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Find all real solutions to the equation ∣3x−8∣=2x−7|3x - 8| = 2x - 7. Check for extraneous solutions and state whether each candidate value is valid.

Answer and steps

Final answer

No real solutions (empty set, ∅\emptyset)

Steps

01

Understand the non-negativity condition

To solve the absolute value equation ∣3x−8∣=2x−7|3x - 8| = 2x - 7, we first determine the candidate solutions by splitting the equation into two cases based on the definition of absolute value, and then check for extraneous solutions.

Since the absolute value of any real expression is always non-negative (∣A∣≥0|A| \ge 0), the right-hand side must also be non-negative for a real solution to exist: 2x−7≥0  ⟹  x≥72=3.52x - 7 \ge 0 \implies x \ge \frac{7}{2} = 3.5 Any candidate value that does not satisfy x≥3.5x \ge 3.5 will produce a negative value on the right-hand side and will be extraneous.

02

Split into two linear cases

By definition, ∣A∣=B|A| = B means A=BA = B or A=−BA = -B.

Case 1: 3x−8=2x−73x - 8 = 2x - 7 Subtract 2x2x from both sides: x−8=−7x - 8 = -7 Add 88 to both sides: x=1x = 1

Case 2: 3x−8=−(2x−7)3x - 8 = -(2x - 7) Distribute the negative sign on the right: 3x−8=−2x+73x - 8 = -2x + 7 Add 2x2x to both sides: 5x−8=75x - 8 = 7 Add 88 to both sides: 5x=155x = 15 Divide by 55: x=3x = 3

Our candidate values are x=1x = 1 and x=3x = 3.

03

Check candidate solutions in the original equation

  • Test x=1x = 1: LHS=∣3(1)−8∣=∣3−8∣=∣−5∣=5\text{LHS} = |3(1) - 8| = |3 - 8| = |-5| = 5 RHS=2(1)−7=2−7=−5\text{RHS} = 2(1) - 7 = 2 - 7 = -5 Since 5≠−55 \ne -5, the candidate x=1x = 1 is extraneous (invalid).
  • Test x=3x = 3: LHS=∣3(3)−8∣=∣9−8∣=∣1∣=1\text{LHS} = |3(3) - 8| = |9 - 8| = |1| = 1 RHS=2(3)−7=6−7=−1\text{RHS} = 2(3) - 7 = 6 - 7 = -1 Since 1≠−11 \ne -1, the candidate x=3x = 3 is extraneous (invalid).

Conclusion

Both candidate values lead to a negative value on the right-hand side, contradicting the non-negative nature of the absolute value function.

Therefore, there are no real solutions, and the solution set is the empty set: ∅\emptyset

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 A student solves the algebraic cases correctly but forgets that the absolute value output must be non-negative, incorrectly accepting both candidate values.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    3x - 8 = 2x - 7   or\text{or}   3x - 8 = -(2x - 7)

  2. Step 2

    3x - 2x = 8 - 7   ⟹  \implies x = 1

  3. Step 3

    3x - 8 = -2x + 7   ⟹  \implies 5x = 15   ⟹  \implies x = 3

  4. Step 4

    Both candidate values \text{Both candidate values } x = 1  and \text{ and } x = 3  are valid solutions.\text{ are valid solutions.}

    Correct

    Check \text{Check } x = 1: |3(1) - 8| = 5 \ne -5.   Check \text{Check } x = 3: |3(3) - 8| = 1 \ne -1.   ⟹  \implies No real solutions \text{No real solutions } (\emptyset)

    Because an absolute value cannot equal a negative number, the right-hand side 2x−72x - 7 must be greater than or equal to 00 (x≥3.5x \ge 3.5). Substituting x=1x = 1 yields ∣−5∣=−5|-5| = -5 and substituting x=3x = 3 yields ∣1∣=−1|1| = -1, both of which are false. Both candidates are extraneous.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly set up the two algebraic cases using the definition of absolute value.

