Solve for v 2/3|4v+6|-2<=10

$\frac{2}{3} | 4 v + 6 | - 2 \leq 10$

Step 1

Write $\frac{2}{3} \cdot | 4 v + 6 | - 2 \leq 10$ as a piecewise.

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

$4 v + 6 \geq 0$

Solve the inequality.

Subtract $6$ from both sides of the inequality.

$4 v \geq - 6$

Divide each term in $4 v \geq - 6$ by $4$ and simplify.

Divide each term in $4 v \geq - 6$ by $4$ .

$\frac{4 v}{4} \geq \frac{- 6}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 v}{4} \geq \frac{- 6}{4}$

Divide $v$ by $1$ .

$v \geq \frac{- 6}{4}$

Simplify the right side.

Cancel the common factor of $- 6$ and $4$ .

Factor $2$ out of $- 6$ .

$v \geq \frac{2 (- 3)}{4}$

Cancel the common factors.

Factor $2$ out of $4$ .

$v \geq \frac{2 \cdot - 3}{2 \cdot 2}$

Cancel the common factor.

$v \geq \frac{2 \cdot - 3}{2 \cdot 2}$

Rewrite the expression.

$v \geq \frac{- 3}{2}$

Move the negative in front of the fraction.

$v \geq - \frac{3}{2}$

In the piece where $4 v + 6$ is non-negative, remove the absolute value.

$\frac{2}{3} \cdot (4 v + 6) - 2 \leq 10$

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

$4 v + 6 < 0$

Solve the inequality.

Subtract $6$ from both sides of the inequality.

$4 v < - 6$

Divide each term in $4 v < - 6$ by $4$ and simplify.

Divide each term in $4 v < - 6$ by $4$ .

$\frac{4 v}{4} < \frac{- 6}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 v}{4} < \frac{- 6}{4}$

Divide $v$ by $1$ .

$v < \frac{- 6}{4}$

Simplify the right side.

Cancel the common factor of $- 6$ and $4$ .

Factor $2$ out of $- 6$ .

$v < \frac{2 (- 3)}{4}$

Cancel the common factors.

Factor $2$ out of $4$ .

$v < \frac{2 \cdot - 3}{2 \cdot 2}$

Cancel the common factor.

$v < \frac{2 \cdot - 3}{2 \cdot 2}$

Rewrite the expression.

$v < \frac{- 3}{2}$

Move the negative in front of the fraction.

$v < - \frac{3}{2}$

In the piece where $4 v + 6$ is negative, remove the absolute value and multiply by $- 1$ .

$\frac{2}{3} \cdot (- (4 v + 6)) - 2 \leq 10$

Write as a piecewise.

${\begin{cases} \frac{2}{3} \cdot (4 v + 6) - 2 \leq 10 & v \geq - \frac{3}{2} \\ \frac{2}{3} \cdot (- (4 v + 6)) - 2 \leq 10 & v < - \frac{3}{2} \end{cases}$

Simplify $\frac{2}{3} \cdot (4 v + 6) - 2 \leq 10$ .

Simplify each term.

Apply the distributive property.

${\begin{cases} \frac{2}{3} (4 v) + \frac{2}{3} \cdot 6 - 2 \leq 10 & v \geq - \frac{3}{2} \\ \frac{2}{3} \cdot (- (4 v + 6)) - 2 \leq 10 & v < - \frac{3}{2} \end{cases}$

Multiply $\frac{2}{3} (4 v)$ .

Combine $4$ and $\frac{2}{3}$ .

${\begin{cases} \frac{4 \cdot 2}{3} v + \frac{2}{3} \cdot 6 - 2 \leq 10 & v \geq - \frac{3}{2} \\ \frac{2}{3} \cdot (- (4 v + 6)) - 2 \leq 10 & v < - \frac{3}{2} \end{cases}$

Multiply $4$ by $2$ .

${\begin{cases} \frac{8}{3} v + \frac{2}{3} \cdot 6 - 2 \leq 10 & v \geq - \frac{3}{2} \\ \frac{2}{3} \cdot (- (4 v + 6)) - 2 \leq 10 & v < - \frac{3}{2} \end{cases}$

Combine $\frac{8}{3}$ and $v$ .

${\begin{cases} \frac{8 v}{3} + \frac{2}{3} \cdot 6 - 2 \leq 10 & v \geq - \frac{3}{2} \\ \frac{2}{3} \cdot (- (4 v + 6)) - 2 \leq 10 & v < - \frac{3}{2} \end{cases}$

Cancel the common factor of $3$ .

Factor $3$ out of $6$ .

