Solve the equation $|2x - 6| + |x + 1| = 7$ for all real values of $x$ by partitioning the real line into distinct intervals to eliminate the absolute value bars.

Final answer

$x = 0$ or $x = 4$

Steps

Find the critical points

To solve the absolute value equation $|2x - 6| + |x + 1| = 7$ , we first identify the critical points where the expressions inside the absolute values change signs.

Set each expression inside the absolute value bars equal to zero: $2x - 6 = 0 \implies x = 3$ $x + 1 = 0 \implies x = -1$

These two critical points divide the real number line into three non-overlapping intervals:

$x < -1$
$-1 \le x < 3$
$x \ge 3$

Solve the equation in each interval

Case 1: $x < -1$

In this interval:

$2x - 6 < 0$ , so $|2x - 6| = -(2x - 6) = -2x + 6$
$x + 1 < 0$ , so $|x + 1| = -(x + 1) = -x - 1$

Substitute these into the original equation: $(-2x + 6) + (-x - 1) = 7$ $-3x + 5 = 7$ $-3x = 2$ $x = -\frac{2}{3}$

Now, check if $x = -\frac{2}{3}$ satisfies the condition for this case ( $x < -1$ ): Since $-\frac{2}{3} \approx -0.67 > -1$ , this value does not lie in the interval $x < -1$ . Therefore, there is no solution in this interval.

Case 2: $-1 \le x < 3$

In this interval:

$2x - 6 < 0$ , so $|2x - 6| = -(2x - 6) = -2x + 6$
$x + 1 \ge 0$ , so $|x + 1| = x + 1$

Substitute these into the equation: $(-2x + 6) + (x + 1) = 7$ $-x + 7 = 7$ $-x = 0$ $x = 0$

Check if $x = 0$ satisfies the condition $-1 \le x < 3$ : Since $-1 \le 0 < 3$ , this value is valid. Thus, $x = 0$ is a solution.

Case 3: $x \ge 3$

In this interval:

$2x - 6 \ge 0$ , so $|2x - 6| = 2x - 6$
$x + 1 > 0$ , so $|x + 1| = x + 1$

Substitute these into the equation: $(2x - 6) + (x + 1) = 7$ $3x - 5 = 7$ $3x = 12$ $x = 4$

Check if $x = 4$ satisfies the condition $x \ge 3$ : Since $4 \ge 3$ , this value is valid. Thus, $x = 4$ is a solution.

Combine all valid solutions

Collecting the valid solutions from all three cases gives: $x = 0 \quad \text{or} \quad x = 4$

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student solves each piecewise linear equation correctly but fails to check whether the resulting values lie within their respective defining intervals, erroneously including an extraneous solution.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
Critical points: $x = -1, x = 3$
Step 2
Interval 1: $x < -1 \implies -(2x - 6) - (x + 1) = 7 \implies -3x + 5 = 7 \implies -3x = 2 \implies x = -\frac{2}{3}$
Correct
For $x < -1$ : $-(2x - 6) - (x + 1) = 7 \implies -3x = 2 \implies x = -\frac{2}{3}$ . Since $-\frac{2}{3} > -1$ , this value does not lie in $x < -1$ , so there is no solution in this interval.
You solved the linear equation correctly, but you did not check whether $x = -\frac{2}{3}$ satisfies the defining interval condition $x < -1$ . Because $-\frac{2}{3} > -1$ , it must be rejected as extraneous.
Step 3
Interval 2: $-1 \le x < 3 \implies -(2x - 6) + (x + 1) = 7 \implies -x + 7 = 7 \implies x = 0$
Step 4
Interval 3: $x \ge 3 \implies (2x - 6) + (x + 1) = 7 \implies 3x - 5 = 7 \implies 3x = 12 \implies x = 4$
Step 5
x = - $\frac{2}{3}$ ,; x = 0,; x = 4
Correct
$x = 0$ or $x = 4$
Because $x = -\frac{2}{3}$ was not discarded in Interval 1, it was mistakenly carried into your final set of solutions.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly identified the boundary points where the expressions inside the absolute values equal zero.

You partitioned the real number line into the correct three intervals.

You accurately determined the sign of each expression to remove the absolute value bars in all three cases.

You executed the linear algebraic steps without any computational slips.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Extraneous Solution from Interval Domain Mismatch

none

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You treated each piecewise linear equation as an unrestricted equation, omitting the essential step of verifying whether the candidate solution lies inside the interval that defined that piecewise case.

REASONING PATH

Your ACEpath teacher’s suggested approach

Interval Verification Check: Any solution derived under a piecewise condition $x \in [a, b)$ is valid if and only if that value satisfies the condition $a \le x < b$ .](streamdown:incomplete-link)
Immediately compare the computed $x$ -value against the interval inequality before moving on to the next case. If it falls outside the interval, label it as extraneous and discard it.

