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Difficulty: HardSystems of Linear Equations

In the system of equations below, kk is a constant.

kx−3y=4kx - 3y = 4
(k−2)x−5y=8(k-2)x - 5y = 8

If the system has no solution, what is the value of kk?

  1. A
    -1
  2. B
    5
  3. -3Answer
  4. D
    3

Answer

-3
To find the value of kk for which the system of linear equations has no solution, we determine when the two lines represented by the equations are parallel and distinct. Parallel lines have equal slopes. We can express each equation in slope-intercept form, y=mx+by = mx + b. For the first equation, kx−3y=4kx - 3y = 4 simplifies to y=k3x−43y = \frac{k}{3}x - \frac{4}{3}, giving a slope of k3\frac{k}{3}. For the second equation, (k−2)x−5y=8(k-2)x - 5y = 8 simplifies to y=k−25x−85y = \frac{k-2}{5}x - \frac{8}{5}, giving a slope of k−25\frac{k-2}{5}. Equating the two slopes yields k3=k−25\frac{k}{3} = \frac{k-2}{5}. Cross-multiplying gives 5k=3(k−2)5k = 3(k-2), which simplifies to 5k=3k−65k = 3k - 6. Subtracting 3k3k from both sides gives 2k=−62k = -6, and dividing by 22 results in k=−3k = -3. Since the y-intercepts −43-\frac{4}{3} and −85-\frac{8}{5} are distinct, the lines are parallel and do not intersect, meaning there is no solution.

Step-by-Step Solution

1
Express both equations in slope-intercept form (y=mx+by = mx + b) to find their slopes.
For the first equation, kx−3y=4  ⟹  3y=kx−4  ⟹  y=k3x−43kx - 3y = 4 \implies 3y = kx - 4 \implies y = \frac{k}{3}x - \frac{4}{3} (slope is k3\frac{k}{3}). For the second equation, (k−2)x−5y=8  ⟹  5y=(k−2)x−8  ⟹  y=k−25x−85(k-2)x - 5y = 8 \implies 5y = (k-2)x - 8 \implies y = \frac{k-2}{5}x - \frac{8}{5} (slope is k−25\frac{k-2}{5}).
A system of two linear equations has no solution if the lines are parallel (slopes are equal) and their y-intercepts are different.
2
Set the slopes equal to each other.
k3=k−25\frac{k}{3} = \frac{k-2}{5}
Since the lines must be parallel, their slopes must be equivalent.
3
Solve for kk by cross-multiplying.
5k=3(k−2)  ⟹  5k=3k−6  ⟹  2k=−6  ⟹  k=−35k = 3(k-2) \implies 5k = 3k - 6 \implies 2k = -6 \implies k = -3.
Cross-multiplication isolates the variable kk. Since the y-intercepts (−43-\frac{4}{3} and −85-\frac{8}{5}) are different, this value of kk guarantees the lines are parallel and distinct.

Key Concept

Determining parameters for a system of linear equations with no solution

Alternative Method

Alternatively, align the coefficients of yy to eliminate the yy terms. Multiply the first equation by 55 and the second equation by −3-3:

5(kx−3y)=5(4)  ⟹  5kx−15y=205(kx - 3y) = 5(4) \implies 5kx - 15y = 20
−3((k−2)x−5y)=−3(8)  ⟹  −3(k−2)x+15y=−24-3((k-2)x - 5y) = -3(8) \implies -3(k-2)x + 15y = -24

Adding these two equations yields:
(5k−3(k−2))x=−4(5k - 3(k-2))x = -4
(2k+6)x=−4(2k + 6)x = -4

For the system to have no solution, the coefficient of xx must be 00 while the constant on the right side is non-zero (which it is, −4-4). Setting the coefficient of xx to 00:
2k+6=0  ⟹  k=−32k + 6 = 0 \implies k = -3
Estimated Time:2m 0s
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Topics

Systems of Linear Equations

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