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Consider the equation ∣4x−7∣=13|4x - 7| = 13 with solutions x1x_1 and x2x_2 such that x1<x2x_1 < x_2. Solve for the midpoint M=x1+x22M = \frac{x_1 + x_2}{2} and the absolute distance D=∣x2−x1∣D = |x_2 - x_1| between the two solutions.

Answer and steps

Final answer

M=74M = \frac{7}{4}, D=132D = \frac{13}{2}

Steps

To solve the absolute value equation and find the required midpoint and distance, we can split the equation into two linear cases or use the geometric meaning of absolute value.

Method 1: Algebraic Solution

Step 1: Set up the two linear equations. The equation ∣A∣=c|A| = c (where c>0c > 0) means that A=cA = c or A=−cA = -c. Applying this to ∣4x−7∣=13|4x - 7| = 13:

  1. 4x−7=134x - 7 = 13
  2. 4x−7=−134x - 7 = -13

Step 2: Solve each linear equation.

  • For the first equation: 4x−7=134x - 7 = 13 4x=204x = 20 x=5x = 5
  • For the second equation: 4x−7=−134x - 7 = -13 4x=−64x = -6 x=−64=−32x = -\frac{6}{4} = -\frac{3}{2}

Given that x1<x2x_1 < x_2, we have: x1=−32,x2=5x_1 = -\frac{3}{2}, \quad x_2 = 5

Step 3: Calculate the midpoint MM. The midpoint formula is M=x1+x22M = \frac{x_1 + x_2}{2}: M=−32+52=722=74M = \frac{-\frac{3}{2} + 5}{2} = \frac{\frac{7}{2}}{2} = \frac{7}{4}

Step 4: Calculate the absolute distance DD. The distance between the two solutions is D=∣x2−x1∣D = |x_2 - x_1|: D=5−(−32)=102+32=132D = 5 - \left(-\frac{3}{2}\right) = \frac{10}{2} + \frac{3}{2} = \frac{13}{2}

Alternative Geometric Insight

Factoring 44 out of the absolute value expression gives: 4∣x−74∣=13  ⟹  ∣x−74∣=1344 \left|x - \frac{7}{4}\right| = 13 \implies \left|x - \frac{7}{4}\right| = \frac{13}{4}

In geometric terms, ∣x−c∣=r|x - c| = r represents points xx at a distance of rr from a center cc on the number line:

  • The center is the midpoint: M=74M = \frac{7}{4}.
  • The total distance between the two points is 2r=2⋅134=1322r = 2 \cdot \frac{13}{4} = \frac{13}{2}.
Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets to negate the right-hand side for the negative case and instead negates only the constant term inside the absolute value, leading to incorrect solutions.

Sample Student Work

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

    4x - 7 = 13   ⟹  \implies 4x = 20   ⟹  \implies x = 5

  2. Step 2

    4x + 7 = 13   ⟹  \implies 4x = 6   ⟹  \implies x = 32\frac{3}{2}

    Correct

    4x - 7 = -13   ⟹  \implies 4x = -6   ⟹  \implies x = -32\frac{3}{2}

    To solve ∣A∣=c|A| = c, the negative branch must negate the entire expression, −(4x−7)=13-(4x - 7) = 13, or equivalently negate the constant on the right, 4x−7=−134x - 7 = -13. Writing 4x+7=134x + 7 = 13 only changed the sign of the constant term inside the bars, which gave an incorrect root x=32x = \frac{3}{2} instead of x=−32x = -\frac{3}{2}.

  3. Step 3

    x_1 = 32\frac{3}{2},\ x_2 = 5   ⟹  \implies M = 32+52\frac{\frac{3}{2} + 5}{2} = 1322\frac{\frac{13}{2}}{2} = 134\frac{13}{4}

    Correct

    x_1 = -32\frac{3}{2},\ x_2 = 5   ⟹  \implies M = −32+52\frac{-\frac{3}{2} + 5}{2} = 722\frac{\frac{7}{2}}{2} = 74\frac{7}{4}

    Your arithmetic for the midpoint was executed correctly, but the result was affected by the incorrect value of x1x_1.