You accurately solved both linear equations to find the candidate values x=1x = 1 and x=3x = 3.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Omission of Extraneous Solution Check

You stopped after finding the algebraic solutions without checking whether they make the right-hand side negative.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Treating ∣A∣=B|A| = B as strictly equivalent to A=B or A=−BA = B \text{ or } A = -B without enforcing the domain restriction B≥0B \ge 0, leading to the acceptance of extraneous roots.

REASONING PATH
Your ACEpath teacher’s suggested approach
Non-negativity requirement

  • For any real expression, ∣A∣≥0|A| \ge 0, which implies that in ∣A∣=B|A| = B, a real solution exists only if B≥0B \ge 0.

  • Before or after solving, evaluate the sign of the non-absolute-value side for each candidate.

Direct substitution check

  • Every candidate value obtained from splitting an absolute value equation must satisfy the original equation.

  • Substitute each candidate back into both sides of the original equation to verify equality.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Evaluate the right-hand side condition: 2x−7≥0  ⟹  x≥72=3.52x - 7 \ge 0 \implies x \ge \frac{7}{2} = 3.5.

  2. Compare candidates to the condition: neither 1≥3.51 \ge 3.5 nor 3≥3.53 \ge 3.5 holds.

  3. Substitute to confirm: for x=1x = 1, LHS=5\text{LHS} = 5 and RHS=−5\text{RHS} = -5; for x=3x = 3, LHS=1\text{LHS} = 1 and RHS=−1\text{RHS} = -1.

Final answer

No real solutions (empty set, ∅\emptyset)

NEXT STEPS
Final Takeaways from Your Teacher

  • State the domain requirement B≥0B \ge 0 at the very beginning when given an equation of the form ∣A∣=B(x)|A| = B(x).

  • Always substitute every candidate back into the original expression to catch negative outputs.

OverviewACEpath Teacher's Diagnosis

02 A student incorrectly evaluates the right-hand side when testing x = 3 by dropping the negative sign, leading them to falsely conclude x = 3 is a valid solution.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    3x - 8 = 2x - 7   ⟹  \implies x = 1  and\text{and}  3x - 8 = -(2x - 7)   ⟹  \implies 5x = 15   ⟹  \implies x = 3

  2. Step 2

    Check \text{Check } x = 1:\ |3(1) - 8| = 5,\ 2(1) - 7 = -5,  so \text{ so } x = 1  is extraneous\text{ is extraneous}

  3. Step 3

    Check \text{Check } x = 3:\ |3(3) - 8| = 1,\ 2(3) - 7 = 1,\ |1| = 1,  so \text{ so } x = 3  is a valid solution.\text{ is a valid solution.}

    Correct

    Check \text{Check } x = 3:\ |3(3) - 8| = |1| = 1,\ 2(3) - 7 = 6 - 7 = -1.  Since \text{ Since } 1 \ne -1, x = 3  is extraneous, leaving no real solutions \text{ is extraneous, leaving no real solutions } (\emptyset).

    When checking x=3x = 3 on the right-hand side, 2(3)−7=6−7=−12(3) - 7 = 6 - 7 = -1, not 11. Because the left-hand side is 11 and the right-hand side is −1-1, x=3x = 3 is also extraneous, meaning the equation has no real solutions.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly split the absolute value equation into the two algebraic cases A=BA = B and A=−BA = -B.

You solved both linear equations accurately to find the candidate values x=1x = 1 and x=3x = 3.

You remembered to check your candidate solutions back in the original equation to test for extraneous roots.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Arithmetic Evaluation Error in Extraneous Check

You computed 2(3)−72(3) - 7 as 11 instead of 6−7=−16 - 7 = -1, likely subtracting 66 from 77 rather than evaluating 6−76 - 7.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

A basic integer subtraction slip when evaluating the right-hand side caused a false match (1=11 = 1), obscuring the fact that the right-hand side is negative and therefore cannot equal an absolute value.

REASONING PATH
Your ACEpath teacher’s suggested approach
Integer Subtraction Sign

  • When subtracting a larger number from a smaller number, the result is always negative: a−b=−(b−a)a - b = -(b - a) when b>ab > a.

