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Difficulty: Very hardLinear Equations in One and Two Variables
Consider the system of linear equations in xx and yy:
kx+6y=18kx + 6y = 18
(k+1)x+9y=27(k + 1)x + 9y = 27
where kk is a constant. If this system has infinitely many solutions (x,y)(x, y), what is the sum of all real values of xx that satisfy the equation ∣kx−10∣=3x−4|kx - 10| = 3x - 4?
  1. 145\frac{14}{5}Answer
  2. B
    −165-\frac{16}{5}
  3. C
    −6-6
  4. D
    1010
  5. E
    65\frac{6}{5}

Answer

The sum of all real values of xx satisfying the equation is 145\frac{14}{5} (or 2.82.8).
The system of linear equations has infinitely many solutions when the ratio of coefficients is constant: kk+1=69=1827=23\frac{k}{k+1} = \frac{6}{9} = \frac{18}{27} = \frac{2}{3}, which yields k=2k = 2. Substituting k=2k = 2 gives ∣2x−10∣=3x−4|2x - 10| = 3x - 4. Breaking this into cases yields 2x−10=3x−4  ⟹  x=−62x - 10 = 3x - 4 \implies x = -6 and −(2x−10)=3x−4  ⟹  x=145-(2x - 10) = 3x - 4 \implies x = \frac{14}{5}. Testing x=−6x = -6 in the original equation gives ∣−22∣=−22|-22| = -22, which is false, making x=−6x = -6 an extraneous solution. Testing x=145x = \frac{14}{5} gives ∣−4.4∣=4.4|-4.4| = 4.4, which is true. Thus, the only real solution is 145\frac{14}{5}, making the option equal to 145\frac{14}{5} correct.

Step-by-Step Solution

1
Determine the value of the constant kk from the system of linear equations.
k=2k = 2
For a system of two linear equations in two variables to have infinitely many solutions, the ratio of coefficients of xx, yy, and the constant terms must all be equal: kk+1=69=1827\frac{k}{k+1} = \frac{6}{9} = \frac{18}{27}. Since 69=23\frac{6}{9} = \frac{2}{3}, setting kk+1=23\frac{k}{k+1} = \frac{2}{3} gives 3k=2k+2  ⟹  k=23k = 2k + 2 \implies k = 2.
2
Substitute k=2k = 2 into the absolute value equation.
∣2x−10∣=3x−4|2x - 10| = 3x - 4
Replacing kk with 22 sets up the single-variable linear absolute value equation to solve.
3
Solve Case 1 where the expression inside the absolute value is non-negative (2x−10≥02x - 10 \ge 0).
x=−6x = -6 (Extraneous)
Setting 2x−10=3x−42x - 10 = 3x - 4 yields x=−6x = -6. However, substituting x=−6x = -6 back into the right side gives 3(−6)−4=−223(-6) - 4 = -22. Since an absolute value cannot equal a negative number, x=−6x = -6 is an extraneous solution.
4
Solve Case 2 where the expression inside the absolute value is negative (2x−10<02x - 10 < 0).
x=145x = \frac{14}{5} (Valid)
Setting −(2x−10)=3x−4-(2x - 10) = 3x - 4 gives −2x+10=3x−4  ⟹  5x=14  ⟹  x=145-2x + 10 = 3x - 4 \implies 5x = 14 \implies x = \frac{14}{5}. Substituting x=2.8x = 2.8 into the right side yields 3(2.8)−4=4.4≥03(2.8) - 4 = 4.4 \ge 0, and ∣2(2.8)−10∣=∣−4.4∣=4.4|2(2.8) - 10| = |-4.4| = 4.4. Thus, x=145x = \frac{14}{5} is valid.
5
Sum all valid real solutions.
145\frac{14}{5}
Since x=145x = \frac{14}{5} is the only valid solution, the sum of all solutions is 145\frac{14}{5}.

Key Concept

Systems of linear equations with infinitely many solutions require identical proportional coefficients. Absolute value equations of the form ∣A∣=B|A| = B require checking candidate solutions in B≥0B \ge 0 to eliminate extraneous roots.
Estimated Time:3m 0s
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Topics

Linear Equations in One and Two Variables

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