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Determine the number of solutions to the following linear equation: 4(3x−5)−2x=2(5x−7)−64(3x - 5) - 2x = 2(5x - 7) - 6. If it has one solution, find the value of xx.

Answer and steps

Final answer

Infinitely many solutions (the equation is an identity, reducing to 10x−20=10x−2010x - 20 = 10x - 20, which holds true for all real numbers xx).

Steps

To determine the number of solutions, simplify both sides of the equation by applying the distributive property and combining like terms.

Step 1: Simplify the left-hand side (LHS) Apply the distributive property to expand the expression, then combine the terms containing xx: 4(3x−5)−2x=12x−20−2x4(3x - 5) - 2x = 12x - 20 - 2x =10x−20= 10x - 20

Step 2: Simplify the right-hand side (RHS) Apply the distributive property and combine the constant terms: 2(5x−7)−6=10x−14−62(5x - 7) - 6 = 10x - 14 - 6 =10x−20= 10x - 20

Step 3: Compare both sides of the equation Set the simplified LHS equal to the simplified RHS: 10x−20=10x−2010x - 20 = 10x - 20

Subtracting 10x10x from both sides gives: −20=−20-20 = -20

Adding 2020 to both sides results in: 0=00 = 0

Conclusion: Because the variable cancels out completely and leaves a statement that is always true (0=00 = 0, or −20=−20-20 = -20), the equation is an identity. This means the equation holds true for any real value of xx, so there are infinitely many solutions.

Common Pitfall: Do not confuse an equation yielding 0=00 = 0 with an equation having x=0x = 0 as a single solution. When the variables cancel and leave a true equality, any real number works; if the variables cancel and leave a false statement (such as 0=50 = 5), there is no solution.

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student misapplies the distributive property by forgetting to multiply the second term inside the parentheses by 4, leading to a false single solution of x = 1.

Sample Student Work

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

    4(3x - 5) - 2x = 2(5x - 7) - 6   ⟹  \implies 12x - 5 - 2x = 10x - 14 - 6

    Correct

    4(3x - 5) - 2x = 2(5x - 7) - 6   ⟹  \implies 12x - 20 - 2x = 10x - 14 - 6

    You distributed 44 to 3x3x to get 12x12x, but you did not multiply 44 by −5-5, writing −5-5 instead of −20-20.

  2. Step 2

    10x - 5 = 10x - 20

    Correct

    10x - 20 = 10x - 20

    You combined like terms correctly on both sides, but this expression carries forward the distribution error from the previous step.

  3. Step 3

    -5 = -20

    Correct

    -20 = -20   ⟹  \implies 0 = 0

    You subtracted 10x10x from both sides accurately, but the resulting false equality arose entirely from the earlier distribution slip.

  4. Step 4

    No solution\text{No solution}

    Correct

    Infinitely many solutions\text{Infinitely many solutions}

    Because the correct simplified equation is 10x−20=10x−2010x - 20 = 10x - 20, the statement is always true (0=00 = 0), meaning the equation has infinitely many solutions.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly expanded the right-hand side using the distributive property to obtain 10x−14−610x - 14 - 6.

You accurately combined linear and constant terms on both sides of the equation.

You correctly understood that an equation reducing to a false statement like −5=−20-5 = -20 means no solution.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Incomplete Distribution

Forgetting to multiply the outside factor by the constant term inside the parentheses.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Incomplete application of the distributive property, multiplying only the first term inside parentheses while leaving the second term unmultiplied.

REASONING PATH
Your ACEpath teacher’s suggested approach
Distribute to Every Term

  • For any expression a(b+c)a(b + c), the factor aa must multiply both bb and cc, giving ab+acab + ac.

  • Draw distribution arrows from the outside coefficient to every term inside the parentheses before multiplying.

Classify Equation Outcomes

  • If variables cancel and leave a true statement (e.g., 0=00 = 0), there are infinitely many solutions; if they leave a false statement (e.g., 0=−150 = -15), there is no solution.

  • Check whether the two simplified sides are identical expressions before concluding the number of solutions.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Combine like terms on both sides: 10x−20=10x−2010x - 20 = 10x - 20.

  2. Subtract 10x10x from both sides: −20=−20-20 = -20.

  3. Add 2020 to both sides: 0=00 = 0, which is an identity true for all real values of xx.

Final answer

Infinitely many solutions

NEXT STEPS
Final Takeaways from Your Teacher

  • Always draw multiplication arcs from the outer number to every term inside parentheses to ensure no term is missed.

  • Compare the left-hand and right-hand expressions after simplifying: identical linear and constant terms immediately indicate an identity with infinitely many solutions.

OverviewACEpath Teacher's Diagnosis

02 Upon arriving at 0x = 0, the student confuses 0x = 0 with having no solution instead of infinitely many solutions.

Sample Student Work

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

    12x - 20 - 2x = 10x - 14 - 6   ⟹  \implies 10x - 20 = 10x - 20

  2. Step 2

    10x - 10x = -20 + 20   ⟹  \implies 0 = 0

  3. Step 3

    No solution\text{No solution}

    Correct

    Infinitely many solutions\text{Infinitely many solutions}

    When an equation simplifies to an identity like 0=00 = 0 (a statement that is always true for any value of xx), the equation has infinitely many solutions, not no solution.