CORRECTED PATH

ACEpath Teacher's Demonstration

Check whether $x = -\frac{2}{3}$ satisfies $x < -1$ .
Notice that $-\frac{2}{3} \approx -0.67$ , which is strictly greater than $-1$ , so $x = -\frac{2}{3}$ lies outside Interval 1.
Discard $x = -\frac{2}{3}$ as an extraneous root, yielding no solution from Interval 1.
Combine the valid solutions from Interval 2 ( $x = 0$ ) and Interval 3 ( $x = 4$ ).

Final answer

$x = 0$ or $x = 4$

NEXT STEPS

Final Takeaways from Your Teacher

Draw each candidate solution on a number line marked with your interval boundaries to visually confirm whether it lands within the active region.
Practice immediately writing a verification statement such as 'Check: does $x$ lie in the interval?' at the end of every piecewise branch.

OverviewACEpath Teacher's Diagnosis

02 The student misidentifies the sign of (2x - 6) on the middle interval -1 ≤ x < 3 as positive, leading to an incorrect equation and invalid solution.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
Critical points: $x = -1$ and $x = 3$
Step 2
For $x < -1$ : $-(2x - 6) - (x + 1) = 7 \implies -3x + 5 = 7 \implies x = -\frac{2}{3}$ (not in $x < -1$ , extraneous)
Step 3
For $x \ge 3$ : $(2x - 6) + (x + 1) = 7 \implies 3x - 5 = 7 \implies x = 4$
Step 4
For $-1 \le x < 3$ : $(2x - 6) + (x + 1) = 7 \implies 3x - 5 = 7 \implies 3x = 12 \implies x = 4$ (not in $[-1, 3)$ , extraneous)](streamdown:incomplete-link)
Correct
$-(2x - 6) + (x + 1) = 7 \implies -x + 7 = 7 \implies x = 0$ ; valid since $0 \in [-1, 3)$ ](streamdown:incomplete-link)
Because $x < 3$ , the expression $2x - 6$ is strictly negative on this interval. Removing the absolute value bars requires taking its opposite, $-(2x - 6) = -2x + 6$ , rather than leaving it as $2x - 6$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly found the critical boundary values where the absolute value expressions change sign.

You set up and evaluated the sign changes properly in the two outer intervals $x < -1$ and $x \ge 3$ .

You checked candidate solutions against interval boundary constraints to identify extraneous values.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Incorrect sign in piecewise absolute value removal

You treated $2x - 6$ as positive on $-1 \le x < 3$ , likely confusing which boundary point controls its sign.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Failing to test a test point or systematically check whether $x$ lies to the left or right of a specific critical value when evaluating piecewise signs across multiple intervals.

REASONING PATH

Your ACEpath teacher’s suggested approach

Evaluate Expression Signs with Test Values: For any interval $(a, b)$ , pick a convenient test number $x_0$ inside the interval and evaluate the sign of each linear term.
On $[-1, 3)$ , pick $x = 0$ . Since $2(0) - 6 = -6 < 0$ , substitute $|2x - 6| = -(2x - 6) = -2x + 6$ .](streamdown:incomplete-link)

CORRECTED PATH

ACEpath Teacher's Demonstration

For $x \in [-1, 3)$ , we have $x + 1 \ge 0$ and $2x - 6 < 0$ .](streamdown:incomplete-link)
Replace the absolute values accordingly: $-(2x - 6) + (x + 1) = 7$ .
Expand and simplify: $-2x + 6 + x + 1 = 7 \implies -x + 7 = 7$ .
Solve for $x$ : $-x = 0 \implies x = 0$ .
Check the condition: $-1 \le 0 < 3$ is true, so $x = 0$ is a valid solution.

Final answer

$x = 0$ or $x = 4$

NEXT STEPS

Final Takeaways from Your Teacher

Draw a sign chart or number line with critical points labeled to track the sign of each individual factor before combining them.
Plug a simple number like $x = 0$ into each term whenever working on an interval containing zero to verify signs before solving.
State the combined solution set explicitly after all individual case intervals have been analyzed.

