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Difficulty: Very hardLinear Equations in One Variable
For a constant aa, consider the following linear equation in one variable xx:
a(x−2)3−2x+14=(a−3)x+512\frac{a(x - 2)}{3} - \frac{2x + 1}{4} = \frac{(a - 3)x + 5}{12}
Which of the following statements are true? Select all such statements.
  1. If a=1a = 1, the equation has no solution for xx.Answer
  2. B
    If a=5a = 5, then x=3x = 3.
  3. If a=−1a = -1, then x=0x = 0.Answer
  4. If x=4x = 4, then a=5a = 5.Answer
  5. E
    If a=0a = 0, the equation has a positive solution for xx.

Answer

The correct statements are the ones asserting that a=1a = 1 leaves the equation with no solution, a=−1a = -1 leads to x=0x = 0, and x=4x = 4 requires a=5a = 5.
Multiplying the equation by 12 and simplifying yields 3(a−1)x=8(a+1)3(a - 1)x = 8(a + 1). When a=1a = 1, the left side becomes 0 while the right side becomes 16, creating an inconsistent equation 0=160 = 16 with no solution. When a=−1a = -1, the equation becomes −6x=0-6x = 0, which gives x=0x = 0. When x=4x = 4, substituting into the simplified equation gives 12(a−1)=8(a+1)12(a - 1) = 8(a + 1), which simplifies to 4a=20  ⟹  a=54a = 20 \implies a = 5.

Step-by-Step Solution

1
Clear denominators by multiplying every term of the equation by the least common multiple, 12.
4a(x−2)−3(2x+1)=(a−3)x+54a(x - 2) - 3(2x + 1) = (a - 3)x + 5
Eliminating fractional coefficients simplifies the distribution and collection of variable terms.
2
Expand products on both sides of the equation.
4ax−8a−6x−3=ax−3x+54ax - 8a - 6x - 3 = ax - 3x + 5
Applying the distributive property isolates individual algebraic terms.
3
Rearrange terms to collect all xx-terms on the left side and constant terms on the right side.
(4a−6−a+3)x=8a+8  ⟹  3(a−1)x=8(a+1)(4a - 6 - a + 3)x = 8a + 8 \implies 3(a - 1)x = 8(a + 1)
Factoring out xx provides the canonical linear form Ax=BA x = B.
4
Evaluate the given conditions for aa and xx against the canonical form 3(a−1)x=8(a+1)3(a - 1)x = 8(a + 1).
For a=1a = 1: 0=160 = 16 (no solution). For a=5a = 5: 12x=48  ⟹  x=412x = 48 \implies x = 4. For a=−1a = -1: −6x=0  ⟹  x=0-6x = 0 \implies x = 0. For x=4x = 4: 12(a−1)=8(a+1)  ⟹  a=512(a - 1) = 8(a + 1) \implies a = 5. For a=0a = 0: −3x=8  ⟹  x=−83-3x = 8 \implies x = -\frac{8}{3} (negative).
Direct substitution verifies which algebraic relationships hold true.

Key Concept

Linear equations in one variable containing symbolic parameters can be analyzed for existence of solutions, zero-roots, and explicit values by reducing to the form Ax=BA x = B.
Estimated Time:2m 30s
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Topics

Linear Equations in One Variable

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