WHAT YOU DID WELL
Compliments from Your Teacher

You expanded both products accurately using the distributive property.

You combined like terms correctly on both sides of the equation.

You subtracted 10x10x and added 2020 to isolate terms, correctly reducing the equation to 0=00 = 0.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Misinterpreting True Identity as No Solution

You recognized that the variable cancelled out, but you wrote "No solution" instead of recognizing 0=00 = 0 as an identity.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Confusing the interpretation of a universally true equality (0=00 = 0, identity) with a contradiction (0=c0 = c where c≠0c \neq 0, inconsistency). A true equality means any real number makes the equation true, which gives infinitely many solutions.

REASONING PATH
Your ACEpath teacher’s suggested approach
Check the final statement's truth value

  • If the variable cancels out, check whether the remaining numerical equation is true or false.

  • If it is always true (e.g., 0=00 = 0), conclude "infinitely many solutions". If it is false (e.g., 0=40 = 4), conclude "no solution".

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Notice that the variable xx has cancelled out completely, leaving 0=00 = 0.

  2. Ask: Is 0=00 = 0 true for all values of xx, or false for all values of xx?

  3. Because 0=00 = 0 is always true regardless of what value xx takes, every real number satisfies the original equation.

Final answer

Infinitely many solutions

NEXT STEPS
Final Takeaways from Your Teacher

  • Remember that "no solution" only occurs when you get an impossible statement like 0=30 = 3 or −5=2-5 = 2.

  • Whenever both sides simplify to the exact same expression, immediately classify the equation as an identity with infinitely many solutions.

OverviewACEpath Teacher's Diagnosis

03 The student interprets the cancellation to 0 = 0 as meaning the single solution is x = 0.

Sample Student Work

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

    12x - 20 - 2x = 10x - 14 - 6   ⟹  \implies 10x - 20 = 10x - 20

  2. Step 2

    10x - 10x = -20 + 20   ⟹  \implies 0x = 0

  3. Step 3

    One solution, x=0\text{One solution, } x = 0

    Correct

    Infinitely many solutions\text{Infinitely many solutions}

    The statement 0x=00x = 0 is true for every real value of xx, not just x=0x = 0. Therefore, the equation is an identity with infinitely many solutions.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly applied the distributive property to both sides of the equation.

You combined like terms accurately on both the left-hand and right-hand sides.

You moved the variable and constant terms to opposite sides without sign errors.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Misinterpreting an Identity as a Single Zero Solution

none

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

You confused the statement 0x=00x = 0 (an identity true for any xx) with the statement x=0x = 0 (a specific single numerical solution).

REASONING PATH
Your ACEpath teacher’s suggested approach
Interpreting 0x = 0

  • An equation of the form 0x=00x = 0 simplifies to 0=00 = 0, which is always true regardless of what value is substituted for xx.

  • When all variable terms cancel and yield 0=00 = 0, conclude that there are infinitely many solutions (an identity).

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Notice that 0⋅x=00 \cdot x = 0 is true for any real number xx (for example, 0(1)=00(1) = 0, 0(−5)=00(-5) = 0, and 0(100)=00(100) = 0).

  2. Because every real number makes the equation true, the equation is an identity.

Final answer

Infinitely many solutions

NEXT STEPS
Final Takeaways from Your Teacher

  • Always ask yourself: 'Does substituting a number other than 00 into 0x=00x = 0 make the equation true?' If yes, all real numbers are solutions.

  • Differentiate between 0x=00x = 0 (infinitely many solutions), 0x=c0x = c where c≠0c \neq 0 (no solution), and cx=0cx = 0 where c≠0c \neq 0 (one solution, x=0x = 0).

Lock it in

Question

State the distributive property for multiplying a single factor across a binomial difference: a(b−c)=?a(b - c) = ?

Answer

a(b−c)=ab−aca(b - c) = ab - ac. The outside multiplier aa must multiply every term inside the parentheses, including the second term.

Similar question

01

Solve the linear equation and determine the number of solutions: 3(2x−4)+6=6x−63(2x - 4) + 6 = 6x - 6.

Answer

Infinitely many solutions. Expanding the LHS gives 6x−12+6=6x−66x - 12 + 6 = 6x - 6. Since LHS and RHS are identical (6x−6=6x−66x - 6 = 6x - 6), subtracting 6x6x gives −6=−6-6 = -6, an identity true for all real xx.

02

Solve the linear equation and determine the number of solutions: 5(2x−1)−4x=2(3x+2)−95(2x - 1) - 4x = 2(3x + 2) - 9.

Answer

Infinitely many solutions. Distribute on both sides: LHS =10x−5−4x=6x−5= 10x - 5 - 4x = 6x - 5; RHS =6x+4−9=6x−5= 6x + 4 - 9 = 6x - 5. Because 6x−5=6x−56x - 5 = 6x - 5, the equation is an identity with infinitely many solutions.

03

Solve the linear equation and determine the number of solutions: 2(4x−3)−3(2x+1)=2x−92(4x - 3) - 3(2x + 1) = 2x - 9.

Answer

Infinitely many solutions. Distribute carefully: LHS =8x−6−6x−3=2x−9= 8x - 6 - 6x - 3 = 2x - 9. The equation simplifies to 2x−9=2x−92x - 9 = 2x - 9, which holds true for all real numbers.

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