  • Write out the intermediate multiplication step explicitly, 6−76 - 7, before subtracting to verify the negative sign.

Non-Negativity Domain Check

  • For ∣A∣=B|A| = B to hold, BB must satisfy B≥0B \ge 0 because absolute values are never negative.

  • Check the domain restriction 2x−7≥0  ⟹  x≥3.52x - 7 \ge 0 \implies x \ge 3.5 at the start to immediately see that neither x=1x = 1 nor x=3x = 3 can be valid.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Evaluate the left-hand side: ∣3(3)−8∣=∣9−8∣=∣1∣=1|3(3) - 8| = |9 - 8| = |1| = 1.

  2. Evaluate the right-hand side: 2(3)−7=6−7=−12(3) - 7 = 6 - 7 = -1.

  3. Compare both sides: 1≠−11 \ne -1, so x=3x = 3 is extraneous.

  4. Since both x=1x = 1 and x=3x = 3 are extraneous, there are no valid real solutions.

Final answer

No real solutions (empty set, ∅\emptyset)

NEXT STEPS
Final Takeaways from Your Teacher

  • Take an extra second to write down intermediate steps during integer subtraction, especially when the subtrahend is larger than the minuend.

  • Establish the non-negativity constraint B≥0B \ge 0 as Step 0 before solving ∣A∣=B|A| = B to quickly filter out invalid candidates.

OverviewACEpath Teacher's Diagnosis

03 A student only negates the first term when setting up the second case as -2x - 7, yielding an incorrect second candidate and faulty validation.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Partially correct
  1. Step 1

    Case 1: 3x−8=2x−7  ⟹  x=13x - 8 = 2x - 7 \implies x = 1

  2. Step 2

    Case 2: 3x−8=−2x−7  ⟹  5x=1  ⟹  x=153x - 8 = -2x - 7 \implies 5x = 1 \implies x = \frac{1}{5}

    Correct

    3x−8=−(2x−7)  ⟹  3x−8=−2x+7  ⟹  5x=15  ⟹  x=33x - 8 = -(2x - 7) \implies 3x - 8 = -2x + 7 \implies 5x = 15 \implies x = 3

    When setting up the negative case for ∣A∣=B|A| = B, you must negate the entire right-hand side using parentheses: −(2x−7)=−2x+7-(2x - 7) = -2x + 7. Because the negative sign was not distributed to −7-7, you solved 5x=15x = 1 and got x=15x = \frac{1}{5} instead of x=3x = 3.

  3. Step 3

    Check x=1x = 1: ∣3−8∣=5|3 - 8| = 5, 2−7=−52 - 7 = -5, extraneous   ⟹  \implies Check x=15x = \frac{1}{5}: ∣35−8∣=∣−375∣=375|\frac{3}{5} - 8| = |-\frac{37}{5}| = \frac{37}{5}, 25−7=−335\frac{2}{5} - 7 = -\frac{33}{5}, extraneous

    Correct

    Check x=1x = 1: ∣3(1)−8∣=5≠−5|3(1) - 8| = 5 \ne -5 (extraneous)   ⟹  \implies Check x=3x = 3: ∣3(3)−8∣=1≠−1|3(3) - 8| = 1 \ne -1 (extraneous)

    Your verification method is completely sound and you correctly identified that x=1x = 1 is extraneous. However, testing x=15x = \frac{1}{5} was affected by the algebraic error in Case 2; you needed to test the true candidate x=3x = 3.

  4. Step 4

    Neither candidate value is valid, no real solutions.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly identified the two-case approach required to solve an absolute value equation.

You solved the linear equation in Case 1 accurately to find x=1x = 1.

You remembered to check candidate solutions back into the original equation to identify extraneous solutions.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Sign distribution error across parentheses

You wrote −2x−7-2x - 7 instead of −(2x−7)-(2x - 7) when setting up the second case.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Failing to treat a multi-term expression as a grouped unit when taking its opposite. In ∣A∣=B|A| = B, the negative branch requires A=−(B)A = -(B), which demands distributing the negative sign across every term in BB.

