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Difficulty: HardLinear Equations in One Variable

If 23(3x−12)−34(2x−13)=16(x+4)−512\frac{2}{3}\left(3x - \frac{1}{2}\right) - \frac{3}{4}\left(2x - \frac{1}{3}\right) = \frac{1}{6}(x + 4) - \frac{5}{12}, what is the value of 12x−512x - 5?

Answer: 7

Answer

7
The correct answer is 7. Expanding the left side of the equation yields 2x−13−32x+14=12x−1122x - \frac{1}{3} - \frac{3}{2}x + \frac{1}{4} = \frac{1}{2}x - \frac{1}{12}. Expanding the right side yields 16x+46−512=16x+14\frac{1}{6}x + \frac{4}{6} - \frac{5}{12} = \frac{1}{6}x + \frac{1}{4}. Setting the two sides equal gives 12x−112=16x+14\frac{1}{2}x - \frac{1}{12} = \frac{1}{6}x + \frac{1}{4}. Multiplying all terms by 12 clears the fractions, resulting in 6x−1=2x+36x - 1 = 2x + 3. Solving for xx gives 4x=44x = 4, which means x=1x = 1. Substituting x=1x = 1 into 12x−512x - 5 yields 12(1)−5=712(1) - 5 = 7.

Step-by-Step Solution

1
Distribute the factors on the left side of the equation: 23(3x−12)\frac{2}{3}\left(3x - \frac{1}{2}\right) and −34(2x−13)-\frac{3}{4}\left(2x - \frac{1}{3}\right).
2x−13−32x+142x - \frac{1}{3} - \frac{3}{2}x + \frac{1}{4}
To eliminate the parentheses and prepare the left side of the equation for combining like terms.
2
Combine the variable terms and the constant terms on the left side: (2x−32x)+(−13+14)\left(2x - \frac{3}{2}x\right) + \left(-\frac{1}{3} + \frac{1}{4}\right).
12x−112\frac{1}{2}x - \frac{1}{12}
To simplify the left side into a single linear expression with a common denominator for the constants.
3
Expand and simplify the right side of the equation: 16(x+4)−512\frac{1}{6}(x + 4) - \frac{5}{12}.
16x+14\frac{1}{6}x + \frac{1}{4}
By distributing 16\frac{1}{6}, we get 16x+46−512\frac{1}{6}x + \frac{4}{6} - \frac{5}{12}. Finding a common denominator of 12 for the constant terms yields 812−512=312=14\frac{8}{12} - \frac{5}{12} = \frac{3}{12} = \frac{1}{4}.
4
Equate the simplified left and right sides: 12x−112=16x+14\frac{1}{2}x - \frac{1}{12} = \frac{1}{6}x + \frac{1}{4}. Multiply both sides of the equation by 12 to clear the fractions.
6x−1=2x+36x - 1 = 2x + 3, which simplifies to 4x=44x = 4, and thus x=1x = 1.
To solve for the variable xx in a simplified integer form.
5
Substitute x=1x = 1 into the requested expression 12x−512x - 5.
12(1)−5=712(1) - 5 = 7
To calculate the final value asked by the question.

Key Concept

Solving multi-step linear equations in one variable with fractional coefficients, distributing negative signs, and evaluating expressions.
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Topics

Linear Equations in One Variable

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