A commercial air conditioning unit is programmed to keep an industrial laboratory's temperature $T$ within the operating range $-4.00^\circ\text{C} \le T \le 15.0^\circ\text{C}$. The thermostat's digital controller operates in degrees Fahrenheit ($^\circ\text{F}$), where $T = \frac{5}{9}(F - 32)$. Solve the compound inequality to find the allowable temperature range in $^\circ\text{F}$, expressing the final bounds to three significant figures in interval notation.

A commercial air conditioning unit is programmed to keep an industrial laboratory's temperature $T$ within the operating range $-4.00^\circ\text{C} \le T \le 15.0^\circ\text{C}$ . The thermostat's digital controller operates in degrees Fahrenheit ( $^\circ\text{F}$ ), where $T = \frac{5}{9}(F - 32)$ . Solve the compound inequality to find the allowable temperature range in $^\circ\text{F}$ , expressing the final bounds to three significant figures in interval notation.

Final answer

$[24.8^\circ\text{F}, 59.0^\circ\text{F}]$

Steps

Substitute the conversion formula into the compound inequality

To find the allowable temperature range in degrees Fahrenheit ( $^\circ\text{F}$ ), we set up and solve a compound inequality.

The laboratory's temperature $T$ must satisfy: $-4.00 \le T \le 15.0$

Substitute the relationship $T = \frac{5}{9}(F - 32)$ into the inequality: $-4.00 \le \frac{5}{9}(F - 32) \le 15.0$

Clear the fraction

Multiply each part of the compound inequality by $\frac{9}{5}$ (the reciprocal of $\frac{5}{9}$ ). Since $\frac{9}{5}$ is positive, the inequality signs remain unchanged: $-4.00 \cdot \frac{9}{5} \le F - 32 \le 15.0 \cdot \frac{9}{5}$

Calculate the left and right bounds: $-7.20 \le F - 32 \le 27.0$

Isolate $F$

Add $32$ to all three parts of the inequality: $-7.20 + 32 \le F \le 27.0 + 32$ $24.8 \le F \le 59.0$

Write the solution in interval notation

Each bound is given to three significant figures, matching the precision of the initial values ( $-4.00^\circ\text{C}$ and $15.0^\circ\text{C}$ ).

Because the inequality includes the endpoints ( $\le$ ), we use square brackets to write the range in interval notation: $[24.8^\circ\text{F}, 59.0^\circ\text{F}]$

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student inverted the conversion factor when isolating F, multiplying by 5/9 instead of 9/5.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
-4.00 $\le$ $\left(\frac{5}{9}\right)$ (F - 32) $\le$ 15.0
Step 2
$-4.00 * \left(\frac{5}{9}\right) \le F - 32 \le 15.0 * \left(\frac{5}{9}\right) \implies -2.22 \le F - 32 \le 8.33$
Correct
-4.00 $\cdot$ $\frac{9}{5}$ $\le$ F - 32 $\le$ 15.0 $\cdot$ $\frac{9}{5}$ $\implies$ -7.20 $\le$ F - 32 $\le$ 27.0
To eliminate the factor of $\frac{5}{9}$ , you must multiply every part of the inequality by its reciprocal, $\frac{9}{5}$ (or divide by $\frac{5}{9}$ ), rather than multiplying by $\frac{5}{9}$ .
Step 3
29.8 $\le$ F $\le$ 40.3
Correct
-7.20 + 32 $\le$ F $\le$ 27.0 + 32 $\implies$ 24.8 $\le$ F $\le$ 59.0
Your addition of $32$ across all three parts was executed properly, but the numerical bounds were affected by the earlier reciprocal error.
Step 4
[29.8, 40.3]
Correct
[24.8^ $\circ$$\text{F}$ , 59.0^ $\circ$$\text{F}$ ]
Your interval notation formatting was correct, but the final interval inherited the incorrect boundary values.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly substituted the temperature conversion formula into the given compound inequality.

You correctly added $32$ to all parts of the compound inequality to isolate $F$ .

You expressed your final solution using proper closed interval notation with three significant figures.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Reciprocal Multiplication Error

You accidentally multiplied each side by the fraction $\frac{5}{9}$ rather than its reciprocal $\frac{9}{5}$ when undoing multiplication.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing how to invert a fractional coefficient: dividing by $\frac{a}{b}$ requires multiplying all sides by its reciprocal $\frac{b}{a}$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Reciprocal Check: To cancel a multiplying fraction $\frac{a}{b}$ , multiply both sides by the reciprocal $\frac{b}{a}$ , because $\frac{b}{a} \cdot \frac{a}{b} = 1$ .
Whenever you see $\frac{a}{b}(x - c)$ , check that the number you multiply on each side has the numerator and denominator swapped.

