Solve the equation for $x$: $3|2x - 5| - 7 = 14$. Express all solutions in exact form.

Solve the equation for $x$ : $3|2x - 5| - 7 = 14$ . Express all solutions in exact form.

Final answer

$x = 6$ or $x = -1$

Steps

Isolate the absolute value expression

To solve an equation containing an absolute value expression, first isolate the absolute value term on one side of the equation, then split it into two linear cases based on the definition of absolute value.

Add $7$ to both sides of the equation: $3|2x - 5| - 7 + 7 = 14 + 7$ $3|2x - 5| = 21$

Next, divide both sides by $3$ : $\frac{3|2x - 5|}{3} = \frac{21}{3}$ $|2x - 5| = 7$

Common Pitfall: Do not distribute a number into the absolute value bars, such as writing $|6x - 15|$ . Always isolate the expression $|ax + b|$ completely before splitting into cases.

Split into two linear cases

By definition, if $|u| = c$ where $c \ge 0$ , then $u = c$ or $u = -c$ , because numbers with an absolute value of $7$ lie exactly $7$ units away from $0$ on a number line (namely, $7$ and $-7$ ).

Case 1: $2x - 5 = 7$ $2x = 7 + 5$ $2x = 12$ $x = 6$
Case 2: $2x - 5 = -7$ $2x = -7 + 5$ $2x = -2$ $x = -1$

Verify the solutions

For $x = 6$ : $3|2(6) - 5| - 7 = 3|12 - 5| - 7 = 3(7) - 7 = 21 - 7 = 14 \quad \checkmark$
For $x = -1$ : $3|2(-1) - 5| - 7 = 3|-2 - 5| - 7 = 3|-7| - 7 = 3(7) - 7 = 14 \quad \checkmark$

Thus, the solutions are $x = 6$ or $x = -1$ .

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student ignores the negative case when splitting the absolute value equation, finding only the positive branch solution x = 6.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
3|2x - 5| = 21 $\implies$ |2x - 5| = 7
Step 2
2x - 5 = 7 $\implies$ 2x = 12
Correct
2x - 5 = 7 $\text{or}$ 2x - 5 = -7
When removing the absolute value bars from $|u| = c$ ( $c > 0$ ), you must set up two separate linear equations: $u = c$ and $u = -c$ . You only set up the positive branch $2x - 5 = 7$ and missed $2x - 5 = -7$ .
Step 3
x = 6
Correct
x = 6 $\text{or}$ x = -1
Because the negative case was omitted in the previous step, your final solution set only contains $x = 6$ instead of both $x = 6$ and $x = -1$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You isolated the absolute value term accurately by reversing the operations outside the bars.

You solved the linear equation for the positive branch with correct arithmetic to find $x = 6$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Omitted Negative Case in Absolute Value Equation

You dropped the absolute value bars directly without setting up the negative branch equation.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Absolute value represents geometric distance from zero on a number line. An expression whose distance is $7$ can equal either $7$ or $-7$ . Forgetting this geometric definition often leads to treating absolute value bars like standard grouping parentheses.

REASONING PATH

Your ACEpath teacher’s suggested approach

Two-Branch Rule: For any positive real number $c$ , $|u| = c \iff u = c \text{ or } u = -c$ .
As soon as the absolute value term is isolated, immediately write down two distinct equations linked by 'or' before solving either one.

CORRECTED PATH

ACEpath Teacher's Demonstration

Split the equation into its two possible cases: $2x - 5 = 7$ or $2x - 5 = -7$ .
Solve the positive case: $2x = 12 \implies x = 6$ .
Solve the negative case: $2x = -7 + 5 \implies 2x = -2 \implies x = -1$ .

Final answer

$x = 6 \text{ or } x = -1$

NEXT STEPS

Final Takeaways from Your Teacher

Review the number-line interpretation of absolute value to reinforce why two solutions typically exist.
Practice splitting $|u| = c$ immediately into both branches as a reflex step before performing any algebra on the inner expression.
Substitute both candidate solutions back into the original equation to verify that both make the statement true.

OverviewACEpath Teacher's Diagnosis

02 The student distributes the coefficient 3 into the absolute value without isolating it and incorrectly negates only the constant inside when setting up the second case.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
3|2x - 5| = 21
Step 2
|6x - 15| = 21
Step 3
6x - 15 = 21 $\text{ or }$ 6x + 15 = 21
Correct
6x - 15 = 21 $\text{ or }$ 6x - 15 = -21
By definition, $|u| = c$ splits into $u = c$ or $u = -c$ . You negated only the constant term $-15$ inside rather than the entire expression $-(6x - 15)$ or the target value $-21$ .
Step 4
6x = 36 $\text{ or }$ 6x = 6 $\implies$ x = 6 $\text{ or }$ x = 1
Correct
6x = 36 $\text{ or }$ 6x = -6 $\implies$ x = 6 $\text{ or }$ x = -1
You correctly solved the linear equations you wrote, but the second solution is incorrect due to the sign error in the setup.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly added $7$ to both sides as the first algebraic step.

