Middle School Math Quiz: Classify Solutions To Linear Equations
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Classify Solutions To Linear EquationsQuestion 1 of 20

Sarah wrote three equations on the board. Which equation has exactly one solution?

I. 5x7=5x+35x - 7 = 5x + 3

II. 4(x+1)=4x+44(x + 1) = 4x + 4

III. 3x+2=7x103x + 2 = 7x - 10

Only equation I has exactly one solution
Only equation II has exactly one solution
Only equation III has exactly one solution
Both equations I and III have exactly one solution
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Middle School Math Quiz

Middle School Math Quiz: Classify Solutions To Linear Equations

Practice Classify Solutions To Linear Equations in Middle School Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Classify Solutions To Linear Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for Middle School Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Sarah wrote three equations on the board. Which equation has exactly one solution?

I. 5x7=5x+35x - 7 = 5x + 3

II. 4(x+1)=4x+44(x + 1) = 4x + 4

III. 3x+2=7x103x + 2 = 7x - 10

  1. Only equation I has exactly one solution
  2. Only equation II has exactly one solution
  3. Only equation III has exactly one solution (correct answer)
  4. Both equations I and III have exactly one solution
Explanation: Equation I: 5x7=5x+35x - 7 = 5x + 3 simplifies to 7=3-7 = 3 (no solution). Equation II: 4(x+1)=4x+44(x + 1) = 4x + 4 simplifies to 4x+4=4x+44x + 4 = 4x + 4 or 4=44 = 4 (infinitely many solutions). Equation III: 3x+2=7x103x + 2 = 7x - 10 gives 4x=12-4x = -12, so x=3x = 3 (one solution). Choice A misidentifies equation I as having one solution. Choice B misidentifies equation II as having one solution. Choice D incorrectly includes equation I.

Question 2

A student writes an equation for the number of stickers in two boxes: 3x+2=3(x+1)+13x+2=3(x+1)+1. After simplifying, how many solutions does the equation have?

  1. Infinitely many solutions
  2. Cannot be determined without a value of xx
  3. One solution
  4. No solutions (correct answer)
Explanation: This question tests classifying linear equations by solution count: one (x=a), infinitely many (identity a=a), or none (contradiction a≠b). Solving reveals type: distribute and collect terms (2(x+3)=2x+6 → 2x+6=2x+6 → 6=6 identity), result shows classification (x=5 means one solution 5, 3=3 means all x work infinitely many, 4=7 means no x works no solution); type depends on whether variables cancel: if x remains with unique value (one solution), if variables cancel to truth (infinite), if cancel to falsehood (none). For 3x+2=3(x+1)+1, distribute to get 3x+2=3x+3+1, simplify to 3x+2=3x+4, subtract 3x to reach 2=4, a false statement. This is a contradiction, so it has no solutions because no x makes it true. A common error is forgetting to add the +1 and claiming infinite solutions. Strategy: (1) distribute parentheses, (2) collect like terms (combine x's, combine constants), (3) move variables to one side, (4) observe result (x=number → one, number=number → infinite, number=different → none), (5) verify (substitute x back if one solution, try multiple x values if claiming infinite). Mistakes: stopping too early (not simplifying fully), misinterpreting 2=4 as true, claiming one solution, distributing incorrectly changing the type.

Question 3

A student is simplifying 4x6=2(2x1)4x-6=2(2x-1). After distributing and combining like terms, what type of solution does the equation have: one solution, infinitely many solutions, or no solution?

