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Difficulty: Very hardLinear Equations in One Variable
For real constants aa and bb, consider the linear equation in one variable xx:
a(x−2)3−2x+14=(a−3)x+b12\frac{a(x - 2)}{3} - \frac{2x + 1}{4} = \frac{(a - 3)x + b}{12}
If this equation has infinitely many solutions for xx, which of the following statements must be true? Select all such statements.
  1. a+b=−10a + b = -10Answer
  2. ab<0ab < 0Answer
  3. 2a−b=132a - b = 13Answer
  4. D
    a−b=−10a - b = -10
  5. E
    ab=11ab = 11

Answer

The statements a+b=−10a + b = -10, ab<0ab < 0, and 2a−b=132a - b = 13 are all true.
Clearing denominators gives 4a(x−2)−3(2x+1)=(a−3)x+b4a(x - 2) - 3(2x + 1) = (a - 3)x + b. Expanding both sides yields (4a−6)x−(8a+3)=(a−3)x+b(4a - 6)x - (8a + 3) = (a - 3)x + b, which rearranges to (3a−3)x=8a+b+3(3a - 3)x = 8a + b + 3. For a linear equation in one variable to have infinitely many solutions, both the coefficient of xx and the constant term must be zero (0x=00x = 0). Setting 3a−3=03a - 3 = 0 gives a=1a = 1, and substituting a=1a = 1 into 8a+b+3=08a + b + 3 = 0 yields b=−11b = -11. Evaluating the statements with a=1a = 1 and b=−11b = -11 shows that a+b=1+(−11)=−10a + b = 1 + (-11) = -10 is true, ab=−11<0ab = -11 < 0 is true, and 2a−b=2(1)−(−11)=132a - b = 2(1) - (-11) = 13 is true.

Step-by-Step Solution

1
Clear the denominators by multiplying both sides of the equation by 12.
4a(x−2)−3(2x+1)=(a−3)x+b4a(x - 2) - 3(2x + 1) = (a - 3)x + b
Eliminating fractions simplifies expanding terms and combining like variables.
2
Expand all terms and group terms containing xx on the left-hand side.
4ax−8a−6x−3=(a−3)x+b  ⟹  (4a−6)x−(8a+3)=(a−3)x+b4ax - 8a - 6x - 3 = (a - 3)x + b \implies (4a - 6)x - (8a + 3) = (a - 3)x + b
Preparing the linear equation to be expressed in standard form Ax=BAx = B.
3
Rearrange into standard form (A)x=B(A)x = B.
[(4a−6)−(a−3)]x=8a+b+3  ⟹  (3a−3)x=8a+b+3[(4a - 6) - (a - 3)]x = 8a + b + 3 \implies (3a - 3)x = 8a + b + 3
A linear equation has infinitely many solutions if and only if A=0A = 0 and B=0B = 0 simultaneously.
4
Set the coefficient of xx and the constant term equal to zero to determine aa and bb.
3a−3=0  ⟹  a=13a - 3 = 0 \implies a = 1; then 8(1)+b+3=0  ⟹  b+11=0  ⟹  b=−118(1) + b + 3 = 0 \implies b + 11 = 0 \implies b = -11
The equation reduces to 0⋅x=00 \cdot x = 0, which is satisfied by every real number xx.
5
Evaluate the given statements using a=1a = 1 and b=−11b = -11.
a+b=1+(−11)=−10a + b = 1 + (-11) = -10 (True); ab=(1)(−11)=−11<0ab = (1)(-11) = -11 < 0 (True); 2a−b=2(1)−(−11)=132a - b = 2(1) - (-11) = 13 (True).
Direct substitution confirms which statements hold true.

Key Concept

Conditions for a linear equation in one variable to have infinitely many solutions
Estimated Time:2m 30s
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Topics

Linear Equations in One Variable

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