You recognized that an absolute value equation splits into two distinct linear cases.

You solved the linear equation for the positive case $6x - 15 = 21$ accurately to find $x = 6$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Incorrect Case Splitting in Absolute Value Equation

You negated only the constant inside the bars ( $+15$ ) instead of setting the entire expression equal to the negative value ( $-21$ ).

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing the negation of an expression with altering a single sign inside the expression, rather than applying the definition $|u| = c \iff u = c \text{ or } u = -c$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Isolate First: Always isolate $|ax + b| = c$ completely by dividing out any outer multiplier before splitting.
Divide both sides of $3|2x - 5| = 21$ by $3$ to get $|2x - 5| = 7$ .
Split the Entire Quantity: If $|u| = c$ with $c \ge 0$ , the two branches are $u = c$ and $u = -c$ .
Keep the inside expression identical and negate the right-hand constant: $2x - 5 = 7$ or $2x - 5 = -7$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Set up the two branches using $|u| = c \iff u = c \text{ or } u = -c$ :
Case 1: $2x - 5 = 7 \implies 2x = 12 \implies x = 6$
Case 2: $2x - 5 = -7 \implies 2x = -2 \implies x = -1$

Final answer

$x = 6$ or $x = -1$

NEXT STEPS

Final Takeaways from Your Teacher

Divide by the outer coefficient before removing the absolute value bars to keep the numbers smaller and prevent sign slips.
When splitting into two cases, leave the expression that was inside the bars completely unchanged on both branches, and apply the $\pm$ sign exclusively to the right-hand side.
Substitute each final solution back into the original equation $3|2x - 5| - 7 = 14$ to verify.

OverviewACEpath Teacher's Diagnosis

03 The student removes the absolute value bars before isolating them by dropping the bars directly in the initial equation.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
3(2x - 5) - 7 = 14
Correct
3|2x - 5| = 21 $\implies$ |2x - 5| = 7
Absolute value bars cannot be directly replaced with parentheses. You need to isolate the absolute value expression $|2x - 5|$ first by adding $7$ to both sides and then dividing by $3$ .
Step 2
6x - 15 - 7 = 14 $\implies$ 6x - 22 = 14 $\implies$ 6x = 36
Correct
2x - 5 = 7 $\text{or}$ 2x - 5 = -7
Because the absolute value was removed prematurely, your algebraic simplification followed the wrong equation and missed the second negative case entirely.
Step 3
x = 6
Correct
x = 6 $\text{or}$ x = -1
Your final answer only captures one of the two solutions because the second case was omitted.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly expanded terms and solved a standard multi-step linear equation.

You performed arithmetic operations accurately when moving constants across the equals sign.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

dropping_absolute_value_bars

You treated the absolute value bars as ordinary parentheses instead of an operation measuring distance.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Conflating absolute value notation $|\cdot|$ with grouping parentheses $(\cdot)$ , which bypasses isolating the absolute value term and misses the negative branch of the definition $|u| = c \implies u = c \text{ or } u = -c$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Isolate Before Removing: Never distribute into or remove absolute value bars until $|ax + b|$ is alone on one side.
Use inverse operations to isolate $|2x - 5|$ so that $|2x - 5| = 7$ .
Split into Dual Cases: If $|u| = c$ and $c > 0$ , the equation splits into $u = c$ and $u = -c$ .
Set up and solve both $2x - 5 = 7$ and $2x - 5 = -7$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Isolate the absolute value term by adding $7$ to both sides: $3|2x - 5| = 21$ .
Divide both sides by $3$ : $|2x - 5| = 7$ .
Split into two cases: $2x - 5 = 7$ or $2x - 5 = -7$ .
Solve Case 1: $2x = 12 \implies x = 6$ .
Solve Case 2: $2x = -2 \implies x = -1$ .

Final answer

x = 6 $\text{ or }$ x = -1

NEXT STEPS

Final Takeaways from Your Teacher

Always isolate the absolute value term $|ax + b|$ completely before opening it up.
Remember that an isolated positive absolute value equation $|u| = c$ always produces two linear branches: $u = c$ and $u = -c$ .

Lock it in

Question

When an equation is in the form $|u| = c$ (with $c > 0$ ), what are the two algebraic equations you must write?

Answer

$u = c$ or $u = -c$ .

Principle: Absolute value measures distance from zero, so the inner expression can be located $c$ units in either the positive or negative direction.

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student ignores the negative case when splitting the absolute value equation, finding only the positive branch solution x = 6.

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 distributes the coefficient 3 into the absolute value without isolating it and incorrectly negates only the constant inside when setting up the second case.

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 removes the absolute value bars before isolating them by dropping the bars directly in the initial equation.

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

Similar question