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Difficulty: MediumLinear Equations in One Variable

In the equation 3(2x−a)−4(x+1)=2x+53(2x - a) - 4(x + 1) = 2x + 5, aa is a constant. If the equation has infinitely many solutions, what is the value of aa?

  1. A
    −13-\frac{1}{3}
  2. −3-3Answer
  3. C
    33
  4. D
    −9-9

Answer

The value of aa must be −3-3.
The value of −3-3 is correct. Expanding and simplifying the left side of the equation yields 2x−3a−4=2x+52x - 3a - 4 = 2x + 5. For a linear equation to have infinitely many solutions, the constant terms on both sides must be identical. Setting −3a−4=5-3a - 4 = 5 and solving for aa gives −3a=9-3a = 9, which simplifies to a=−3a = -3.

Step-by-Step Solution

1
Distribute the coefficients to the terms inside the parentheses on the left side of the equation.
6x−3a−4x−4=2x+56x - 3a - 4x - 4 = 2x + 5
Distributing eliminates the parentheses, allowing like terms to be combined.
2
Combine the variable terms on the left side of the equation.
2x−3a−4=2x+52x - 3a - 4 = 2x + 5
Grouping the xx terms (6x−4x=2x6x - 4x = 2x) simplifies the structure of the equation.
3
Set up the condition for the equation to have infinitely many solutions.
−3a−4=5-3a - 4 = 5
For a linear equation to have infinitely many solutions, the coefficients of xx on both sides must be equal, and the constant terms must also be equal. The xx coefficients are already equal (2=22 = 2), so we equate the constants.
4
Isolate the constant aa by adding 44 to both sides and then dividing by −3-3.
a=−3a = -3
Adding 44 yields −3a=9-3a = 9, and dividing by −3-3 isolates the variable to find the solution.

Key Concept

Linear Equations in One Variable

Alternative Method

Instead of fully simplifying the equation, substitute x=0x = 0 directly into the original equation, since a statement with infinitely many solutions must hold true for all values of xx. Substituting x=0x = 0 gives 3(−a)−4(1)=53(-a) - 4(1) = 5, which simplifies directly to −3a−4=5-3a - 4 = 5, yielding a=−3a = -3.
Estimated Time:1m 15s
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Topics

Linear Equations in One Variable

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