${\begin{cases} \frac{8 v}{3} + \frac{2}{3} \cdot (3 (2)) - 2 \leq 10 & v \geq - \frac{3}{2} \\ \frac{2}{3} \cdot (- (4 v + 6)) - 2 \leq 10 & v < - \frac{3}{2} \end{cases}$

Cancel the common factor.

${\begin{cases} \frac{8 v}{3} + \frac{2}{3} \cdot (3 \cdot 2) - 2 \leq 10 & v \geq - \frac{3}{2} \\ \frac{2}{3} \cdot (- (4 v + 6)) - 2 \leq 10 & v < - \frac{3}{2} \end{cases}$

Rewrite the expression.

${\begin{cases} \frac{8 v}{3} + 2 \cdot 2 - 2 \leq 10 & v \geq - \frac{3}{2} \\ \frac{2}{3} \cdot (- (4 v + 6)) - 2 \leq 10 & v < - \frac{3}{2} \end{cases}$

Multiply $2$ by $2$ .

${\begin{cases} \frac{8 v}{3} + 4 - 2 \leq 10 & v \geq - \frac{3}{2} \\ \frac{2}{3} \cdot (- (4 v + 6)) - 2 \leq 10 & v < - \frac{3}{2} \end{cases}$

Subtract $2$ from $4$ .

${\begin{cases} \frac{8 v}{3} + 2 \leq 10 & v \geq - \frac{3}{2} \\ \frac{2}{3} \cdot (- (4 v + 6)) - 2 \leq 10 & v < - \frac{3}{2} \end{cases}$

Simplify $\frac{2}{3} \cdot (- (4 v + 6)) - 2 \leq 10$ .

Simplify each term.

Apply the distributive property.

${\begin{cases} \frac{8 v}{3} + 2 \leq 10 & v \geq - \frac{3}{2} \\ \frac{2}{3} \cdot (- (4 v) - 1 \cdot 6) - 2 \leq 10 & v < - \frac{3}{2} \end{cases}$

Multiply $4$ by $- 1$ .

${\begin{cases} \frac{8 v}{3} + 2 \leq 10 & v \geq - \frac{3}{2} \\ \frac{2}{3} \cdot (- 4 v - 1 \cdot 6) - 2 \leq 10 & v < - \frac{3}{2} \end{cases}$

Multiply $- 1$ by $6$ .

${\begin{cases} \frac{8 v}{3} + 2 \leq 10 & v \geq - \frac{3}{2} \\ \frac{2}{3} \cdot (- 4 v - 6) - 2 \leq 10 & v < - \frac{3}{2} \end{cases}$

Apply the distributive property.

${\begin{cases} \frac{8 v}{3} + 2 \leq 10 & v \geq - \frac{3}{2} \\ \frac{2}{3} (- 4 v) + \frac{2}{3} \cdot - 6 - 2 \leq 10 & v < - \frac{3}{2} \end{cases}$

Multiply $\frac{2}{3} (- 4 v)$ .

Combine $- 4$ and $\frac{2}{3}$ .

${\begin{cases} \frac{8 v}{3} + 2 \leq 10 & v \geq - \frac{3}{2} \\ \frac{- 4 \cdot 2}{3} v + \frac{2}{3} \cdot - 6 - 2 \leq 10 & v < - \frac{3}{2} \end{cases}$

Multiply $- 4$ by $2$ .

${\begin{cases} \frac{8 v}{3} + 2 \leq 10 & v \geq - \frac{3}{2} \\ \frac{- 8}{3} v + \frac{2}{3} \cdot - 6 - 2 \leq 10 & v < - \frac{3}{2} \end{cases}$

Combine $\frac{- 8}{3}$ and $v$ .

${\begin{cases} \frac{8 v}{3} + 2 \leq 10 & v \geq - \frac{3}{2} \\ \frac{- 8 v}{3} + \frac{2}{3} \cdot - 6 - 2 \leq 10 & v < - \frac{3}{2} \end{cases}$

Cancel the common factor of $3$ .

Factor $3$ out of $- 6$ .

${\begin{cases} \frac{8 v}{3} + 2 \leq 10 & v \geq - \frac{3}{2} \\ \frac{- 8 v}{3} + \frac{2}{3} \cdot (3 (- 2)) - 2 \leq 10 & v < - \frac{3}{2} \end{cases}$

Cancel the common factor.

${\begin{cases} \frac{8 v}{3} + 2 \leq 10 & v \geq - \frac{3}{2} \\ \frac{- 8 v}{3} + \frac{2}{3} \cdot (3 \cdot - 2) - 2 \leq 10 & v < - \frac{3}{2} \end{cases}$

Rewrite the expression.

