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The linear equation 2x−k3−3(x+2)4=x−56\frac{2x - k}{3} - \frac{3(x + 2)}{4} = \frac{x - 5}{6} has the solution x=−2x = -2. Determine the original constant value of kk.

Answer and steps

Final answer

k=3k = 3

Steps

01

Substitute the Given Root into the Equation

Since x=−2x = -2 is a solution to the equation, substituting x=−2x = -2 into the equation must satisfy the equality:

2(−2)−k3−3((−2)+2)4=(−2)−56\frac{2(-2) - k}{3} - \frac{3((-2) + 2)}{4} = \frac{(-2) - 5}{6}

02

Simplify Each Component

Evaluate each term individually using the order of operations:

  1. First term: 2(−2)−k=−4−k  ⟹  −4−k32(-2) - k = -4 - k \implies \frac{-4 - k}{3}
  2. Second term: Notice that the factor inside the parentheses becomes zero: −2+2=0-2 + 2 = 0 Therefore: 3(0)4=0\frac{3(0)}{4} = 0 The entire middle term vanishes.
  3. Right-hand side: −2−5=−7-2 - 5 = -7
03

Solve for the Constant kk

With the middle term eliminated, equate the resulting numerator containing the unknown parameter kk to the corresponding evaluated numerator −7-7:

−4−k=−7-4 - k = -7

Add 44 to both sides of the equation:

−k=−7+4-k = -7 + 4 −k=−3-k = -3

Divide both sides by −1-1:

k=3k = 3

Key Takeaway and Pitfall to Avoid

  • Common Pitfall: Expanding and finding a common denominator for all terms before substituting the known value x=−2x = -2 creates unnecessary algebraic work and increases the chance of sign errors.
  • Best Strategy: Always substitute the known root first. Checking for zero factors—such as (x+2)(x + 2) evaluating to 00 when x=−2x = -2—can immediately eliminate complex terms.
Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets to distribute the negative sign across the numerator when clearing denominators, leading to an incorrect constant term and k value of 15/2.

Sample Student Work

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

    2(−2)−k3\frac{2(-2)-k}{3} - 3(−2+2)4\frac{3(-2+2)}{4} = −2−56\frac{-2-5}{6}   ⟹  \implies −4−k3\frac{-4-k}{3} - 0 = −76\frac{-7}{6}

  2. Step 2

    4(-4-k) - 0 = 2(-7)   ⟹  \implies -16 - 4k = -14

  3. Step 3

    -4k = 2

  4. Step 4

    k = -12\frac{1}{2}

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly substituted the known root x=−2x = -2 into every occurrence of xx.

You noticed that the factor (−2+2)(-2 + 2) equals 00, which eliminated the middle term immediately.

You properly cleared denominators across fractions with different denominators rather than simply dropping them.

You maintained sign accuracy throughout your algebraic isolation of kk.

REASONING PATH
Your ACEpath teacher’s suggested approach
Substitute Known Roots Early

  • When an equation contains a variable and an unknown constant and the root is given, substitute the root first.

  • Replace xx with its value to reduce the equation to a single variable in kk.

Clear Denominators Using the LCD

  • When equating fractions such as A3=B6\frac{A}{3} = \frac{B}{6}, multiply through by the LCD (66) so that 2A=B2A = B.

  • Ensure the ratio of denominators is accounted for before equating numerators.

Verify by Back-Substitution

  • Substitute both the root and the calculated constant back into the original equation to ensure equality holds.

  • Check that 2(−2)−(−1/2)3−0=−76\frac{2(-2) - (-1/2)}{3} - 0 = -\frac{7}{6}.

CORRECTED PATH
ACEpath Teacher's Demonstration
Final answer

k=−12k = -\frac{1}{2}

NEXT STEPS
Final Takeaways from Your Teacher

  • Continue substituting known roots before expanding complex algebraic expressions.

  • Keep applying the least common denominator technique when clearing rational expressions.

OverviewACEpath Teacher's Diagnosis

02 The student miscalculates (-2 - 5) as 3 or -3, causing an arithmetic error in the right-hand side that results in an incorrect k value.

Sample Student Work

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

    2(−2)−k3\frac{2(-2) - k}{3} - 3(−2+2)4\frac{3(-2 + 2)}{4} = −2−56\frac{-2 - 5}{6}

  2. Step 2

    −4−k3\frac{-4 - k}{3} - 0 = −36\frac{-3}{6}

    Correct

    −4−k3\frac{-4 - k}{3} - 0 = −76\frac{-7}{6}

    In the numerator of the right-hand side, −2−5-2 - 5 evaluates to −7-7, not −3-3.

  3. Step 3

    2(-4 - k) = -3   ⟹  \implies -8 - 2k = -3   ⟹  \implies -2k = 5   ⟹  \implies k = -52\frac{5}{2}

    Correct

    2(-4 - k) = -7   ⟹  \implies -8 - 2k = -7   ⟹  \implies -2k = 1   ⟹  \implies k = -12\frac{1}{2}  (or \text{ (or } −4−k3\frac{-4 - k}{3} = -76\frac{7}{6}   ⟹  \implies -4 - k = -72\frac{7}{2})

    Your subsequent algebraic steps correctly followed your equation, but the final answer was incorrect due to the arithmetic error in the previous step.

WHAT YOU DID WELL
Compliments from Your Teacher

You substituted the known root x=−2x = -2 into the equation accurately.

You recognized that −2+2=0-2 + 2 = 0, which cleanly eliminated the entire middle fraction.