  1. Infinitely many solutions: simplifies to 6=2-6=-2
  2. No solution: simplifies to 6=2-6=-2 (correct answer)
  3. No solution: simplifies to 6=6-6=-6
  4. One solution: x=2x=2
Explanation: This question tests classifying linear equations by solution count: one (x=a), infinitely many (identity a=a), or none (contradiction a≠b). Solving reveals type: distribute and collect terms (2(x+3)=2x+6 → 2x+6=2x+6 → 6=6 identity), result shows classification (x=5 means one solution 5, 3=3 means all x work infinitely many, 4=7 means no x works no solution); type depends on whether variables cancel: if x remains with unique value (one solution), if variables cancel to truth (infinite), if cancel to falsehood (none). For the equation 4x-6=2(2x-1), distribute to get 4x-6=4x-2, subtract 4x to get -6=-2. This simplifies to -6=-2, a contradiction since -6 ≠ -2, indicating no solution. A common error is claiming infinitely many solutions by misreading -6=-2 as -6=-6, or thinking it's one solution like x=2. Strategy: (1) distribute parentheses, (2) collect like terms (combine x's, combine constants), (3) move variables to one side, (4) observe result (x=number → one, number=number → infinite, number=different → none), (5) verify (substitute x back if one solution, try multiple x values if claiming infinite). Mistakes: stopping too early (not simplifying fully), misinterpreting 0=0 as x=0, claiming 5=8 has solution, distributing incorrectly changing the type.

Question 4

A student is told that an equation has no solution if simplifying leads to a contradiction like 2=72=7. Which equation will simplify to a contradiction?

  1. 3(x2)+5=3x13(x-2)+5=3x-1
  2. 2(x+3)=2x+62(x+3)=2x+6
  3. 5x+4=5x15x+4=5x-1 (correct answer)
  4. 4x8=2x4x-8=2x
Explanation: This question tests classifying linear equations by solution count: one (x=a), infinitely many (identity a=a), or none (contradiction a≠b). Solving reveals type: distribute and collect terms (2(x+3)=2x+6 → 2x+6=2x+6 → 6=6 identity), result shows classification (x=5 means one solution 5, 3=3 means all x work infinitely many, 4=7 means no x works no solution); type depends on whether variables cancel: if x remains with unique value (one solution), if variables cancel to truth (infinite), if cancel to falsehood (none). For option B, 5x+4=5x-1 simplifies by subtracting 5x to get 4=-1, which is a contradiction. This contradiction 4=-1 means no solutions, matching the description of simplifying to something like 2=7, while other options lead to identities or one solution. A common error is confusing a contradiction with an identity, such as thinking 4=-1 could have solutions if not fully simplified. The strategy is to (1) distribute parentheses, (2) collect like terms (combine x's, combine constants), (3) move variables to one side, (4) observe result (x=number → one, number=number → infinite, number=different → none), (5) verify (substitute x back if one solution, try multiple x values if claiming infinite). Mistakes include stopping too early (not simplifying fully), misinterpreting 0=0 as x=0, claiming 5=8 has solution, distributing incorrectly changing the type.

Question 5

A student claims the equation 5(x+1)2=5x+35(x+1)-2=5x+3 has one solution. Solve and classify the equation as one solution, infinitely many solutions, or no solution.

  1. One solution: x=1x=1
  2. No solution: simplifies to 3=53=5
  3. Infinitely many solutions: simplifies to x=xx=x
  4. Infinitely many solutions: simplifies to 0=00=0 (correct answer)
Explanation: This question tests classifying linear equations by solution count: one (x=a), infinitely many (identity a=a), or none (contradiction a≠b). Solving reveals type: distribute and collect terms (2(x+3)=2x+6 → 2x+6=2x+6 → 6=6 identity), result shows classification (x=5 means one solution 5, 3=3 means all x work infinitely many, 4=7 means no x works no solution); type depends on whether variables cancel: if x remains with unique value (one solution), if variables cancel to truth (infinite), if cancel to falsehood (none). For the equation 5(x+1)-2=5x+3, distribute to get 5x+5-2=5x+3, simplify to 5x+3=5x+3, subtract 5x to get 3=3 (or 0=0). This simplifies to 0=0, indicating infinitely many solutions since it's an identity. A common error is thinking it's one solution like x=1 by incorrectly isolating without full simplification, or claiming no solution like 3=5. Strategy: (1) distribute parentheses, (2) collect like terms (combine x's, combine constants), (3) move variables to one side, (4) observe result (x=number → one, number=number → infinite, number=different → none), (5) verify (substitute x back if one solution, try multiple x values if claiming infinite). Mistakes: stopping too early (not simplifying fully), misinterpreting 0=0 as x=0, claiming 5=8 has solution, distributing incorrectly changing the type.

