A standard laboratory balance registers two unknown masses $A$ and $B$. Three units of mass $A$ and four units of mass $B$ together balance a $43\text{ g}$ tare mass. Simultaneously, five units of mass $A$ exceed two units of mass $B$ by exactly $11\text{ g}$. Set up and solve the multi-step linear equation model to determine the individual mass of $A$ and mass of $B$.

A standard laboratory balance registers two unknown masses $A$ and $B$ . Three units of mass $A$ and four units of mass $B$ together balance a $43\text{ g}$ tare mass. Simultaneously, five units of mass $A$ exceed two units of mass $B$ by exactly $11\text{ g}$ . Set up and solve the multi-step linear equation model to determine the individual mass of $A$ and mass of $B$ .

Final answer

$A = 5\text{ g}$ , $B = 7\text{ g}$

Steps

Translate the physical conditions into algebraic equations

Let $A$ represent the mass of one unit of $A$ (in grams), and let $B$ represent the mass of one unit of $B$ (in grams).

"Three units of mass $A$ and four units of mass $B$ balance a $43\text{ g}$ tare mass": $3A + 4B = 43 \quad \text{--- (Equation 1)}$
"Five units of mass $A$ exceed two units of mass $B$ by $11\text{ g}$ ": $5A - 2B = 11 \quad \text{--- (Equation 2)}$

Express one variable to create a single-variable linear equation

To solve this systematically using a multi-step linear equation in one variable, isolate the term containing $B$ in Equation 2: $5A - 11 = 2B$

Multiply both sides by $2$ to find an equivalent expression for $4B$ : $2(5A - 11) = 2(2B)$ $10A - 22 = 4B$

Substitute and solve for $A$

Substitute $(10A - 22)$ in place of $4B$ into Equation 1: $3A + (10A - 22) = 43$

Now, solve this multi-step equation for $A$ :

Combine like terms: $13A - 22 = 43$
Add $22$ to both sides: $13A = 43 + 22$ $13A = 65$
Divide both sides by $13$ : $A = \frac{65}{13} = 5\text{ g}$

Solve for $B$

Substitute $A = 5$ back into the expression for $2B$ : $2B = 5(5) - 11$ $2B = 25 - 11$ $2B = 14$

Divide by $2$ : $B = 7\text{ g}$

Verification and Common Pitfall

Check both values in Equation 1: $3(5) + 4(7) = 15 + 28 = 43\text{ g}$ .
Check both values in Equation 2: $5(5) - 2(7) = 25 - 14 = 11\text{ g}$ .

Common Pitfall: When phrasing "exceeds $2B$ by $11\text{ g}$ ", learners sometimes write $2B - 5A = 11$ . Remember that if $5A$ is larger than $2B$ , the difference is $(5A - 2B) = 11$ .

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student misinterprets the phrase 'five units of mass A exceed two units of mass B by 11 g' as meaning two units of B exceed five units of A, setting up 2B - 5A = 11 instead of 5A - 2B = 11.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
3A + 4B = 43 $\implies$ 2B - 5A = 11
Correct
3A + 4B = 43 $\implies$ 5A - 2B = 11
You translated the first condition correctly as $3A + 4B = 43$ . However, because $5A$ exceeds $2B$ by $11\text{ g}$ , $5A$ is the larger quantity, so the equation must be written as $5A - 2B = 11$ , not $2B - 5A = 11$ .
Step 2
-5A + 2B = 11 $\implies$ -10A + 4B = 22 $\implies$ (3A + 4B) - (-10A + 4B) = 43 - 22 $\implies$ 13A = 21 $\implies$ A = $\frac{21}{13} $\text{ g}$
Correct
10A - 4B = 22 $\implies$ (3A + 4B) + (10A - 4B) = 43 + 22 $\implies$ 13A = 65 $\implies$ A = 5 $\text{ g}$
Your elimination procedure was algebraically sound, but the result was affected by the sign error in the setup of the second equation.
Step 3
2B = 11 + 5 $\left(\frac{21}{13}\right)$ = $\frac{248}{13}$ $\implies$ B = $\frac{124}{13} $\text{ g}$
Correct
2B = 5(5) - 11 = 14 $\implies$ B = 7 $\text{ g}$
Your substitution and fraction operations were executed correctly, but this step inherited the erroneous value of $A$ from the earlier misinterpretation.

WHAT YOU DID WELL

Compliments from Your Teacher

You translated the first balance condition into $3A + 4B = 43$ without error.

You multiplied the equation by 2 to align the coefficients of $B$ for elimination.