OverviewACEpath Teacher's Diagnosis

03 The student incorrectly distributes the negative sign to the binomial (2x - 6), writing -2x - 6 instead of -2x + 6.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
Boundary values: $x = -1, x = 3$
Step 2
Case 1: $x < -1 \implies -2x - 6 - (x + 1) = 7 \implies -3x - 7 = 7 \implies -3x = 14 \implies x = -\frac{14}{3}$ ; $-\frac{14}{3} < -1$ , so $x = -\frac{14}{3}$ is a solution.
Correct
For $x < -1$ , $|2x - 6| = -(2x - 6) = -2x + 6$ and $|x + 1| = -(x + 1) = -x - 1$ , giving $(-2x + 6) + (-x - 1) = 7 \implies -3x + 5 = 7 \implies x = -\frac{2}{3}$ . Since $-\frac{2}{3} \not< -1$ , there is no solution in this interval.
When removing the absolute value bars from $|2x - 6|$ when $2x - 6 < 0$ , you must negate the entire binomial: $-(2x - 6) = -2x + 6$ . Writing $-2x - 6$ led to an extraneous solution.
Step 3
Case 2: $-1 \le x < 3 \implies -2x - 6 + x + 1 = 7 \implies -x - 5 = 7 \implies -x = 12 \implies x = -12$ ; $-12$ is not in $[-1, 3)$ , reject.](streamdown:incomplete-link)
Correct
For $-1 \le x < 3$ , $|2x - 6| = -(2x - 6) = -2x + 6$ and $|x + 1| = x + 1$ , giving $(-2x + 6) + (x + 1) = 7 \implies -x + 7 = 7 \implies x = 0$ . Since $0 \in [-1, 3)$ , $x = 0$ is a valid solution.](streamdown:incomplete-link)
You repeated the same sign error on $|2x - 6|$ , writing $-2x - 6$ instead of $-2x + 6$ . This caused you to miss the valid solution $x = 0$ .
Step 4
Case 3: $x \ge 3 \implies 2x - 6 + x + 1 = 7 \implies 3x - 5 = 7 \implies 3x = 12 \implies x = 4$ ; $4 \ge 3$ , so $x = 4$ is a solution.
Step 5
$x = -\frac{14}{3},\ x = 4$
Correct
$x = 0$ or $x = 4$
Because of the sign distribution errors in Cases 1 and 2, your final solution set included an extraneous solution and missed $x = 0$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly identified the critical transition points $x = -1$ and $x = 3$ .

You set up a sound interval partition method covering all three cases.

You solved Case 3 accurately and verified your solution against the interval condition.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Distributive Sign Error

You wrote $-(2x - 6)$ as $-2x - 6$ by only negating the first term instead of distributing the negative sign across both terms.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Incomplete distribution of the negative sign when expanding $-(a - b)$ . When an expression inside absolute value bars is negative, wrapping it in parentheses before negating prevents forgetting to change the sign of the constant term.

REASONING PATH

Your ACEpath teacher’s suggested approach

Wrap in parentheses first: If an expression $E < 0$ , then $|E| = -(E)$ . Always write parentheses around $E$ before simplifying.
Write $|2x - 6| = -(2x - 6)$ explicitly before expanding it into $-2x + 6$ .
Distribute negative signs to every term: $-(a - b) = -a + b$ .
Check that the negative sign flips the sign of every single term inside the parentheses.

CORRECTED PATH

ACEpath Teacher's Demonstration

In Case 1 ( $x < -1$ ): $(-2x + 6) + (-x - 1) = 7 \implies -3x + 5 = 7 \implies x = -\frac{2}{3}$ . Since $-\frac{2}{3} > -1$ , reject it.
In Case 2 ( $-1 \le x < 3$ ): $(-2x + 6) + (x + 1) = 7 \implies -x + 7 = 7 \implies x = 0$ . Since $-1 \le 0 < 3$ , accept $x = 0$ .

Final answer

$x = 0$ or $x = 4$

NEXT STEPS

Final Takeaways from Your Teacher

Practice rewriting expressions of the form $-(ax - b)$ by explicitly writing out $-ax - (-b) = -ax + b$ .
Always substitute your candidate solutions back into the original absolute value equation to catch arithmetic or sign errors.

Lock it in

Question

Why must you check whether a solution falls inside its defining interval when solving absolute value equations piecewise?

Answer

The linear equation is only equivalent to the absolute value equation within that specific interval. Outside that interval, the expressions have different signs, making the root extraneous.

Resources

Answer and steps

Final answer

Steps

Case 1: x<−1x < -1x<−1

Case 2: −1≤x<3-1 \le x < 3−1≤x<3

Case 3: x≥3x \ge 3x≥3

Typical errors

01 The student solves each piecewise linear equation correctly but fails to check whether the resulting values lie within their respective defining intervals, erroneously including an extraneous 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 misidentifies the sign of (2x - 6) on the middle interval -1 ≤ x < 3 as positive, leading to an incorrect equation and invalid solution.

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 incorrectly distributes the negative sign to the binomial (2x - 6), writing -2x - 6 instead of -2x + 6.

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

Case 1: $x < -1$

Case 2: $-1 \le x < 3$

Case 3: $x \ge 3$