  4. Step 4

    D=∣5−32∣=72D = \left|5 - \frac{3}{2}\right| = \frac{7}{2}

    Correct

    D=∣5−(−32)∣=5+32=132D = \left|5 - \left(-\frac{3}{2}\right)\right| = 5 + \frac{3}{2} = \frac{13}{2}

    Your distance calculation was methodologically correct, but it produced an incorrect numerical result because it relied on x1=32x_1 = \frac{3}{2}.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly solved the linear equation 4x−7=134x - 7 = 13 to obtain x=5x = 5.

You correctly substituted your roots into both the midpoint formula and the distance formula.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Sign error in splitting absolute value

You changed −7-7 to +7+7 inside the expression while leaving 4x4x positive, rather than negating the entire expression −(4x−7)-(4x - 7) or negating the right-hand side to −13-13.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Confusing the absolute value case split ∣A∣=c  ⟺  A=c or A=−c|A| = c \iff A = c \text{ or } A = -c with changing the sign of an individual term within AA.

REASONING PATH
Your ACEpath teacher’s suggested approach
Branch Rule

  • For any real expression AA and constant c≥0c \ge 0, ∣A∣=c|A| = c means A=cA = c or A=−cA = -c.

  • Always keep the entire inner expression intact and place the ±\pm sign on the opposite side of the equation.

Geometric Check

  • In ∣a(x−h)∣=c|a(x - h)| = c, the midpoint between solutions is always hh, and the distance between them is 2c∣a∣\frac{2c}{|a|}.

  • Factor out 44 to write 4∣x−74∣=134\left|x - \frac{7}{4}\right| = 13, immediately revealing the midpoint M=74M = \frac{7}{4}.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Set up the second case by negating the right-hand side: 4x−7=−134x - 7 = -13.

  2. Add 77 to both sides: 4x=−6  ⟹  x=−64=−324x = -6 \implies x = -\frac{6}{4} = -\frac{3}{2}.

  3. Identify x1=−32x_1 = -\frac{3}{2} and x2=5x_2 = 5 since −32<5-\frac{3}{2} < 5.

  4. Compute the midpoint: M=−32+52=74M = \frac{-\frac{3}{2} + 5}{2} = \frac{7}{4}.

  5. Compute the distance: D=∣5−(−32)∣=132D = \left|5 - \left(-\frac{3}{2}\right)\right| = \frac{13}{2}.

Final answer

M = 74\frac{7}{4},\ D = 132\frac{13}{2}

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice splitting absolute value equations by explicitly writing A=cA = c and A=−cA = -c before moving any terms.

  • Use the geometric form ∣x−M∣=D2|x - M| = \frac{D}{2} to quickly verify the midpoint and total distance.

OverviewACEpath Teacher's Diagnosis

02 The student subtracts instead of adding 7 when solving the negative case 4x - 7 = -13, resulting in an erroneous x₁.

Sample Student Work

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

    4x−7=13  ⟹  4x=20  ⟹  x2=54x - 7 = 13 \implies 4x = 20 \implies x_2 = 5

  2. Step 2

    4x−7=−13  ⟹  4x=−20  ⟹  x1=−54x - 7 = -13 \implies 4x = -20 \implies x_1 = -5

    Correct

    4x−7=−13  ⟹  4x=−6  ⟹  x1=−324x - 7 = -13 \implies 4x = -6 \implies x_1 = -\frac{3}{2}

    When isolating 4x4x in 4x−7=−134x - 7 = -13, you must add 77 to both sides. Computing −13+7-13 + 7 gives −6-6, not −20-20.