You cleared denominators and solved the resulting two-step linear equation with correct algebraic technique.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Signed Number Subtraction Error

You computed −2−5-2 - 5 as −3-3 instead of −7-7.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Confusing the subtraction of two numbers with opposite signs versus subtracting a positive number from a negative number. When starting at −2-2 and subtracting 55, you move further left on the number line to −7-7, rather than finding the difference between 55 and 22.

REASONING PATH
Your ACEpath teacher’s suggested approach
Same-Sign Addition Rule

  • −a−b=−(a+b)-a - b = -(a + b)

  • When both values have negative signs, add their absolute values together and keep the negative sign.

Number Line Movement

  • Subtracting a positive value means moving left

  • Visualize starting at −2-2 and moving 55 units to the left, which lands on −7-7.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Simplify the right-hand numerator: −2−5=−7-2 - 5 = -7, giving −4−k3=−76\frac{-4 - k}{3} = \frac{-7}{6}.

  2. Multiply both sides by 66 to clear the denominators: 2(−4−k)=−72(-4 - k) = -7.

  3. Expand the left side: −8−2k=−7-8 - 2k = -7.

  4. Add 88 to both sides: −2k=1-2k = 1.

  5. Divide by −2-2: k=−12k = -\frac{1}{2} (or substitute into the original simplified form directly).

Final answer

k = 3

NEXT STEPS
Final Takeaways from Your Teacher

  • Review adding and subtracting negative numbers, paying special attention to expressions like −a−b-a - b.

  • Double-check basic integer arithmetic on numerators before clearing denominators.

OverviewACEpath Teacher's Diagnosis

03 The student clears fractions first by multiplying by 12 but fails to multiply the right-hand side or neglects the negative sign in the expansion.

Sample Student Work

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

    4(2x - k) - 9(x + 2) = 2(x - 5)

  2. Step 2

    8x - 4k - 9x - 18 = 2x - 10   ⟹  \implies -x - 4k - 18 = 2x - 10

  3. Step 3

    -(-2) - 4k - 18 = 2(-2) - 10   ⟹  \implies 2 - 4k - 18 = -4 - 10   ⟹  \implies -16 - 4k = -14

  4. Step 4

    -4k = 2   ⟹  \implies k = -12\frac{1}{2}

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly multiplied by the least common denominator 1212 across all three rational terms.

You distributed the negative factor −9-9 into (x+2)(x + 2) without losing the negative sign.

You substituted the known root x=−2x = -2 cleanly and executed all integer arithmetic without errors.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
none
REASONING PATH
Your ACEpath teacher’s suggested approach
Strategy Comparison: Clearing Denominators vs. Direct Root Substitution

  • When an equation contains a known root x=cx = c, substituting cc before clearing fractions can often make a term vanish immediately.

  • Notice that when x=−2x = -2, the second term 3(x+2)4\frac{3(x + 2)}{4} evaluates to 3(0)4=0\frac{3(0)}{4} = 0, simplifying the equation directly to −4−k3=−76\frac{-4 - k}{3} = -\frac{7}{6}.

CORRECTED PATH
ACEpath Teacher's Demonstration
Final answer

k = -12\frac{1}{2}

NEXT STEPS
Final Takeaways from Your Teacher

  • Continue practicing multi-step fractional equations with parameters.

  • Practice checking whether substituting a given root first makes any factors vanish to save algebraic steps.

Lock it in

Question

When a linear equation with an unknown parameter kk specifies that x=ax = a is a root, what should be your immediate first step?

Answer

Substitute x=ax = a into the equation. This eliminates xx and creates a single-variable equation for kk.

Similar question

01

The linear equation 3x+k2−x−15=4\frac{3x + k}{2} - \frac{x - 1}{5} = 4 has the solution x=1x = 1. Find the value of the constant kk.

Answer

Substitute x=1x = 1: the second term becomes 1−15=0\frac{1 - 1}{5} = 0, giving 3(1)+k2−0=4\frac{3(1) + k}{2} - 0 = 4. Multiply by 22: 3+k=83 + k = 8, so k=5k = 5.

02

The equation x−2k4+2(x+3)3=k−x6\frac{x - 2k}{4} + \frac{2(x + 3)}{3} = \frac{k - x}{6} has the solution x=−3x = -3. Find the value of kk.

Answer

Substitute x=−3x = -3: the middle term vanishes because −3+3=0-3 + 3 = 0. This leaves −3−2k4=k−(−3)6=k+36\frac{-3 - 2k}{4} = \frac{k - (-3)}{6} = \frac{k + 3}{6}. Multiply by the LCD 1212: 3(−3−2k)=2(k+3)  ⟹  −9−6k=2k+6  ⟹  −8k=15  ⟹  k=−1583(-3 - 2k) = 2(k + 3) \implies -9 - 6k = 2k + 6 \implies -8k = 15 \implies k = -\frac{15}{8}.

03

The equation 2(x−k)5−x2−43=3x+k10\frac{2(x - k)}{5} - \frac{x^2 - 4}{3} = \frac{3x + k}{10} has the solution x=2x = 2. Find the value of kk.

Answer

Substitute x=2x = 2: the middle term vanishes since 22−4=02^2 - 4 = 0, leaving 2(2−k)5=3(2)+k10=6+k10\frac{2(2 - k)}{5} = \frac{3(2) + k}{10} = \frac{6 + k}{10}. Multiply by the LCD 1010: 4(2−k)=6+k  ⟹  8−4k=6+k  ⟹  −5k=−2  ⟹  k=254(2 - k) = 6 + k \implies 8 - 4k = 6 + k \implies -5k = -2 \implies k = \frac{2}{5}.

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