${\begin{cases} \frac{8 v}{3} + 2 \leq 10 & v \geq - \frac{3}{2} \\ \frac{- 8 v}{3} + 2 \cdot - 2 - 2 \leq 10 & v < - \frac{3}{2} \end{cases}$

Multiply $2$ by $- 2$ .

${\begin{cases} \frac{8 v}{3} + 2 \leq 10 & v \geq - \frac{3}{2} \\ \frac{- 8 v}{3} - 4 - 2 \leq 10 & v < - \frac{3}{2} \end{cases}$

Move the negative in front of the fraction.

${\begin{cases} \frac{8 v}{3} + 2 \leq 10 & v \geq - \frac{3}{2} \\ - \frac{8 v}{3} - 4 - 2 \leq 10 & v < - \frac{3}{2} \end{cases}$

Subtract $2$ from $- 4$ .

${\begin{cases} \frac{8 v}{3} + 2 \leq 10 & v \geq - \frac{3}{2} \\ - \frac{8 v}{3} - 6 \leq 10 & v < - \frac{3}{2} \end{cases}$

Step 2

Solve $\frac{8 v}{3} + 2 \leq 10$ when $v \geq - \frac{3}{2}$ .

Solve $\frac{8 v}{3} + 2 \leq 10$ for $v$ .

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

Subtract $2$ from both sides of the inequality.

$\frac{8 v}{3} \leq 10 - 2$

Subtract $2$ from $10$ .

$\frac{8 v}{3} \leq 8$

Multiply both sides by $3$ .

$\frac{8 v}{3} \cdot 3 \leq 8 \cdot 3$

Simplify.

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{8 v}{3} \cdot 3 \leq 8 \cdot 3$

Rewrite the expression.

$8 v \leq 8 \cdot 3$

Simplify the right side.

Multiply $8$ by $3$ .

$8 v \leq 24$

Divide each term in $8 v \leq 24$ by $8$ and simplify.

Divide each term in $8 v \leq 24$ by $8$ .

$\frac{8 v}{8} \leq \frac{24}{8}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 v}{8} \leq \frac{24}{8}$

Divide $v$ by $1$ .

$v \leq \frac{24}{8}$

Simplify the right side.

Divide $24$ by $8$ .

$v \leq 3$

Find the intersection of $v \leq 3$ and $v \geq - \frac{3}{2}$ .

$- \frac{3}{2} \leq v \leq 3$

Step 3

Solve $- \frac{8 v}{3} - 6 \leq 10$ when $v < - \frac{3}{2}$ .

Solve $- \frac{8 v}{3} - 6 \leq 10$ for $v$ .

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

Add $6$ to both sides of the inequality.

$- \frac{8 v}{3} \leq 10 + 6$

Add $10$ and $6$ .

$- \frac{8 v}{3} \leq 16$

Divide each term in $- \frac{8 v}{3} \leq 16$ by $- 1$ and simplify.

Divide each term in $- \frac{8 v}{3} \leq 16$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- \frac{8 v}{3}}{- 1} \geq \frac{16}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{\frac{8 v}{3}}{1} \geq \frac{16}{- 1}$

Divide $\frac{8 v}{3}$ by $1$ .

$\frac{8 v}{3} \geq \frac{16}{- 1}$

Simplify the right side.

Divide $16$ by $- 1$ .

$\frac{8 v}{3} \geq - 16$

Multiply both sides by $3$ .

$\frac{8 v}{3} \cdot 3 \geq - 16 \cdot 3$

Simplify.

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{8 v}{3} \cdot 3 \geq - 16 \cdot 3$

Rewrite the expression.

$8 v \geq - 16 \cdot 3$

Simplify the right side.

Multiply $- 16$ by $3$ .

$8 v \geq - 48$

Divide each term in $8 v \geq - 48$ by $8$ and simplify.

Divide each term in $8 v \geq - 48$ by $8$ .

$\frac{8 v}{8} \geq \frac{- 48}{8}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 v}{8} \geq \frac{- 48}{8}$

Divide $v$ by $1$ .

$v \geq \frac{- 48}{8}$

Simplify the right side.

Divide $- 48$ by $8$ .

$v \geq - 6$

Find the intersection of $v \geq - 6$ and $v < - \frac{3}{2}$ .

$- 6 \leq v < - \frac{3}{2}$

Step 4

Find the union of the solutions.

$- 6 \leq v \leq 3$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$- 6 \leq v \leq 3$

Interval Notation:

$[- 6, 3]$

Step 6

Solve for v 2/3|4v+6|-2<=10

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