You carried out algebraic subtraction and fraction arithmetic with high technical accuracy.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Translating Verbal Comparison Expressions

You wrote the difference as $2B - 5A = 11$ instead of $5A - 2B = 11$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You associated the subtraction order with the order of terms or the prepositional target ('exceed two units of mass B') rather than subtracting the smaller quantity from the larger quantity ('larger - smaller = difference').

REASONING PATH

Your ACEpath teacher’s suggested approach

Identify the Larger Quantity: In the statement ' $X$ exceeds $Y$ by $D$ ', $X$ is larger than $Y$ .
Always write $\text{larger} - \text{smaller} = \text{difference}$ , which yields $X - Y = D$ .
Sign Check via Testing: If $X > Y$ , then $X - Y$ must be a positive number.
Pick intuitive test values (for instance, let $X = 15$ and $Y = 4$ ; $15 - 4 = 11 > 0$ ) to verify which variable comes first.

CORRECTED PATH

ACEpath Teacher's Demonstration

Start from the correct system: $\lbrace 3A + 4B = 43,\quad 5A - 2B = 11$ .
Multiply the second equation by $2$ : $10A - 4B = 22$ .
Add the two equations to eliminate $B$ : $(3A + 4B) + (10A - 4B) = 43 + 22 \implies 13A = 65$ .
Solve for $A$ : $A = \frac{65}{13} = 5\text{ g}$ .
Substitute $A = 5$ into $2B = 5A - 11$ : $2B = 5(5) - 11 = 14 \implies B = 7\text{ g}$ .

Final answer

A = 5 $\text{ g}$ , B = 7 $\text{ g}$

NEXT STEPS

Final Takeaways from Your Teacher

Review the verbal-to-algebraic phrase rules for 'exceeds', 'is greater than', and 'is less than'.
Practice translating comparative statements into the form $\text{larger} - \text{smaller} = \text{difference}$ .
Verify solutions by plugging both values back into the original word problem text rather than into your derived equations.

OverviewACEpath Teacher's Diagnosis

02 When multiplying the equation 5A - 2B = 11 by 2 to eliminate B, the student forgets to multiply the constant term on the right side, using 10A - 4B = 11.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
3A + 4B = 43 $\implies$ 5A - 2B = 11
Step 2
10A - 4B = 11
Correct
2(5A - 2B) = 2(11) $\implies$ 10A - 4B = 22
When scaling an equation, you must apply the multiplication property of equality by multiplying every term on both sides. Multiplying the left side by $2$ requires multiplying the right side by $2$ as well ( $11 \times 2 = 22$ ).
Step 3
13A = 54 $\implies$ A = $\frac{54}{13} $\text{ g}$
Correct
(3A + 4B) + (10A - 4B) = 43 + 22 $\implies$ 13A = 65 $\implies$ A = 5 $\text{ g}$
Your elimination technique of adding the two equations together was carried out correctly based on your previous line, but the result was affected by the unscaled right-hand constant.
Step 4
2B = 5 $\left(\frac{54}{13}\right)$ - 11 = $\frac{270}{13}$ - $\frac{143}{13}$ = $\frac{127}{13}$ $\implies$ B = $\frac{127}{26} $\text{ g}$
Correct
2B = 5(5) - 11 = 14 $\implies$ B = 7 $\text{ g}$
Your substitution and fraction arithmetic were executed accurately, but the resulting value for $B$ inherited the earlier error from $A$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly set up both linear equations from the word problem context.

You chose an effective elimination strategy by matching the coefficients of $B$ .

You accurately executed multi-step fraction arithmetic during back-substitution.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Multiplication Property of Equality Omission

When multiplying $5A - 2B = 11$ by $2$ , you multiplied the variable terms on the left side but left $11$ unchanged on the right side.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Failing to distribute the multiplier across the entire equation, particularly overlooking the constant term on the other side of the equals sign when clearing or scaling coefficients.

REASONING PATH

Your ACEpath teacher’s suggested approach

Balancing Equation Scaling: If you multiply one side of an equation by a non-zero constant $k$ , you must multiply the other side by $k$ : $k \cdot \text{LHS} = k \cdot \text{RHS}$ .
Write explicit parentheses around both complete sides before multiplying: $2(5A - 2B) = 2(11)$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Scale both sides of $5A - 2B = 11$ by $2$ : $2(5A - 2B) = 2(11) \implies 10A - 4B = 22$ .
Add this scaled equation to $3A + 4B = 43$ : $(3A + 10A) + (4B - 4B) = 43 + 22 \implies 13A = 65$ .
Divide both sides by $13$ : $A = \frac{65}{13} = 5\text{ g}$ .
Substitute $A = 5$ into $2B = 5A - 11$ : $2B = 5(5) - 11 = 25 - 11 = 14$ .
Divide by $2$ : $B = 7\text{ g}$ .

