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Difficulty: HardLinear Equations in One Variable

In the equation below, aa and bb are constants.

13(2a−5x)−34(x−b)=−2912x+5\frac{1}{3}(2a - 5x) - \frac{3}{4}(x - b) = -\frac{29}{12}x + 5

If the equation has infinitely many solutions for xx, what is the value of 8a+9b8a + 9b?

Answer: 60

Answer

The value of 8a+9b8a + 9b is 6060.
To find the value of 8a+9b8a + 9b that makes the equation have infinitely many solutions, we first expand and simplify the left side of the equation: 23a−53x−34x+34b=−2912x+5\frac{2}{3}a - \frac{5}{3}x - \frac{3}{4}x + \frac{3}{4}b = -\frac{29}{12}x + 5. Combining the xx terms gives -\frac{29}{12}x + \(\frac{2}{3}a + \frac{3}{4}b\) = -\frac{29}{12}x + 5. Since the coefficients of xx on both sides are equal (−2912-\frac{29}{12}), the equation will have infinitely many solutions if the constant terms on both sides are also equal. This requires 23a+34b=5\frac{2}{3}a + \frac{3}{4}b = 5. Multiplying this entire equation by the least common multiple of the denominators, which is 12, yields 8a+9b=608a + 9b = 60.

Step-by-Step Solution

1
Distribute the constants through the parentheses on the left side of the equation.
23a−53x−34x+34b=−2912x+5\frac{2}{3}a - \frac{5}{3}x - \frac{3}{4}x + \frac{3}{4}b = -\frac{29}{12}x + 5
This separates the variable terms from the constant terms so the equation can be simplified.
2
Combine the coefficients of the xx terms on the left side using 12 as the common denominator.
−2912x+23a+34b=−2912x+5-\frac{29}{12}x + \frac{2}{3}a + \frac{3}{4}b = -\frac{29}{12}x + 5
Simplifying the variable terms allows us to compare the coefficients on both sides of the equation.
3
Equate the constant terms from the left and right sides of the equation.
23a+34b=5\frac{2}{3}a + \frac{3}{4}b = 5
A linear equation has infinitely many solutions when the coefficients of the variable on both sides are equal and the constant terms on both sides are also equal.
4
Multiply both sides of the equation by 12 to eliminate the fractional denominators.
8a+9b=608a + 9b = 60
Multiplying the equation by the common denominator directly evaluates the target expression 8a+9b8a + 9b.

Key Concept

For a linear equation in one variable to have infinitely many solutions, it must be reducible to an identity of the form cx+d=cx+dcx + d = cx + d, where both the variable coefficients and the constant terms on both sides of the equation are equal.
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Topics

Linear Equations in One Variable

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