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Difficulty: HardLinear Equations in One Variable
In the equation below, kk is a constant.
5x−2(3−k)x=9x−15x - 2(3 - k)x = 9x - 1
If the equation has no solution, what is the value of kk?
  1. A
    -5
  2. B
    0
  3. 5Answer
  4. D
    4

Answer

5
To find the value of kk for which the equation has no solution, we simplify the left side of the equation: 5x−2(3−k)x=5x−6x+2kx=−x+2kx=(2k−1)x5x - 2(3 - k)x = 5x - 6x + 2kx = -x + 2kx = (2k - 1)x. This yields the simplified equation (2k−1)x=9x−1(2k - 1)x = 9x - 1. A linear equation in one variable has no solution when the variable terms on both sides cancel each other out (meaning the coefficients of xx are equal) but the constant terms are different. Setting the coefficients of xx equal to each other gives 2k−1=92k - 1 = 9. Adding 11 to both sides gives 2k=102k = 10, and dividing by 22 yields k=5k = 5. Since the constant term on the left side is 00 and the constant term on the right side is −1-1, the constants are different, confirming that the equation has no solution when kk equals 55.

Step-by-Step Solution

1
Distribute the term −2-2 to the expressions inside the parentheses on the left side of the equation.
5x−6x+2kx=9x−15x - 6x + 2kx = 9x - 1
To eliminate the parentheses so that all like terms can be grouped.
2
Combine the xx terms on the left side of the equation.
−x+2kx=9x−1-x + 2kx = 9x - 1, which can be factored as (2k−1)x=9x−1(2k - 1)x = 9x - 1
Grouping the coefficients of the variable xx allows direct comparison of both sides of the linear equation.
3
Set the coefficient of xx on the left side equal to the coefficient of xx on the right side.
2k−1=92k - 1 = 9
For a linear equation in one variable to have no solution, the variable terms on both sides must cancel out (meaning their coefficients must be equal), while the constant terms must remain unequal.
4
Solve the resulting equation for kk.
2k=10  ⟹  k=52k = 10 \implies k = 5
Isolating kk by adding 11 to both sides and then dividing by 22 determines the specific constant value.
5
Verify that the constant terms are different when k=5k = 5.
Substitute k=5k = 5 back into the original equation to get 9x=9x−19x = 9x - 1, which simplifies to 0=−10 = -1.
Since 0=−10 = -1 is a false statement, the equation has no solution, confirming that k=5k = 5 is correct.

Key Concept

Linear Equations in One Variable

Alternative Method

Instead of algebraic simplification, the value of kk can be found by substituting the answer choices into the equation to see which value eliminates the variable xx while leaving an untrue statement. Plugging in 55 for kk gives 5x−2(3−5)x=9x−1  ⟹  5x−2(−2)x=9x−1  ⟹  9x=9x−1  ⟹  0=−15x - 2(3 - 5)x = 9x - 1 \implies 5x - 2(-2)x = 9x - 1 \implies 9x = 9x - 1 \implies 0 = -1. Because this statement is false, the equation has no solution, verifying that 55 is the correct answer.
Estimated Time:2m 0s
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Topics

Linear Equations in One Variable

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