  3. Step 3

    M=−5+52=0M = \frac{-5 + 5}{2} = 0

    Correct

    M=−32+52=74M = \frac{-\frac{3}{2} + 5}{2} = \frac{7}{4}

    Your midpoint formula setup is structurally correct, but the result is incorrect due to your earlier calculation of x1x_1.

  4. Step 4

    D=∣5−(−5)∣=10D = |5 - (-5)| = 10

    Correct

    D=∣5−(−32)∣=132D = \left|5 - \left(-\frac{3}{2}\right)\right| = \frac{13}{2}

    Your distance calculation is structurally correct, but it inherits the erroneous value of x1x_1 from Step 2.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly split the absolute value equation into its positive and negative linear branches.

You solved the positive branch 4x−7=134x - 7 = 13 completely and accurately.

You correctly applied the formulas for the midpoint MM and absolute distance DD using your calculated roots.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Inverse Operation Sign Error

You subtracted 77 from −13-13 instead of adding 77, or duplicated the value 2020 with a negative sign.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Losing track of inverse operations when solving an equation with a negative constant on the right-hand side, leading to performing the same sign operation rather than undoing subtraction with addition.

REASONING PATH
Your ACEpath teacher’s suggested approach
Undo Subtraction with Addition

  • To eliminate −c- c from the left side of an equation, always add +c+ c to both sides: −k+c-k + c, not −k−c-k - c.

  • Write the step explicitly: 4x=−13+7=−64x = -13 + 7 = -6 before dividing.

Geometric Midpoint Verification

  • For ∣ax−b∣=c|ax - b| = c, the center (midpoint) depends only on ax−b=0ax - b = 0, giving M=baM = \frac{b}{a}.

  • Quickly verify that M=74M = \frac{7}{4}; if your roots do not average to 74\frac{7}{4}, recheck your algebra.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Add 77 to both sides: 4x=−13+7=−64x = -13 + 7 = -6.

  2. Divide by 44: x1=−64=−32x_1 = -\frac{6}{4} = -\frac{3}{2}.

  3. Compute the midpoint: M=−32+52=722=74M = \frac{-\frac{3}{2} + 5}{2} = \frac{\frac{7}{2}}{2} = \frac{7}{4}.

  4. Compute the distance: D=5−(−32)=102+32=132D = 5 - \left(-\frac{3}{2}\right) = \frac{10}{2} + \frac{3}{2} = \frac{13}{2}.

Final answer

M=74,D=132M = \frac{7}{4},\quad D = \frac{13}{2}

NEXT STEPS
Final Takeaways from Your Teacher

  • Explicitly write out the addition step when working with negative integers: −13+7=−6-13 + 7 = -6.

  • Use the geometric property ∣x−M∣=r|x - M| = r to check your midpoint directly from 4x−7=0  ⟹  x=744x - 7 = 0 \implies x = \frac{7}{4}.

OverviewACEpath Teacher's Diagnosis

03 The student divides the sum by 4 instead of 2 when computing the midpoint, confusing the midpoint formula with the coefficient of x.

Sample Student Work

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

    4x - 7 = 13  or \text{ or } 4x - 7 = -13   ⟹  \implies 4x = 20  or \text{ or } 4x = -6   ⟹  \implies x_2 = 5,; x_1 = -32\frac{3}{2}

  2. Step 2

    x_1 + x_2 = 72\frac{7}{2}

  3. Step 3

    M=(72)4=78M = \frac{\left(\frac{7}{2}\right)}{4} = \frac{7}{8}

    Correct

    M = x1+x22\frac{x_1 + x_2}{2} = 722\frac{\frac{7}{2}}{2} = 74\frac{7}{4}

    The midpoint of two numbers is their arithmetic average, which requires dividing the sum by 22, not by the leading coefficient 44.

  4. Step 4

    D=∣5−(−32)∣=132D = \left|5 - \left(-\frac{3}{2}\right)\right| = \frac{13}{2}

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly set up and solved both linear cases for the absolute value equation.

You accurately ordered the roots and computed their sum.

