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Difficulty: Very hardLinear Equations in One Variable

In the equation 12(kx−4)−23(x−k)=56x−1\frac{1}{2}(kx - 4) - \frac{2}{3}(x - k) = \frac{5}{6}x - 1, where kk is a constant, the equation has no solution for xx. What is the value of kk?

Answer: 3

Answer

The value of the constant kk for which the equation has no solution is 33.
For a linear equation of the form Ax+B=Cx+DAx + B = Cx + D to have no solution, the coefficients of the variable terms must be equal (A=CA = C) but the constant terms must be unequal (B≠DB \neq D). Distributing the terms on the left side of the given equation and grouping them yields (12k−23)x+(23k−2)=56x−1\left(\frac{1}{2}k - \frac{2}{3}\right)x + \left(\frac{2}{3}k - 2\right) = \frac{5}{6}x - 1. Setting the coefficients of xx equal to each other gives 12k−23=56\frac{1}{2}k - \frac{2}{3} = \frac{5}{6}. Adding 23\frac{2}{3} to both sides yields 12k=96\frac{1}{2}k = \frac{9}{6}, or 12k=32\frac{1}{2}k = \frac{3}{2}, which simplifies to k=3k = 3. Evaluating the constant terms when k=3k = 3 gives 23(3)−2=0\frac{2}{3}(3) - 2 = 0 on the left side and −1-1 on the right side. Since 0≠−10 \neq -1, the equation has no solution when k=3k = 3.

Step-by-Step Solution

1
Distribute the fractional coefficients on the left side of the equation.
12kx−2−23x+23k=56x−1\frac{1}{2}kx - 2 - \frac{2}{3}x + \frac{2}{3}k = \frac{5}{6}x - 1
To separate the variable terms from the constants for further algebraic manipulation.
2
Group the terms on the left side into a coefficient for xx and a single constant term.
(12k−23)x+(23k−2)=56x−1\left(\frac{1}{2}k - \frac{2}{3}\right)x + \left(\frac{2}{3}k - 2\right) = \frac{5}{6}x - 1
To represent the equation in the standard linear format Ax+B=Cx+DAx + B = Cx + D.
3
Set the coefficients of xx from both sides equal to each other.
12k−23=56\frac{1}{2}k - \frac{2}{3} = \frac{5}{6}
A linear equation has no solution when the variable terms on both sides cancel out, requiring their coefficients to be identical.
4
Solve the linear equation to isolate the constant kk.
k=3k = 3
Add 23\frac{2}{3} to both sides to get 12k=96\frac{1}{2}k = \frac{9}{6}, which simplifies to 12k=32\frac{1}{2}k = \frac{3}{2}. Multiplying both sides by 22 yields k=3k = 3.
5
Verify that the constant terms are not equal when substituting k=3k = 3.
The left-side constant is 00 and the right-side constant is −1-1. Since 0≠−10 \neq -1, the condition is satisfied.
To ensure the equation does not simplify to an identity with infinitely many solutions (which occurs when both the variable coefficients and the constant terms are equal).

Key Concept

Determining the conditions under which a linear equation in one variable has no solution.

Alternative Method

To avoid working with fractions, multiply every term in the equation by 66 (the least common multiple of 2,3,62, 3, 6) at the start: 3(kx−4)−4(x−k)=5x−63(kx - 4) - 4(x - k) = 5x - 6. Expand the parentheses to get 3kx−12−4x+4k=5x−63kx - 12 - 4x + 4k = 5x - 6, and group the terms: (3k−4)x+(4k−12)=5x−6(3k - 4)x + (4k - 12) = 5x - 6. For there to be no solution, set the coefficients of xx equal to each other: 3k−4=5  ⟹  3k=9  ⟹  k=33k - 4 = 5 \implies 3k = 9 \implies k = 3. Check the constant terms with k=3k = 3: 4(3)−12=04(3) - 12 = 0, which is unequal to −6-6. This confirms k=3k = 3 is the correct answer.
Estimated Time:2m 0s
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Topics

Linear Equations in One Variable

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