In an analytical chemistry laboratory, a student calibrates a spectrophotometer absorbance reading $A$ using the relationship $|0.850 A - 0.510| = 0.068$, where $A$ denotes the dimensionless absorbance. Determine all valid experimental absorbance values $A$, rounded to three significant figures.

In an analytical chemistry laboratory, a student calibrates a spectrophotometer absorbance reading $A$ using the relationship $|0.850 A - 0.510| = 0.068$ , where $A$ denotes the dimensionless absorbance. Determine all valid experimental absorbance values $A$ , rounded to three significant figures.

Final answer

$A = 0.520$ or $A = 0.680$

Steps

To solve an absolute value equation of the form $|X| = c$ where $c > 0$ , recall that the expression inside the absolute value can equal either $c$ or $-c$ :

$X = c \quad \text{or} \quad X = -c$

Here, let $X = 0.850 A - 0.510$ and $c = 0.068$ . Setting up the two linear equations:

Case 1: Positive branch $0.850 A - 0.510 = 0.068$

Add $0.510$ to both sides: $0.850 A = 0.510 + 0.068$ $0.850 A = 0.578$

Divide both sides by $0.850$ : $A = \frac{0.578}{0.850} = 0.680$

Case 2: Negative branch $0.850 A - 0.510 = -0.068$

Add $0.510$ to both sides: $0.850 A = 0.510 - 0.068$ $0.850 A = 0.442$

Divide both sides by $0.850$ : $A = \frac{0.442}{0.850} = 0.520$

Verification and Significant Figures

Both values $0.520$ and $0.680$ are positive and physically reasonable for spectrophotometer absorbance readings.
Writing each to three significant figures gives $A = 0.520$ and $A = 0.680$ .

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student ignores the absolute value bars and solves only the positive case, missing the second valid solution.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
0.850 A - 0.510 = 0.068
Correct
0.850 A - 0.510 = 0.068 $\text{or}$ 0.850 A - 0.510 = -0.068
An absolute value equation of the form $|X| = c$ (where $c > 0$ ) requires two separate cases: $X = c$ and $X = -c$ . Removing the bars without setting up the negative case discards half of the solution set.
Step 2
0.850 A = 0.578 $\implies$ A = 0.578 / 0.850
Step 3
A = 0.680
Correct
A = 0.520 $\text{or}$ A = 0.680
This numerical value is correct for the positive branch, but the final answer remains incomplete because the second valid value from the negative branch was never evaluated.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly isolated the variable term and carried out the arithmetic for the positive branch.

You correctly reported your value with three significant figures.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Omitted Negative Branch

You treated the absolute value expression as standard parentheses and solved only the positive case.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

The definition of absolute value as distance on a number line was overlooked: $|X| = c$ means $X$ lies a distance of $c$ away from zero in either direction, requiring both $X = c$ and $X = -c$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Split Before Solving: For any equation $|X| = c$ with $c > 0$ , immediately write down two distinct equations: $X = c$ and $X = -c$ .
Whenever you remove absolute value bars, write an explicit 'or' statement with both $+c$ and $-c$ before performing any algebra.
Check Solution Count: A linear absolute value equation $|ax + b| = c$ with $c > 0$ yields two distinct solutions.
Count your final answers; if you only have one value for a non-zero distance, pause and find the missing branch.

CORRECTED PATH

ACEpath Teacher's Demonstration

Set up the two branches from $|0.850 A - 0.510| = 0.068$ : Branch 1 is $0.850 A - 0.510 = 0.068$ , and Branch 2 is $0.850 A - 0.510 = -0.068$ .
Solve Branch 1: $0.850 A = 0.578 \implies A = \frac{0.578}{0.850} = 0.680$ .
Solve Branch 2: $0.850 A = 0.510 - 0.068 = 0.442 \implies A = \frac{0.442}{0.850} = 0.520$ .

Final answer

$A = 0.520$ or $A = 0.680$

NEXT STEPS

Final Takeaways from Your Teacher

Practice rewriting $|X| = c$ as two separate linear equations in the very first line of your work.
Always verify that both positive and negative branches produce physically sensible values in context.

OverviewACEpath Teacher's Diagnosis

02 The student negates only the variable term inside the absolute value rather than the entire expression or the right-hand side, leading to an incorrect negative absorbance value.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
0.850 A - 0.510 = 0.068 $\text{ or }$ -0.850 A - 0.510 = 0.068
Correct
0.850 A - 0.510 = 0.068 $\text{ or }$ -(0.850 A - 0.510) = 0.068 $\iff$ 0.850 A - 0.510 = -0.068
To branch an absolute value equation $|X| = c$ , you must negate the entire enclosed expression $-(X) = c$ , or equivalently negate the constant side $X = -c$ . Negating only the leading term $0.850 A$ while leaving $-0.510$ unchanged is mathematically invalid.
Step 2
$\text{Case 1: }$ 0.850 A = 0.578 $\implies$ A = 0.680
Step 3
$\text{Case 2: }$ -0.850 A = 0.578 $\implies$ A = $\frac{-0.578}{0.850}$ = -0.680
Correct
0.850 A - 0.510 = -0.068 $\implies$ 0.850 A = 0.442 $\implies$ A = 0.520
Your algebraic isolation here was executed consistently with the equation you wrote, but it inherits the setup error from Step 1.
Step 4
A = 0.680 $\text{ and }$ A = -0.680
Correct
A = 0.520 $\text{ or }$ A = 0.680
Because the second branch was set up incorrectly, your final combined set of values includes $-0.680$ instead of $0.520$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly recognized that an absolute value equation splits into two linear branches.