CORRECTED PATH

ACEpath Teacher's Demonstration

Multiply all three parts of $-4.00 \le \frac{5}{9}(F - 32) \le 15.0$ by $\frac{9}{5}$ : $-4.00 \cdot \frac{9}{5} \le F - 32 \le 15.0 \cdot \frac{9}{5}$ , which simplifies to $-7.20 \le F - 32 \le 27.0$ .
Add $32$ to each part: $-7.20 + 32 \le F \le 27.0 + 32$ , which yields $24.8 \le F \le 59.0$ .
Write the allowable temperature range in interval notation with units: $[24.8^\circ\text{F}, 59.0^\circ\text{F}]$ .

Final answer

[24.8^ $\circ$$\text{F}$ , 59.0^ $\circ$$\text{F}$ ]

NEXT STEPS

Final Takeaways from Your Teacher

Practice isolating variables with fractional coefficients by explicitly writing out multiplication by the reciprocal $\frac{b}{a}$ .
Verify your converted endpoints with a quick sanity check: $0^\circ\text{C} = 32^\circ\text{F}$ , so $-4^\circ\text{C}$ must be below $32^\circ\text{F}$ and $15^\circ\text{C}$ must be well above $32^\circ\text{F}$ .

OverviewACEpath Teacher's Diagnosis

02 The student added 32 before clearing the factor of 5/9 without distributing, violating the order of operations.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
-4.00 $\le$ $\left(\frac{5}{9}\right)$ (F - 32) $\le$ 15.0
Step 2
$-4.00 + 32 \le \left(\frac{5}{9}\right)F \le 15.0 + 32 \implies 28.0 \le \left(\frac{5}{9}\right)F \le 47.0$
Correct
-4.00 $\cdot$ $\frac{9}{5}$ $\le$ F - 32 $\le$ 15.0 $\cdot$ $\frac{9}{5}$ $\implies$ -7.20 $\le$ F - 32 $\le$ 27.0
The term $-32$ is inside grouping parentheses multiplied by $\frac{5}{9}$ . Adding $32$ does not eliminate $-32$ from $\frac{5}{9}(F - 32)$ because $\frac{5}{9}(F - 32) + 32 = \frac{5}{9}F + \frac{128}{9} \ne \frac{5}{9}F$ . You must multiply all parts by $\frac{9}{5}$ first, or distribute $\frac{5}{9}$ across the terms.
Step 3
$28.0 \cdot \left(\frac{9}{5}\right) \le F \le 47.0 \cdot \left(\frac{9}{5}\right) \implies 50.4 \le F \le 84.6$
Correct
-7.20 + 32 $\le$ F $\le$ 27.0 + 32 $\implies$ 24.8 $\le$ F $\le$ 59.0
Your arithmetic multiplying by the reciprocal $\frac{9}{5}$ is executed correctly, but the values are shifted due to the earlier error.
Step 4
[50.4, 84.6]
Correct
[24.8^ $\circ$$\text{F}$ , 59.0^ $\circ$$\text{F}$ ]
You used the correct interval notation with square brackets, but the numerical bounds reflect the upstream error.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly set up the initial compound inequality using the temperature conversion formula.

You correctly identified and applied the reciprocal $\frac{9}{5}$ to eliminate the coefficient of $F$ .

You properly interpreted the non-strict inequality signs by using closed brackets in interval notation.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Order of Operations with Grouped Factors

You attempted to cancel the $-32$ inside parentheses before dealing with the multiplying factor $\frac{5}{9}$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Misinterpreting the scope of grouping symbols: terms inside parentheses cannot be eliminated by standard addition or subtraction until the factor outside the parentheses is either cleared by multiplying/dividing or distributed across the entire expression.

REASONING PATH

Your ACEpath teacher’s suggested approach

Clear outer factors first: When an entire side or middle expression has the form $a(x + b)$ , undo the outer multiplication by $a$ before attempting to isolate $x$ by adding or subtracting $b$ .
Multiply all three parts of $L \le \frac{5}{9}(F - 32) \le R$ by $\frac{9}{5}$ to leave $F - 32$ with a coefficient of $1$ .
Alternatively, distribute completely: If you choose to expand parentheses before isolating the variable, you must multiply every term inside the parentheses by the outside factor: $a(x + b) = ax + ab$ .
Write $\frac{5}{9}(F - 32)$ as $\frac{5}{9}F - \frac{160}{9}$ , then add $\frac{160}{9}$ to all parts of the inequality.