Question 6

Simplify 5x3x+4=2x+45x-3x+4=2x+4. How many solutions does this equation have?

  1. Exactly two solutions
  2. No solutions
  3. Infinitely many solutions (correct answer)
  4. Exactly one solution
Explanation: Combine like terms on the left side: 5x3x+4=2x+45x-3x+4=2x+4. Subtracting 2x2x from both sides gives 4=44=4, which is always true regardless of the value of xx. Since the variable cancels out and leaves a true statement, the equation has infinitely many solutions. This differs from an equation like 2x+4=2x+72x+4=2x+7, where canceling the variable would leave a false statement (no solutions), or an equation like 3x+4=103x+4=10, which has exactly one solution.

Question 7

Solve and classify: 62(x+1)=2x+46-2(x+1)=2x+4. Which statement is correct?

  1. 0=4; no solution
  2. x=1x=1; one solution
  3. 2=62=6; no solutions
  4. x=0x=0; one solution (correct answer)
Explanation: Distribute: 62(x+1)=2x+46-2(x+1)=2x+4 becomes 62x2=2x+46-2x-2=2x+4, which simplifies to 42x=2x+44-2x=2x+4. Adding 2x2x to both sides gives 4=4x+44=4x+4, and subtracting 4 gives 0=4x0=4x, so x=0x=0. Since this gives one specific value, the equation has exactly one solution. Checking: 62(0+1)=46-2(0+1)=4 and 2(0)+4=42(0)+4=4, which match. Choices A and C claim the equation reduces to a false statement with no solution, but it actually reduces to a true statement about a specific x-value. Choice B gives the wrong value for x; solving carefully shows x=0, not x=1.

Question 8

A student claims the equation 9+2x=2x+99+2x=2x+9 has the solution x=0x=0. Which classification is correct?

  1. No solutions
  2. Infinitely many solutions (correct answer)
  3. One solution (x=0x=0)
  4. One solution (x=9x=9)
Explanation: This question tests classifying linear equations by solution count: one (x=a), infinitely many (identity a=a), or none (contradiction a≠b). Solving reveals type: distribute and collect terms (2(x+3)=2x+6 → 2x+6=2x+6 → 6=6 identity), result shows classification (x=5 means one solution 5, 3=3 means all x work infinitely many, 4=7 means no x works no solution); type depends on whether variables cancel: if x remains with unique value (one solution), if variables cancel to truth (infinite), if cancel to falsehood (none). For 9+2x=2x+9, subtract 2x to get 9=9, a true identity. This is an identity, so it has infinitely many solutions because it's true for all x, not just x=0 as claimed. A common error is assuming x=0 is the only solution since variables cancel, or thinking it's one solution. Strategy: (1) distribute parentheses, (2) collect like terms (combine x's, combine constants), (3) move variables to one side, (4) observe result (x=number → one, number=number → infinite, number=different → none), (5) verify (substitute x back if one solution, try multiple x values if claiming infinite). Mistakes: stopping too early (not simplifying fully), misinterpreting 9=9 as x=0, claiming no solutions, no distribution needed here but arithmetic errors possible.

Question 9

Which equation simplifies to a true statement like 0=00=0 and therefore has infinitely many solutions?