You correctly computed the absolute distance between the solutions.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Formula Substitution Error

You placed 44 in the denominator of the midpoint calculation instead of 22.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

You confused the constant denominator 22 in the definition of a midpoint with the leading coefficient 44 of the original equation.

REASONING PATH
Your ACEpath teacher’s suggested approach
Midpoint Definition Check

  • The midpoint of any two numbers aa and bb is always given by a+b2\frac{a + b}{2}, regardless of equation coefficients.

  • Always write the generic formula M=x1+x22M = \frac{x_1 + x_2}{2} first before plugging in numbers.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Substitute the sum into the standard midpoint formula: M=x1+x22=722M = \frac{x_1 + x_2}{2} = \frac{\frac{7}{2}}{2}

  2. Simplify the compound fraction: M=74M = \frac{7}{4}

Final answer

M = 74\frac{7}{4},  D = 132\frac{13}{2}

NEXT STEPS
Final Takeaways from Your Teacher

  • Review the geometric definition of a midpoint as the arithmetic mean of two coordinates.

  • Double-check that variable coefficients from the original equation do not inadvertently replace fixed constants in standard formulas.

Lock it in

Question

How do you split an equation of the form ∣A∣=c|A| = c (where c>0c > 0) into two linear equations?

Answer

Set A=cA = c or A=−cA = -c. Keep the entire inner expression AA exactly as it is and apply the negative sign to the other side.

Similar question

01

Solve the equation ∣3x−5∣=16|3x - 5| = 16 for its two solutions x1x_1 and x2x_2 (x1<x2x_1 < x_2), then find their midpoint MM and distance DD.

Answer

Case 1: 3x−5=16  ⟹  3x=21  ⟹  x=73x - 5 = 16 \implies 3x = 21 \implies x = 7. Case 2: 3x−5=−16  ⟹  3x=−11  ⟹  x=−1133x - 5 = -16 \implies 3x = -11 \implies x = -\frac{11}{3}. With x1=−113x_1 = -\frac{11}{3} and x2=7x_2 = 7: M=−113+72=53M = \frac{-\frac{11}{3} + 7}{2} = \frac{5}{3} D=7−(−113)=323D = 7 - \left(-\frac{11}{3}\right) = \frac{32}{3}.

02

Solve the equation ∣2−5x∣=18|2 - 5x| = 18 for its two solutions x1x_1 and x2x_2 (x1<x2x_1 < x_2), then find their midpoint MM and distance DD.

Answer

Case 1: 2−5x=18  ⟹  −5x=16  ⟹  x=−165=−3.22 - 5x = 18 \implies -5x = 16 \implies x = -\frac{16}{5} = -3.2. Case 2: 2−5x=−18  ⟹  −5x=−20  ⟹  x=42 - 5x = -18 \implies -5x = -20 \implies x = 4. With x1=−165x_1 = -\frac{16}{5} and x2=4x_2 = 4: M=−165+42=25M = \frac{-\frac{16}{5} + 4}{2} = \frac{2}{5} D=4−(−165)=365D = 4 - \left(-\frac{16}{5}\right) = \frac{36}{5}.

03

An absolute value equation has solutions x1=−1x_1 = -1 and x2=8x_2 = 8. Write the equation in the form ∣ax−b∣=c|ax - b| = c where a=2a = 2, and state the midpoint MM and distance DD between the roots.

Answer

The midpoint is M=−1+82=72M = \frac{-1 + 8}{2} = \frac{7}{2}. The distance is D=8−(−1)=9D = 8 - (-1) = 9. The radius is r=D2=92r = \frac{D}{2} = \frac{9}{2}. In center-radius form: ∣x−72∣=92\left|x - \frac{7}{2}\right| = \frac{9}{2}. Multiply both sides by a=2a = 2: ∣2x−7∣=9|2x - 7| = 9. Thus, the equation is ∣2x−7∣=9|2x - 7| = 9 with M=72M = \frac{7}{2} and D=9D = 9.

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