Question

Difficulty: HardLinear Equations in One Variable

For what value of the constant kk does the equation 12(2kx6)23(93x)=5x+4\frac{1}{2}(2kx - 6) - \frac{2}{3}(9 - 3x) = 5x + 4 have no solution for xx?

  1. 3Answer
  2. B
    7
  3. C
    1.5
  4. D
    -7

Answer

The correct value of kk is 33.
To find the value of kk for which the equation has no solution, we first distribute the coefficients on the left side of the equation to get kx36+2x=5x+4kx - 3 - 6 + 2x = 5x + 4. Combining the like terms on the left side gives (k+2)x9=5x+4(k + 2)x - 9 = 5x + 4. Moving all the xx terms to one side yields (k3)x=13(k - 3)x = 13. For a linear equation of the form Ax=BAx = B to have no solution, the coefficient of the variable AA must equal 00 while the constant BB must be non-zero. Setting k3=0k - 3 = 0 gives the correct value of 33.

Step-by-Step Solution

1
Distribute the fractional coefficients on the left side of the equation.
The term 12(2kx6)\frac{1}{2}(2kx - 6) simplifies to kx3kx - 3, and the term 23(93x)-\frac{2}{3}(9 - 3x) simplifies to 6+2x-6 + 2x.
Applying the distributive property simplifies the parenthetical expressions.
2
Substitute the simplified expressions back into the equation and group like terms.
The equation becomes kx36+2x=5x+4kx - 3 - 6 + 2x = 5x + 4, which simplifies to (k+2)x9=5x+4(k + 2)x - 9 = 5x + 4.
Grouping terms allows us to isolate the variable.
3
Isolate the terms containing xx on one side of the equation.
Subtract 5x5x and add 99 to both sides, yielding (k3)x=13(k - 3)x = 13.
This puts the equation into the standard linear form Ax=BAx = B.
4
Set the coefficient of xx to 00 to find the value of kk that yields no solution.
Setting k3=0k - 3 = 0 gives k=3k = 3. Since 13013 \neq 0, the equation 0x=130x = 13 has no solution.
A linear equation of the form Ax=BAx = B has no solution if and only if A=0A = 0 and B0B \neq 0.

Key Concept

Determining the parameter value for which a linear equation in one variable has no solution.
Estimated Time:2m 0s
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