Solve the Inequality for v 8+|4v-7|>=17

$8 + | 4 v - 7 | \geq 17$

Write $8 + | 4 v - 7 | \geq 17$ as a piecewise.

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

$4 v - 7 \geq 0$

Solve the inequality.

Add $7$ to both sides of the inequality.

$4 v \geq 7$

Divide each term in $4 v \geq 7$ by $4$ and simplify.

Divide each term in $4 v \geq 7$ by $4$ .

$\frac{4 v}{4} \geq \frac{7}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 v}{4} \geq \frac{7}{4}$

Divide $v$ by $1$ .

$v \geq \frac{7}{4}$

In the piece where $4 v - 7$ is non-negative, remove the absolute value.

$8 + 4 v - 7 \geq 17$

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

$4 v - 7 < 0$

Solve the inequality.

Add $7$ to both sides of the inequality.

$4 v < 7$

Divide each term in $4 v < 7$ by $4$ and simplify.

Divide each term in $4 v < 7$ by $4$ .

$\frac{4 v}{4} < \frac{7}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 v}{4} < \frac{7}{4}$

Divide $v$ by $1$ .

$v < \frac{7}{4}$

In the piece where $4 v - 7$ is negative, remove the absolute value and multiply by $- 1$ .

$8 - (4 v - 7) \geq 17$

Write as a piecewise.

${\begin{cases} 8 + 4 v - 7 \geq 17 & v \geq \frac{7}{4} \\ 8 - (4 v - 7) \geq 17 & v < \frac{7}{4} \end{cases}$

Subtract $7$ from $8$ .

${\begin{cases} 4 v + 1 \geq 17 & v \geq \frac{7}{4} \\ 8 - (4 v - 7) \geq 17 & v < \frac{7}{4} \end{cases}$

Simplify $8 - (4 v - 7) \geq 17$ .

Simplify each term.

Apply the distributive property.

${\begin{cases} 4 v + 1 \geq 17 & v \geq \frac{7}{4} \\ 8 - (4 v) - - 7 \geq 17 & v < \frac{7}{4} \end{cases}$

Multiply $4$ by $- 1$ .

${\begin{cases} 4 v + 1 \geq 17 & v \geq \frac{7}{4} \\ 8 - 4 v - - 7 \geq 17 & v < \frac{7}{4} \end{cases}$

Multiply $- 1$ by $- 7$ .

${\begin{cases} 4 v + 1 \geq 17 & v \geq \frac{7}{4} \\ 8 - 4 v + 7 \geq 17 & v < \frac{7}{4} \end{cases}$

Add $8$ and $7$ .

${\begin{cases} 4 v + 1 \geq 17 & v \geq \frac{7}{4} \\ - 4 v + 15 \geq 17 & v < \frac{7}{4} \end{cases}$

Solve $4 v + 1 \geq 17$ for $v$ .

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

Subtract $1$ from both sides of the inequality.

$4 v \geq 17 - 1$

Subtract $1$ from $17$ .

$4 v \geq 16$

Divide each term in $4 v \geq 16$ by $4$ and simplify.

Divide each term in $4 v \geq 16$ by $4$ .

$\frac{4 v}{4} \geq \frac{16}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 v}{4} \geq \frac{16}{4}$

Divide $v$ by $1$ .

$v \geq \frac{16}{4}$

Simplify the right side.

Divide $16$ by $4$ .

$v \geq 4$

Solve $- 4 v + 15 \geq 17$ for $v$ .

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

Subtract $15$ from both sides of the inequality.

$- 4 v \geq 17 - 15$

Subtract $15$ from $17$ .

$- 4 v \geq 2$

Divide each term in $- 4 v \geq 2$ by $- 4$ and simplify.

Divide each term in $- 4 v \geq 2$ by $- 4$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 4 v}{- 4} \leq \frac{2}{- 4}$

Simplify the left side.

Cancel the common factor of $- 4$ .

Cancel the common factor.

$\frac{- 4 v}{- 4} \leq \frac{2}{- 4}$

Divide $v$ by $1$ .

$v \leq \frac{2}{- 4}$

Simplify the right side.

Cancel the common factor of $2$ and $- 4$ .

Factor $2$ out of $2$ .

$v \leq \frac{2 (1)}{- 4}$

Cancel the common factors.

Factor $2$ out of $- 4$ .

$v \leq \frac{2 \cdot 1}{2 \cdot - 2}$

Cancel the common factor.

$v \leq \frac{2 \cdot 1}{2 \cdot - 2}$

Rewrite the expression.

$v \leq \frac{1}{- 2}$

Move the negative in front of the fraction.

$v \leq - \frac{1}{2}$

Find the union of the solutions.

$v \leq - \frac{1}{2}$ or $v \geq 4$

The result can be shown in multiple forms.

Inequality Form:

$v \leq - \frac{1}{2} or v \geq 4$

Interval Notation:

$(- \infty, - \frac{1}{2}] \cup [4, \infty)$

Solve the Inequality for v 8+|4v-7|>=17

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