  1. 4x+8=2x+104x+8=2x+10
  2. 5x+1=5x+45x+1=5x+4
  3. 2(x3)=2x62(x-3)=2x-6 (correct answer)
  4. 3x7=3x23x-7=3x-2
Explanation: This question tests classifying linear equations by solution count: one (x=a), infinitely many (identity a=a), or none (contradiction a≠b). Solving reveals type: distribute and collect terms (2(x+3)=2x+6 → 2x+6=2x+6 → 6=6 identity), result shows classification (x=5 means one solution 5, 3=3 means all x work infinitely many, 4=7 means no x works no solution); type depends on whether variables cancel: if x remains with unique value (one solution), if variables cancel to truth (infinite), if cancel to falsehood (none). For 2(x-3)=2x-6, distribute to get 2x-6=2x-6, subtract 2x to reach -6=-6, a true identity. This is an identity, so it has infinitely many solutions because it's true for all x. A common error is thinking only x=0 works or classifying as one solution. Strategy: (1) distribute parentheses, (2) collect like terms (combine x's, combine constants), (3) move variables to one side, (4) observe result (x=number → one, number=number → infinite, number=different → none), (5) verify (substitute x back if one solution, try multiple x values if claiming infinite). Mistakes: stopping too early (not simplifying fully), misinterpreting -6=-6 as false, claiming no solutions, distributing incorrectly changing the type.

Question 10

Consider the equation 8+2(x3)=2x+58+2(x-3)=2x+5. Based on simplifying the equation, what can you conclude about the solutions?

  1. There is exactly one solution for xx.
  2. There are infinitely many solutions.
  3. There is no solution for xx. (correct answer)
  4. The solution is x=3x=3.
Explanation: Distribute and simplify: 8+2(x3)=2x+58+2(x-3)=2x+5 becomes 8+2x6=2x+58+2x-6=2x+5, which simplifies to 2x+2=2x+52x+2=2x+5. Subtracting 2x2x from both sides gives 2=52=5, which is never true no matter what xx is. Since the equation reduces to a false statement, it has no solution. Choice A is wrong because the variable terms cancel out entirely, so there's no single value of xx that works. Choice B is wrong because the equation doesn't reduce to a true statement like 5=55=5; it reduces to a false one. Choice D is wrong because x=3x=3 doesn't actually satisfy the original equation; no value of xx does.

Question 11

A science lab group records a relationship that can be written as 2x+8=2(x+4)2x+8=2(x+4), where xx is the number of minutes. Simplify to determine whether the equation has one solution, infinitely many solutions, or no solution.

  1. One solution: x=0x=0
  2. No solution: simplifies to 8=48=4
  3. Infinitely many solutions: simplifies to 8=88=8 (correct answer)
  4. One solution: x=4x=4
Explanation: This question tests classifying linear equations by solution count: one (x=a), infinitely many (identity a=a), or none (contradiction a≠b). Solving reveals type: distribute and collect terms (2(x+3)=2x+6 → 2x+6=2x+6 → 6=6 identity), result shows classification (x=5 means one solution 5, 3=3 means all x work infinitely many, 4=7 means no x works no solution); type depends on whether variables cancel: if x remains with unique value (one solution), if variables cancel to truth (infinite), if cancel to falsehood (none). For the equation 2x+8=2(x+4), distribute to get 2x+8=2x+8, subtract 2x to get 8=8. This simplifies to 8=8, indicating infinitely many solutions since it's an identity. A common error is thinking it's one solution like x=4 by incorrectly solving without noticing the identity, or claiming no solution like 8=4. Strategy: (1) distribute parentheses, (2) collect like terms (combine x's, combine constants), (3) move variables to one side, (4) observe result (x=number → one, number=number → infinite, number=different → none), (5) verify (substitute x back if one solution, try multiple x values if claiming infinite). Mistakes: stopping too early (not simplifying fully), misinterpreting 0=0 as x=0, claiming 5=8 has solution, distributing incorrectly changing the type.

Question 12

Marcus is solving the equation 3(x+2)5=3x+k3(x + 2) - 5 = 3x + k. For what value of kk will this equation have infinitely many solutions?

