Solve the System of Inequalities -20<=5/9(F-32)<=-15

$- 20 \leq \frac{5}{9} (F - 32) \leq - 15$

Step 1

Solve $- 20 \leq \frac{5}{9} \cdot (F - 32)$ for $F$ .

Rewrite so $F$ is on the left side of the inequality.

$\frac{5}{9} \cdot (F - 32) \geq - 20$

Multiply each term in $\frac{5}{9} \cdot (F - 32) \geq - 20$ by $\frac{9}{5}$ to eliminate the fractions.

Multiply each term in $\frac{5}{9} \cdot (F - 32) \geq - 20$ by $\frac{9}{5}$ .

$\frac{5}{9} \cdot (F - 32) \frac{9}{5} \geq - 20 (\frac{9}{5})$

Simplify the left side.

Apply the distributive property.

$(\frac{5}{9} F + \frac{5}{9} \cdot - 32) \frac{9}{5} \geq - 20 (\frac{9}{5})$

Combine $\frac{5}{9}$ and $F$ .

$(\frac{5 F}{9} + \frac{5}{9} \cdot - 32) \frac{9}{5} \geq - 20 (\frac{9}{5})$

Multiply $\frac{5}{9} \cdot - 32$ .

Combine $\frac{5}{9}$ and $- 32$ .

$(\frac{5 F}{9} + \frac{5 \cdot - 32}{9}) \frac{9}{5} \geq - 20 (\frac{9}{5})$

Multiply $5$ by $- 32$ .

$(\frac{5 F}{9} + \frac{- 160}{9}) \frac{9}{5} \geq - 20 (\frac{9}{5})$

Move the negative in front of the fraction.

$(\frac{5 F}{9} - \frac{160}{9}) \frac{9}{5} \geq - 20 (\frac{9}{5})$

Apply the distributive property.

$\frac{5 F}{9} \cdot \frac{9}{5} - \frac{160}{9} \cdot \frac{9}{5} \geq - 20 (\frac{9}{5})$

Cancel the common factor of $5$ .

Factor $5$ out of $5 F$ .

$\frac{5 (F)}{9} \cdot \frac{9}{5} - \frac{160}{9} \cdot \frac{9}{5} \geq - 20 (\frac{9}{5})$

Cancel the common factor.

$\frac{5 F}{9} \cdot \frac{9}{5} - \frac{160}{9} \cdot \frac{9}{5} \geq - 20 (\frac{9}{5})$

Rewrite the expression.

$\frac{F}{9} \cdot 9 - \frac{160}{9} \cdot \frac{9}{5} \geq - 20 (\frac{9}{5})$

Cancel the common factor of $9$ .

Cancel the common factor.

$\frac{F}{9} \cdot 9 - \frac{160}{9} \cdot \frac{9}{5} \geq - 20 (\frac{9}{5})$

Rewrite the expression.

$F - \frac{160}{9} \cdot \frac{9}{5} \geq - 20 (\frac{9}{5})$

Cancel the common factor of $5$ .

Move the leading negative in $- \frac{160}{9}$ into the numerator.

$F + \frac{- 160}{9} \cdot \frac{9}{5} \geq - 20 (\frac{9}{5})$

Factor $5$ out of $- 160$ .

$F + \frac{5 (- 32)}{9} \cdot \frac{9}{5} \geq - 20 (\frac{9}{5})$

Cancel the common factor.

$F + \frac{5 \cdot - 32}{9} \cdot \frac{9}{5} \geq - 20 (\frac{9}{5})$

Rewrite the expression.

$F + \frac{- 32}{9} \cdot 9 \geq - 20 (\frac{9}{5})$

Cancel the common factor of $9$ .

Cancel the common factor.

$F + \frac{- 32}{9} \cdot 9 \geq - 20 (\frac{9}{5})$

Rewrite the expression.

$F - 32 \geq - 20 (\frac{9}{5})$

Simplify the right side.

Cancel the common factor of $5$ .

Factor $5$ out of $- 20$ .

$F - 32 \geq 5 (- 4) \frac{9}{5}$

Cancel the common factor.

$F - 32 \geq 5 \cdot - 4 \frac{9}{5}$

Rewrite the expression.

$F - 32 \geq - 4 \cdot 9$

Multiply $- 4$ by $9$ .

$F - 32 \geq - 36$

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

Add $32$ to both sides of the inequality.

$F \geq - 36 + 32$

Add $- 36$ and $32$ .

$F \geq - 4$

Step 2

Solve $\frac{5}{9} \cdot (F - 32) \leq - 15$ for $F$ .

