Solve for f 6|-f+3|+7>7

$6 | - f + 3 | + 7 > 7$

Write $6 | - f + 3 | + 7 > 7$ as a piecewise.

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

$- f + 3 \geq 0$

Solve the inequality.

Subtract $3$ from both sides of the inequality.

$- f \geq - 3$

Divide each term in $- f \geq - 3$ by $- 1$ and simplify.

Divide each term in $- f \geq - 3$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- f}{- 1} \leq \frac{- 3}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{f}{1} \leq \frac{- 3}{- 1}$

Divide $f$ by $1$ .

$f \leq \frac{- 3}{- 1}$

Simplify the right side.

Divide $- 3$ by $- 1$ .

$f \leq 3$

In the piece where $- f + 3$ is non-negative, remove the absolute value.

$6 (- f + 3) + 7 > 7$

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

$- f + 3 < 0$

Solve the inequality.

Subtract $3$ from both sides of the inequality.

$- f < - 3$

Divide each term in $- f < - 3$ by $- 1$ and simplify.

Divide each term in $- f < - 3$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- f}{- 1} > \frac{- 3}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{f}{1} > \frac{- 3}{- 1}$

Divide $f$ by $1$ .

$f > \frac{- 3}{- 1}$

Simplify the right side.

Divide $- 3$ by $- 1$ .

$f > 3$

In the piece where $- f + 3$ is negative, remove the absolute value and multiply by $- 1$ .

$6 (- (- f + 3)) + 7 > 7$

Write as a piecewise.

${\begin{cases} 6 (- f + 3) + 7 > 7 & f \leq 3 \\ 6 (- (- f + 3)) + 7 > 7 & f > 3 \end{cases}$

Simplify $6 (- f + 3) + 7 > 7$ .

Simplify each term.

Apply the distributive property.

${\begin{cases} 6 (- f) + 6 \cdot 3 + 7 > 7 & f \leq 3 \\ 6 (- (- f + 3)) + 7 > 7 & f > 3 \end{cases}$

Multiply $- 1$ by $6$ .

${\begin{cases} - 6 f + 6 \cdot 3 + 7 > 7 & f \leq 3 \\ 6 (- (- f + 3)) + 7 > 7 & f > 3 \end{cases}$

Multiply $6$ by $3$ .

${\begin{cases} - 6 f + 18 + 7 > 7 & f \leq 3 \\ 6 (- (- f + 3)) + 7 > 7 & f > 3 \end{cases}$

Add $18$ and $7$ .

${\begin{cases} - 6 f + 25 > 7 & f \leq 3 \\ 6 (- (- f + 3)) + 7 > 7 & f > 3 \end{cases}$

Simplify $6 (- (- f + 3)) + 7 > 7$ .

Simplify each term.

Apply the distributive property.

${\begin{cases} - 6 f + 25 > 7 & f \leq 3 \\ 6 (- - f - 1 \cdot 3) + 7 > 7 & f > 3 \end{cases}$

Multiply $- - f$ .

Multiply $- 1$ by $- 1$ .

${\begin{cases} - 6 f + 25 > 7 & f \leq 3 \\ 6 (1 f - 1 \cdot 3) + 7 > 7 & f > 3 \end{cases}$

Multiply $f$ by $1$ .

${\begin{cases} - 6 f + 25 > 7 & f \leq 3 \\ 6 (f - 1 \cdot 3) + 7 > 7 & f > 3 \end{cases}$

Multiply $- 1$ by $3$ .

${\begin{cases} - 6 f + 25 > 7 & f \leq 3 \\ 6 (f - 3) + 7 > 7 & f > 3 \end{cases}$

Apply the distributive property.

${\begin{cases} - 6 f + 25 > 7 & f \leq 3 \\ 6 f + 6 \cdot - 3 + 7 > 7 & f > 3 \end{cases}$

Multiply $6$ by $- 3$ .

${\begin{cases} - 6 f + 25 > 7 & f \leq 3 \\ 6 f - 18 + 7 > 7 & f > 3 \end{cases}$

Add $- 18$ and $7$ .

${\begin{cases} - 6 f + 25 > 7 & f \leq 3 \\ 6 f - 11 > 7 & f > 3 \end{cases}$

Solve $- 6 f + 25 > 7$ for $f$ .

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

Subtract $25$ from both sides of the inequality.

$- 6 f > 7 - 25$

Subtract $25$ from $7$ .

$- 6 f > - 18$

Divide each term in $- 6 f > - 18$ by $- 6$ and simplify.

Divide each term in $- 6 f > - 18$ by $- 6$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 6 f}{- 6} < \frac{- 18}{- 6}$

Simplify the left side.

Cancel the common factor of $- 6$ .

Cancel the common factor.

$\frac{- 6 f}{- 6} < \frac{- 18}{- 6}$

Divide $f$ by $1$ .

$f < \frac{- 18}{- 6}$

Simplify the right side.

Divide $- 18$ by $- 6$ .

$f < 3$

Solve $6 f - 11 > 7$ for $f$ .

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

Add $11$ to both sides of the inequality.

$6 f > 7 + 11$

Add $7$ and $11$ .

$6 f > 18$

Divide each term in $6 f > 18$ by $6$ and simplify.

Divide each term in $6 f > 18$ by $6$ .

$\frac{6 f}{6} > \frac{18}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 f}{6} > \frac{18}{6}$

Divide $f$ by $1$ .

$f > \frac{18}{6}$

Simplify the right side.

Divide $18$ by $6$ .

$f > 3$

Find the union of the solutions.

$f < 3$ or $f > 3$

The result can be shown in multiple forms.

Inequality Form:

$f < 3 or f > 3$

Interval Notation:

$(- \infty, 3) \cup (3, \infty)$

Solve for f 6|-f+3|+7>7

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