  1. k=1k = 1 (correct answer)
  2. k=6k = 6
  3. k=11k = 11
  4. k=1k = -1
Explanation: To find when the equation has infinitely many solutions, we need to simplify and get an identity of the form a=aa = a. Expanding the left side: 3(x+2)5=3x+65=3x+13(x + 2) - 5 = 3x + 6 - 5 = 3x + 1. So the equation becomes 3x+1=3x+k3x + 1 = 3x + k. For infinitely many solutions, we need 1=k1 = k, so k=1k = 1. Choice B gives k=6k = 6 (incorrect transformation). Choice C gives k=11k = 11 (adding instead of subtracting). Choice D gives k=1k = -1 (sign error in simplification).

Question 13

Which of the following equations is equivalent to 4x+8=4(x+2)4x + 8 = 4(x + 2) and demonstrates that the original equation has infinitely many solutions?

  1. 4x+8=4x+84x + 8 = 4x + 8, which shows 8=88 = 8 (correct answer)
  2. 4x=4x+84x = 4x + 8, which shows 0=80 = 8
  3. x+2=x+2x + 2 = x + 2, which shows 2=22 = 2
  4. 4x+8=4x+24x + 8 = 4x + 2, which shows 8=28 = 2
Explanation: Expanding the right side: 4(x+2)=4x+84(x + 2) = 4x + 8. The equation becomes 4x+8=4x+84x + 8 = 4x + 8. Subtracting 4x4x from both sides gives 8=88 = 8, which is always true, confirming infinitely many solutions. Choice B incorrectly expands 4(x+2)4(x + 2) as 4x+84x + 8 but then writes 4x=4x+84x = 4x + 8. Choice C divides by 44 correctly but doesn't show the final identity clearly. Choice D incorrectly expands 4(x+2)4(x + 2) as 4x+24x + 2.

Question 14

A student writes the equation for the number of points xx they scored in a game: 3(x+4)6=3x+63(x+4)-6=3x+6. After simplifying, how many solutions does the equation have (one, infinitely many, or no solutions)?

  1. Cannot be determined
  2. No solutions
  3. Infinitely many solutions (correct answer)
  4. One solution
Explanation: This question tests classifying linear equations by solution count: one (x=a), infinitely many (identity a=a), or none (contradiction a≠b). Solving reveals type: distribute and collect terms (2(x+3)=2x+6 → 2x+6=2x+6 → 6=6 identity), result shows classification (x=5 means one solution 5, 3=3 means all x work infinitely many, 4=7 means no x works no solution); type depends on whether variables cancel: if x remains with unique value (one solution), if variables cancel to truth (infinite), if cancel to falsehood (none). For 3(x+4)-6=3x+6, distribute to get 3x+12-6=3x+6, simplify to 3x+6=3x+6, then subtract 3x to reach 6=6, a true identity. This is an identity, so it has infinitely many solutions because it's true for all x. A common error is thinking it's one solution like x=0 since variables cancel, but actually any x works. Strategy: (1) distribute parentheses, (2) collect like terms (combine x's, combine constants), (3) move variables to one side, (4) observe result (x=number → one, number=number → infinite, number=different → none), (5) verify (substitute x back if one solution, try multiple x values if claiming infinite). Mistakes: stopping too early (not simplifying fully), misinterpreting 6=6 as no solution, claiming it has only one solution, distributing incorrectly changing the type.

Question 15

Which equation has exactly one solution?