CORRECTED PATH

ACEpath Teacher's Demonstration

Multiply all three sides by $\frac{9}{5}$ : $\left(-4.00\right)\cdot\frac{9}{5} \le F - 32 \le 15.0\cdot\frac{9}{5}$
Simplify the outer bounds: $-7.20 \le F - 32 \le 27.0$
Add $32$ to all three parts: $-7.20 + 32 \le F \le 27.0 + 32$
Simplify: $24.8 \le F \le 59.0$

Final answer

[24.8^ $\circ$$\text{F}$ , 59.0^ $\circ$$\text{F}$ ]

NEXT STEPS

Final Takeaways from Your Teacher

Check grouping symbols before adding or subtracting across an inequality.
Practice isolating grouped variables by clearing fractional coefficients with reciprocal multiplication first.
Verify your final endpoints by plugging them back into the original formula to check whether they match $-4.00^\circ\text{C}$ and $15.0^\circ\text{C}$ .

OverviewACEpath Teacher's Diagnosis

03 The student substituted the bounds directly into F instead of solving for F in terms of T, confusing the dependent and independent variables.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
F = $\left(\frac{5}{9}\right)$ (T - 32)
Correct
-4.00 $\le$ $\frac{5}{9}$ (F - 32) $\le$ 15.0
The problem provides the conversion $T = \frac{5}{9}(F - 32)$ . You swapped the variables and wrote $F = \frac{5}{9}(T - 32)$ , treating $F$ as the dependent variable rather than isolating it from $T$ .
Step 2
$\left(\frac{5}{9}\right)(-4.00 - 32) \le F \le \left(\frac{5}{9}\right)(15.0 - 32) \implies \left(\frac{5}{9}\right)(-36.0) \le F \le \left(\frac{5}{9}\right)(-17.0) \implies -20.0 \le F \le -9.44$
Correct
-4.00 $\cdot$ $\frac{9}{5}$ $\le$ F - 32 $\le$ 15.0 $\cdot$ $\frac{9}{5}$ $\implies$ -7.20 + 32 $\le$ F $\le$ 27.0 + 32 $\implies$ 24.8 $\le$ F $\le$ 59.0
Your arithmetic evaluation and inequality simplification were executed accurately, but the values were derived from the swapped formula in Step 1.
Step 3
[-20.0, -9.44]
Correct
[24.8^ $\circ$$\text{F}$ , 59.0^ $\circ$$\text{F}$ ]
You correctly converted your computed inequality into interval notation with square brackets, but the interval bounds are incorrect due to the initial variable swap.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly maintained the compound inequality format throughout your work.

You accurately performed the arithmetic steps involving negative values.

You expressed your final solution in standard interval notation using square brackets.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Variable Confusion in Conversion Formula

You wrote $F = \frac{5}{9}(T - 32)$ instead of using the given formula $T = \frac{5}{9}(F - 32)$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You treated the given formula as if it were already solved for the unknown $F$ , substituting the Celsius values directly into the right-hand side rather than replacing $T$ in the compound inequality and solving for $F$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Variable Role Check: Identify which variable represents the given range and which variable needs to be solved.
Since the range is given for $T$ , substitute the entire expression $\frac{5}{9}(F - 32)$ in place of $T$ inside $-4.00 \le T \le 15.0$ .
Physical Reality Check: Water freezes at $0^\circ\text{C} = 32^\circ\text{F}$ and boils at $100^\circ\text{C} = 212^\circ\text{F}$ .
Check whether your result makes physical sense: a temperature around $0^\circ\text{C}$ must correspond to approximately $32^\circ\text{F}$ , not negative Fahrenheit values.

CORRECTED PATH

ACEpath Teacher's Demonstration

Substitute $T = \frac{5}{9}(F - 32)$ into $-4.00 \le T \le 15.0$ : $-4.00 \le \frac{5}{9}(F - 32) \le 15.0$
Multiply all three parts by $\frac{9}{5}$ : $-7.20 \le F - 32 \le 27.0$
Add $32$ to all three parts: $24.8 \le F \le 59.0$

Final answer

[24.8^ $\circ$$\text{F}$ , 59.0^ $\circ$$\text{F}$ ]

NEXT STEPS

Final Takeaways from Your Teacher

Double-check that the given formula matches the problem statement verbatim before writing it down.
When finding the range of an inner variable, substitute the conversion expression directly into the compound inequality and isolate the variable using inverse operations.
Perform a quick benchmark check (such as $0^\circ\text{C} = 32^\circ\text{F}$ ) to ensure the converted values fall in the expected numerical range.

Lock it in

Question

How do you eliminate a fractional coefficient $\frac{a}{b}$ multiplying an expression in an equation or inequality?

Answer

Multiply all parts by its reciprocal $\frac{b}{a}$ , because $\frac{b}{a} \cdot \frac{a}{b} = 1$ .

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student inverted the conversion factor when isolating F, multiplying by 5/9 instead of 9/5.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

02 The student added 32 before clearing the factor of 5/9 without distributing, violating the order of operations.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

03 The student substituted the bounds directly into F instead of solving for F in terms of T, confusing the dependent and independent variables.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

Lock it in

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