REASONING PATH
Your ACEpath teacher’s suggested approach
Enclose Expressions in Parentheses

  • When negating an algebraic expression with multiple terms, always wrap it in parentheses: −(ax+b)=−ax−b-(ax + b) = -ax - b.

  • Write out the parentheses explicitly in Case 2 before simplifying.

Check the Non-Negativity Domain

  • Because ∣A∣≥0|A| \ge 0, any real solution must satisfy B≥0B \ge 0.

  • Check whether candidate values satisfy 2x−7≥0  ⟹  x≥3.52x - 7 \ge 0 \implies x \ge 3.5 as a quick filter for extraneous solutions.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Set up Case 2 with parentheses: 3x−8=−(2x−7)3x - 8 = -(2x - 7).

  2. Distribute the negative sign: 3x−8=−2x+73x - 8 = -2x + 7.

  3. Add 2x2x and 88 to both sides: 5x=15  ⟹  x=35x = 15 \implies x = 3.

  4. Check x=3x = 3 in the original equation: ∣3(3)−8∣=∣1∣=1|3(3) - 8| = |1| = 1, but 2(3)−7=−12(3) - 7 = -1. Since 1≠−11 \ne -1, x=3x = 3 is extraneous.

Final answer

No real solutions (∅\emptyset)

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice distributing negative signs across binomials with explicit grouping parentheses.

  • Form the habit of establishing the domain condition B≥0B \ge 0 whenever solving ∣A∣=B|A| = B.

Lock it in

Question

For an equation of the form ∣A∣=B|A| = B, what condition must BB satisfy for any real solution to exist?

Answer

B≥0B \ge 0. Because the absolute value of any real number is non-negative, the expression it equals cannot be negative.

Similar question

01

Find all real solutions to ∣2x−5∣=x−4|2x - 5| = x - 4. Check for extraneous solutions.

Answer

x=1x = 1 and x=3x = 3 are extraneous candidate values; there are no real solutions (∅\emptyset). Condition: x−4≥0  ⟹  x≥4x - 4 \ge 0 \implies x \ge 4. Case 1: 2x−5=x−4  ⟹  x=1<42x - 5 = x - 4 \implies x = 1 < 4. Case 2: 2x−5=−(x−4)  ⟹  3x=9  ⟹  x=3<42x - 5 = -(x - 4) \implies 3x = 9 \implies x = 3 < 4.

02

Find all real solutions to ∣4x+3∣=3x+8|4x + 3| = 3x + 8. Check for extraneous solutions.

Answer

x=5x = 5 and x=−117x = -\frac{11}{7}. Both candidates are valid solutions. Condition: 3x+8≥0  ⟹  x≥−83≈−2.673x + 8 \ge 0 \implies x \ge -\frac{8}{3} \approx -2.67. Case 1: 4x+3=3x+8  ⟹  x=54x + 3 = 3x + 8 \implies x = 5. Case 2: 4x+3=−3x−8  ⟹  7x=−11  ⟹  x=−117≈−1.574x + 3 = -3x - 8 \implies 7x = -11 \implies x = -\frac{11}{7} \approx -1.57. Both satisfy x≥−2.67x \ge -2.67.

03

Find all real solutions to ∣x2−6∣=5x|x^2 - 6| = 5x. Check for extraneous solutions.

Answer

x=1x = 1 and x=6x = 6 are the only valid real solutions. Condition: 5x≥0  ⟹  x≥05x \ge 0 \implies x \ge 0. Case 1: x2−5x−6=0  ⟹  (x−6)(x+1)=0  ⟹  x=6x^2 - 5x - 6 = 0 \implies (x - 6)(x + 1) = 0 \implies x = 6 (valid) or x=−1x = -1 (extraneous since −1<0-1 < 0). Case 2: x2+5x−6=0  ⟹  (x+6)(x−1)=0  ⟹  x=1x^2 + 5x - 6 = 0 \implies (x + 6)(x - 1) = 0 \implies x = 1 (valid) or x=−6x = -6 (extraneous since −6<0-6 < 0).

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