Final answer

$A = 5\text{ g}$ , $B = 7\text{ g}$

NEXT STEPS

Final Takeaways from Your Teacher

Always write grouping parentheses around both sides of an equation before applying a scalar multiplier.
Perform a quick parity or whole-number check: academic word problems about physical masses usually yield clean integer values unless stated otherwise.
Verify your calculated values in both original equations before finalizing your answer.

OverviewACEpath Teacher's Diagnosis

03 When adding the two equations 3A + 4B = 43 and 10A - 4B = 22 to eliminate B, the student subtracts the equations instead, incorrectly obtaining -7A = 21.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
3A + 4B = 43 $\implies$ 5A - 2B = 11 $\implies$ 10A - 4B = 22
Step 2
(3A + 4B) - (10A - 4B) = 43 - 22 $\implies$ -7A = 21 $\implies$ A = -3 $\text{ g}$
Correct
(3A + 4B) + (10A - 4B) = 43 + 22 $\implies$ 13A = 65 $\implies$ A = 5 $\text{ g}$
Because the terms $+4B$ and $-4B$ have opposite signs, you need to add the two equations to eliminate $B$ . Subtracting them yields $(3A - 10A) + (4B - (-4B)) = -7A + 8B$ , which does not eliminate $B$ .
Step 3
5(-3) - 2B = 11 $\implies$ -15 - 2B = 11 $\implies$ -2B = 26 $\implies$ B = -13 $\text{ g}$
Correct
5(5) - 2B = 11 $\implies$ 25 - 2B = 11 $\implies$ -2B = -14 $\implies$ B = 7 $\text{ g}$
Your algebraic steps to solve for $B$ were carried out consistently with your value of $A$ , but the final value was affected by the earlier elimination error.

WHAT YOU DID WELL

Compliments from Your Teacher

You translated both word clues into accurate linear equations.

You recognized that multiplying the second equation by $2$ creates matching magnitudes for the $B$ terms.

You solved for the second variable systematically using substitution.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Elimination Sign Error

You subtracted the equations instead of adding them when the target variable had opposite signs ( $+4B$ and $-4B$ ).

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing when to add versus subtract in the elimination method: terms with opposite signs cancel out under addition, whereas subtraction requires terms with identical signs.

REASONING PATH

Your ACEpath teacher’s suggested approach

Check Opposite Signs: If the coefficients of the target variable are opposites (e.g., $+k$ and $-k$ ), add the equations: $(+k) + (-k) = 0$ .
Look at the signs before combining: identical signs require subtraction, opposite signs require addition.
Physical Reality Check: Mass, length, and count quantities in standard word problems cannot be negative.
If your solution produces negative values for a physical mass such as $A = -3\text{ g}$ , pause immediately to re-check your elimination step.

CORRECTED PATH

ACEpath Teacher's Demonstration

Write the aligned equations: $3A + 4B = 43$ and $10A - 4B = 22$ .
Add the two equations: $(3A + 10A) + (4B - 4B) = 43 + 22 \implies 13A = 65$ .
Divide by $13$ : $A = 5\text{ g}$ .
Substitute $A = 5$ into $5A - 2B = 11$ : $5(5) - 2B = 11 \implies 25 - 2B = 11$ .
Subtract $25$ and divide by $-2$ : $-2B = -14 \implies B = 7\text{ g}$ .

Final answer

A = 5 $\text{ g}$ , B = 7 $\text{ g}$

NEXT STEPS

Final Takeaways from Your Teacher

Review the rule for elimination: add when signs are opposite, subtract when signs are identical.
Practice distributing negative signs through parentheses when subtracting polynomials.
Perform a sanity check on word problem answers to ensure physical quantities remain positive.

Lock it in

Question

How do you algebraically translate the verbal phrase "Quantity $X$ exceeds Quantity $Y$ by $k$ "?

Answer

$X - Y = k$ (or $X = Y + k$ ). Principle: $X$ is larger than $Y$ , so $\text{larger} - \text{smaller} = \text{difference}$ .

Resources

Answer and steps

Final answer

Steps

Verification and Common Pitfall

Typical errors

01 The student misinterprets the phrase 'five units of mass A exceed two units of mass B by 11 g' as meaning two units of B exceed five units of A, setting up 2B - 5A = 11 instead of 5A - 2B = 11.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

02 When multiplying the equation 5A - 2B = 11 by 2 to eliminate B, the student forgets to multiply the constant term on the right side, using 10A - 4B = 11.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

03 When adding the two equations 3A + 4B = 43 and 10A - 4B = 22 to eliminate B, the student subtracts the equations instead, incorrectly obtaining -7A = 21.

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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