You accurately solved the linear equation for the positive branch to find $A = 0.680$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Partial Negation of Absolute Value Argument

You applied the negative sign to only the first term $0.850 A$ instead of distributing it across the entire binomial $(0.850 A - 0.510)$ or applying it to the right-hand constant $0.068$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing the negation of an entire algebraic expression with changing the sign of just the leading variable term. In $|A - B| = c$ , the opposite branch is $-(A - B) = c$ , which equals $-A + B = c$ , or more simply $A - B = -c$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Rule for Branching $|X| = c$: For any real expression $X$ and constant $c \ge 0$ , $|X| = c$ splits into $X = c$ or $X = -c$ .
Keep the entire expression inside the absolute value unchanged on the left side, and place the $\pm$ sign on the single number on the right side.
Physical Reality Check: Spectrophotometer absorbance $A = -\log_{10}(T)$ is typically non-negative under standard calibration conditions.
When you obtain a negative absorbance like $-0.680$ , pause and verify whether you correctly handled the negative branch.

CORRECTED PATH

ACEpath Teacher's Demonstration

Rewrite Case 2 by placing the negative sign on the constant: $0.850 A - 0.510 = -0.068$ .
Add $0.510$ to both sides: $0.850 A = 0.510 - 0.068 = 0.442$ .
Divide by $0.850$ : $A = \frac{0.442}{0.850} = 0.520$ .

Final answer

$A = 0.520$ or $A = 0.680$

NEXT STEPS

Final Takeaways from Your Teacher

Always branch $|\text{expression}| = c$ into $\text{expression} = c$ and $\text{expression} = -c$ to avoid bracket distribution slips.
If you choose to negate the expression side instead, always wrap it in parentheses: $-(0.850 A - 0.510) = 0.068$ .
Perform a quick substitution check of both solutions into the original equation before writing your final answer.

OverviewACEpath Teacher's Diagnosis

03 When solving the negative case, the student adds 0.510 incorrectly by computing 0.068 - 0.510 instead of -0.068 + 0.510, arriving at an erroneous absorbance value.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
$0.850 A - 0.510 = 0.068 \text{ or } 0.850 A - 0.510 = -0.068$
Step 2
$0.850 A = 0.578 \implies A = 0.680$
Step 3
$0.850 A = 0.068 - 0.510 = -0.442$
Correct
$0.850 A = -0.068 + 0.510 = 0.442$
When adding $0.510$ to both sides of $0.850 A - 0.510 = -0.068$ , the right-hand side is $-0.068 + 0.510 = 0.442$ . You dropped the negative sign on $0.068$ and subtracted $0.510$ instead.
Step 4
$A = -0.442 / 0.850 = -0.520$
Correct
$A = \frac{0.442}{0.850} = 0.520$
Your division arithmetic was executed consistently, but the value carried the incorrect negative sign from the previous line.
Step 5
$A = 0.680 \text{ and } A = -0.520$
Correct
$A = 0.520 \quad \text{or} \quad A = 0.680$
The final pair of solutions contains the propagated sign error from the negative branch.

WHAT YOU DID WELL

Compliments from Your Teacher

You split the absolute value equation into the correct positive and negative cases.

You solved the positive branch completely and rounded properly to three significant figures.

You carried out the division step with correct arithmetic.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Transposition Sign Error in Negative Branch

When isolating $0.850 A$ in the second branch, you wrote $0.068 - 0.510$ rather than keeping $-0.068$ and adding $0.510$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Losing track of existing negative signs on the right-hand side when applying inverse operations to the constant term on the left.

REASONING PATH

Your ACEpath teacher’s suggested approach

Keep Existing Signs Explicit: When solving $m x - b = -c$ , adding $b$ to both sides yields $m x = -c + b$ , which equals $b - c$ .
Always write out $+b$ explicitly on both sides before combining terms.
Physical Reality Check: In spectrophotometry, standard absorbance values $A = -\log_{10}(T)$ are non-negative for transmitting samples.
Pause and re-check algebraic signs whenever an intermediate or final physical quantity yields an unexpected negative value.

CORRECTED PATH

ACEpath Teacher's Demonstration

Add $0.510$ to both sides: $0.850 A = -0.068 + 0.510 = 0.442$ .
Divide both sides by $0.850$ : $A = \frac{0.442}{0.850} = 0.520$ .

Final answer

$A = 0.520$ or $A = 0.680$

NEXT STEPS

Final Takeaways from Your Teacher

Practice isolating variables when the constant on the right-hand side is already negative.
Review the three targeted flashcards to reinforce sign management during transposition.
Solve the follow-up practice exercises to solidify accuracy on decimal absolute value equations.

Lock it in

Question

What two equations must you write when solving $|u| = d$ , where $d > 0$ ?

Answer

$u = d$ or $u = -d$ . Absolute value represents distance from zero, which can lie in either the positive or negative direction.

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student ignores the absolute value bars and solves only the positive case, missing the second valid solution.

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 negates only the variable term inside the absolute value rather than the entire expression or the right-hand side, leading to an incorrect negative absorbance value.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

03 When solving the negative case, the student adds 0.510 incorrectly by computing 0.068 - 0.510 instead of -0.068 + 0.510, arriving at an erroneous absorbance value.

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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