  1. 3(x+1)=3x3-3(x+1)=-3x-3
  2. 2(x5)+1=x92(x-5)+1=x-9
  3. 9x3=9x+19x-3=9x+1 (correct answer)
  4. 4(x+2)=4x+84(x+2)=4x+8
Explanation: This question tests classifying linear equations by solution count: one (x=a), infinitely many (identity a=a), or none (contradiction a≠b). Solving reveals type: distribute and collect terms (2(x+3)=2x+6 → 2x+6=2x+6 → 6=6 identity), result shows classification (x=5 means one solution 5, 3=3 means all x work infinitely many, 4=7 means no x works no solution); type depends on whether variables cancel: if x remains with unique value (one solution), if variables cancel to truth (infinite), if cancel to falsehood (none). For option C, 2(x-5)+1=x-9 distributes to 2x-10+1=x-9, simplifies to 2x-9=x-9, subtract x to get x-9=-9, add 9 to get x=0. This results in x=0, which is one unique solution, while other options lead to identities or contradictions. A common error is not fully simplifying, like stopping at 2x-9=x-9 and claiming no solution instead of solving for x=0. The strategy is to (1) distribute parentheses, (2) collect like terms (combine x's, combine constants), (3) move variables to one side, (4) observe result (x=number → one, number=number → infinite, number=different → none), (5) verify (substitute x back if one solution, try multiple x values if claiming infinite). Mistakes include stopping too early (not simplifying fully), misinterpreting 0=0 as x=0, claiming 5=8 has solution, distributing incorrectly changing the type.

Question 16

Classify the equation 93(x2)=153x9-3(x-2)=15-3x by simplifying it to the form x=ax=a, a=aa=a, or a=ba=b (with aba\ne b).

  1. No solution: simplifies to 0=60=6
  2. Infinitely many solutions: simplifies to 15=1515=15 (correct answer)
  3. One solution: x=2x=2
  4. One solution: x=2x=-2
Explanation: This question tests classifying linear equations by solution count: one (x=a), infinitely many (identity a=a), or none (contradiction a≠b). Solving reveals type: distribute and collect terms (2(x+3)=2x+6 → 2x+6=2x+6 → 6=6 identity), result shows classification (x=5 means one solution 5, 3=3 means all x work infinitely many, 4=7 means no x works no solution); type depends on whether variables cancel: if x remains with unique value (one solution), if variables cancel to truth (infinite), if cancel to falsehood (none). For the equation 9-3(x-2)=15-3x, distribute to get 9-3x+6=15-3x, simplify to 15-3x=15-3x, add 3x to get 15=15. This simplifies to 15=15, indicating infinitely many solutions since it's an identity. A common error is misdistributing -3(x-2) as -3x-6 instead of -3x+6, leading to a false no solution like 0=6. Strategy: (1) distribute parentheses, (2) collect like terms (combine x's, combine constants), (3) move variables to one side, (4) observe result (x=number → one, number=number → infinite, number=different → none), (5) verify (substitute x back if one solution, try multiple x values if claiming infinite). Mistakes: stopping too early (not simplifying fully), misinterpreting 0=0 as x=0, claiming 5=8 has solution, distributing incorrectly changing the type.

Question 17

Which statement correctly describes the equation 2x+63=2x3+2\frac{2x + 6}{3} = \frac{2x}{3} + 2?

  1. It has one solution because solving gives x=0x = 0
  2. It has no solution because 632\frac{6}{3} \neq 2
  3. It has no solution because the variable terms don't cancel completely
  4. It has infinitely many solutions because the equation is an identity. (correct answer)
Explanation: When you encounter an equation where the same expression appears on both sides, you're dealing with a special case that requires careful analysis of how many solutions exist. Let's solve this equation step by step. Starting with 2x+63=2x3+2\frac{2x + 6}{3} = \frac{2x}{3} + 2, you can split the left side using the distributive property: 2x3+63=2x3+2\frac{2x}{3} + \frac{6}{3} = \frac{2x}{3} + 2. This simplifies to 2x3+2=2x3+2\frac{2x}{3} + 2 = \frac{2x}{3} + 2. Now subtract 2x3\frac{2x}{3} from both sides: 2=22 = 2. Since this statement is always true regardless of the value of xx, the equation has infinitely many solutions. Any real number you substitute for xx will make the original equation true. Choice A incorrectly assumes there's a unique solution and mistakes the process of eliminating variables. Choice B makes an error by comparing 63\frac{6}{3} and 22 as if they're unequal, when 63=2\frac{6}{3} = 2 exactly. Choice C misunderstands what happens when variable terms cancel—this cancellation is precisely what reveals the equation's true nature. The correct answer is D because the equation simplifies to the identity 2=22 = 2. Study tip: When solving linear equations, pay attention to what happens after you eliminate variables. If you get a true statement like 2=22 = 2, you have infinitely many solutions. If you get a false statement like 2=32 = 3, you have no solution.

