

${\begin{cases} - \frac{2 x}{5} - 1 > \frac{3}{5} & x \geq - \frac{15}{2} \\ 2 - (- \frac{2 x}{5} - 3) > \frac{3}{5} & x < - \frac{15}{2} \end{cases}$

Apply the distributive property.

${\begin{cases} - \frac{2 x}{5} - 1 > \frac{3}{5} & x \geq - \frac{15}{2} \\ 2 - - \frac{2 x}{5} - - 3 > \frac{3}{5} & x < - \frac{15}{2} \end{cases}$

Multiply $- - \frac{2 x}{5}$ .

Multiply $- 1$ by $- 1$ .

${\begin{cases} - \frac{2 x}{5} - 1 > \frac{3}{5} & x \geq - \frac{15}{2} \\ 2 + 1 \frac{2 x}{5} - - 3 > \frac{3}{5} & x < - \frac{15}{2} \end{cases}$

Multiply $\frac{2 x}{5}$ by $1$ .

${\begin{cases} - \frac{2 x}{5} - 1 > \frac{3}{5} & x \geq - \frac{15}{2} \\ 2 + \frac{2 x}{5} - - 3 > \frac{3}{5} & x < - \frac{15}{2} \end{cases}$

Multiply $- 1$ by $- 3$ .

${\begin{cases} - \frac{2 x}{5} - 1 > \frac{3}{5} & x \geq - \frac{15}{2} \\ 2 + \frac{2 x}{5} + 3 > \frac{3}{5} & x < - \frac{15}{2} \end{cases}$

Add $2$ and $3$ .

${\begin{cases} - \frac{2 x}{5} - 1 > \frac{3}{5} & x \geq - \frac{15}{2} \\ \frac{2 x}{5} + 5 > \frac{3}{5} & x < - \frac{15}{2} \end{cases}$

Solve $- \frac{2 x}{5} - 1 > \frac{3}{5}$ when $x \geq - \frac{15}{2}$ .

Solve $- \frac{2 x}{5} - 1 > \frac{3}{5}$ for $x$ .

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

Add $1$ to both sides of the inequality.

$- \frac{2 x}{5} > \frac{3}{5} + 1$

Write $1$ as a fraction with a common denominator.

$- \frac{2 x}{5} > \frac{3}{5} + \frac{5}{5}$

Combine the numerators over the common denominator.

$- \frac{2 x}{5} > \frac{3 + 5}{5}$

Add $3$ and $5$ .

$- \frac{2 x}{5} > \frac{8}{5}$

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$- 2 x > 8$

Divide each term in $- 2 x > 8$ by $- 2$ and simplify.

Divide each term in $- 2 x > 8$ by $- 2$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 2 x}{- 2} < \frac{8}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} < \frac{8}{- 2}$

Divide $x$ by $1$ .

$x < \frac{8}{- 2}$

Simplify the right side.

Divide $8$ by $- 2$ .

$x < - 4$

Find the intersection of $x < - 4$ and $x \geq - \frac{15}{2}$ .

$- \frac{15}{2} \leq x < - 4$

Solve $\frac{2 x}{5} + 5 > \frac{3}{5}$ when $x < - \frac{15}{2}$ .

Solve $\frac{2 x}{5} + 5 > \frac{3}{5}$ for $x$ .

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

Subtract $5$ from both sides of the inequality.

$\frac{2 x}{5} > \frac{3}{5} - 5$

To write $- 5$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$\frac{2 x}{5} > \frac{3}{5} - 5 \cdot \frac{5}{5}$

Combine $- 5$ and $\frac{5}{5}$ .

$\frac{2 x}{5} > \frac{3}{5} + \frac{- 5 \cdot 5}{5}$

Combine the numerators over the common denominator.

$\frac{2 x}{5} > \frac{3 - 5 \cdot 5}{5}$

Simplify the numerator.

Multiply $- 5$ by $5$ .

$\frac{2 x}{5} > \frac{3 - 25}{5}$

Subtract $25$ from $3$ .

$\frac{2 x}{5} > \frac{- 22}{5}$

Move the negative in front of the fraction.

$\frac{2 x}{5} > - \frac{22}{5}$

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$2 x > - 22$

Divide each term in $2 x > - 22$ by $2$ and simplify.

Divide each term in $2 x > - 22$ by $2$ .

$\frac{2 x}{2} > \frac{- 22}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} > \frac{- 22}{2}$

Divide $x$ by $1$ .

$x > \frac{- 22}{2}$

Simplify the right side.

Divide $- 22$ by $2$ .

$x > - 11$

Find the intersection of $x > - 11$ and $x < - \frac{15}{2}$ .

$- 11 < x < - \frac{15}{2}$

Find the union of the solutions.

$- 11 < x < - 4$

The result can be shown in multiple forms.

Inequality Form:

$- 11 < x < - 4$

Interval Notation:

$(- 11, - 4)$

Solve the Inequality for x 2-|2/5x+3|>3/5

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