Multiply each term in $\frac{5}{9} \cdot (F - 32) \leq - 15$ by $\frac{9}{5}$ to eliminate the fractions.

Multiply each term in $\frac{5}{9} \cdot (F - 32) \leq - 15$ by $\frac{9}{5}$ .

$\frac{5}{9} \cdot (F - 32) \frac{9}{5} \leq - 15 (\frac{9}{5})$

Simplify the left side.

Apply the distributive property.

$(\frac{5}{9} F + \frac{5}{9} \cdot - 32) \frac{9}{5} \leq - 15 (\frac{9}{5})$

Combine $\frac{5}{9}$ and $F$ .

$(\frac{5 F}{9} + \frac{5}{9} \cdot - 32) \frac{9}{5} \leq - 15 (\frac{9}{5})$

Multiply $\frac{5}{9} \cdot - 32$ .

Combine $\frac{5}{9}$ and $- 32$ .

$(\frac{5 F}{9} + \frac{5 \cdot - 32}{9}) \frac{9}{5} \leq - 15 (\frac{9}{5})$

Multiply $5$ by $- 32$ .

$(\frac{5 F}{9} + \frac{- 160}{9}) \frac{9}{5} \leq - 15 (\frac{9}{5})$

Move the negative in front of the fraction.

$(\frac{5 F}{9} - \frac{160}{9}) \frac{9}{5} \leq - 15 (\frac{9}{5})$

Apply the distributive property.

$\frac{5 F}{9} \cdot \frac{9}{5} - \frac{160}{9} \cdot \frac{9}{5} \leq - 15 (\frac{9}{5})$

Cancel the common factor of $5$ .

Factor $5$ out of $5 F$ .

$\frac{5 (F)}{9} \cdot \frac{9}{5} - \frac{160}{9} \cdot \frac{9}{5} \leq - 15 (\frac{9}{5})$

Cancel the common factor.

$\frac{5 F}{9} \cdot \frac{9}{5} - \frac{160}{9} \cdot \frac{9}{5} \leq - 15 (\frac{9}{5})$

Rewrite the expression.

$\frac{F}{9} \cdot 9 - \frac{160}{9} \cdot \frac{9}{5} \leq - 15 (\frac{9}{5})$

Cancel the common factor of $9$ .

Cancel the common factor.

$\frac{F}{9} \cdot 9 - \frac{160}{9} \cdot \frac{9}{5} \leq - 15 (\frac{9}{5})$

Rewrite the expression.

$F - \frac{160}{9} \cdot \frac{9}{5} \leq - 15 (\frac{9}{5})$

Cancel the common factor of $5$ .

Move the leading negative in $- \frac{160}{9}$ into the numerator.

$F + \frac{- 160}{9} \cdot \frac{9}{5} \leq - 15 (\frac{9}{5})$

Factor $5$ out of $- 160$ .

$F + \frac{5 (- 32)}{9} \cdot \frac{9}{5} \leq - 15 (\frac{9}{5})$

Cancel the common factor.

$F + \frac{5 \cdot - 32}{9} \cdot \frac{9}{5} \leq - 15 (\frac{9}{5})$

Rewrite the expression.

$F + \frac{- 32}{9} \cdot 9 \leq - 15 (\frac{9}{5})$

Cancel the common factor of $9$ .

Cancel the common factor.

$F + \frac{- 32}{9} \cdot 9 \leq - 15 (\frac{9}{5})$

Rewrite the expression.

$F - 32 \leq - 15 (\frac{9}{5})$

Simplify the right side.

Cancel the common factor of $5$ .

Factor $5$ out of $- 15$ .

$F - 32 \leq 5 (- 3) \frac{9}{5}$

Cancel the common factor.

$F - 32 \leq 5 \cdot - 3 \frac{9}{5}$

Rewrite the expression.

$F - 32 \leq - 3 \cdot 9$

Multiply $- 3$ by $9$ .

$F - 32 \leq - 27$

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

Add $32$ to both sides of the inequality.

$F \leq - 27 + 32$

Add $- 27$ and $32$ .

$F \leq 5$

Step 3

Find the intersection of $F \geq - 4$ and $F \leq 5$ .

$- 4 \leq F \leq 5$

Step 4

The result can be shown in multiple forms.

Inequality Form:

$- 4 \leq F \leq 5$

Interval Notation:

$[- 4, 5]$

Step 5

Solve the System of Inequalities -20<=5/9(F-32)<=-15

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