Question 18

A student simplifies the equation 4(x3)+6=2x24(x-3)+6=2x-2. How many solutions does the equation have?

  1. No solutions
  2. Infinitely many solutions
  3. Exactly two solutions
  4. Exactly one solution (correct answer)
Explanation: Distribute and combine like terms: 4(x3)+6=4x12+6=4x64(x-3)+6 = 4x-12+6 = 4x-6, so the equation becomes 4x6=2x24x-6=2x-2. Subtracting 2x2x from both sides gives 2x6=22x-6=-2, and adding 6 gives 2x=42x=4, so x=2x=2 — exactly one solution. Choice A (no solutions) only happens when the variable terms cancel and leave a false statement like 5=85=8; here 4x2x=2x4x-2x=2x remains, so they don't cancel. Choice B (infinitely many) only happens when both sides become identical, like 6=66=6, which isn't the case. Choice C is impossible: a linear equation in one variable has at most one solution, so it can never have exactly two.

Question 19

When solving 12(4x+6)=2x+n\frac{1}{2}(4x + 6) = 2x + n, Alex gets the result 3=n3 = n. What can you conclude about the original equation?

  1. The equation has one solution when n=3n = 3
  2. The equation has infinitely many solutions when n=3n = 3 (correct answer)
  3. The equation has no solution when n=3n = 3
  4. Alex made an error; the result should be 6=n6 = n
Explanation: Simplifying the left side: 12(4x+6)=2x+3\frac{1}{2}(4x + 6) = 2x + 3. The equation becomes 2x+3=2x+n2x + 3 = 2x + n. Subtracting 2x2x from both sides gives 3=n3 = n. When n=3n = 3, we have 3=33 = 3, which is always true, indicating infinitely many solutions. Choice A misinterprets the identity as a unique solution. Choice C would be correct if we had 3=n3 = n where n3n \neq 3. Choice D incorrectly distributes 126=6\frac{1}{2} \cdot 6 = 6 instead of 33.

Question 20

Solve and classify the equation 6(x+1)3=6x+36(x+1)-3=6x+3. Which statement is true?

  1. It has exactly one solution, x=1x=1.
  2. It has infinitely many solutions because it simplifies to a true statement. (correct answer)
  3. It has exactly one solution, x=0x=0.
  4. It has no solutions because it simplifies to a false statement.
Explanation: This question tests classifying linear equations by solution count: one (x=a), infinitely many (identity a=a), or none (contradiction a≠b). Solving reveals type: distribute and collect terms (2(x+3)=2x+6 → 2x+6=2x+6 → 6=6 identity), result shows classification (x=5 means one solution 5, 3=3 means all x work infinitely many, 4=7 means no x works no solution); type depends on whether variables cancel: if x remains with unique value (one solution), if variables cancel to truth (infinite), if cancel to falsehood (none). For the equation 6(x+1)-3=6x+3, distribute to get 6x+6-3=6x+3, simplify to 6x+3=6x+3, subtract 6x to get 3=3. This results in 3=3, an identity that is always true, so the equation has infinitely many solutions. A common error is thinking an identity like 3=3 means x=0 or one solution, or mistakenly claiming it has no solutions like a contradiction. The strategy is to (1) distribute parentheses, (2) collect like terms (combine x's, combine constants), (3) move variables to one side, (4) observe result (x=number → one, number=number → infinite, number=different → none), (5) verify (substitute x back if one solution, try multiple x values if claiming infinite). Mistakes include stopping too early (not simplifying fully), misinterpreting 0=0 as x=0, claiming 5=8 has solution, distributing incorrectly changing the type.