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Zorluk: ZorSystems of Linear Equations

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

kx−3y=12(k−5)x−2y=10\begin{aligned} kx - 3y &= 12 \\ (k-5)x - 2y &= 10 \end{aligned}

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

  1. A
    -15
  2. B
    -10
  3. C
    5
  4. 15Cevap

Cevap

15
The correct answer is 1515. For a system of two linear equations to have no solution, the equations must represent parallel lines. This occurs when the coefficients of the variables are proportional but the constant terms are not. In this system, the ratio of the xx-coefficients, kk−5\frac{k}{k-5}, must equal the ratio of the yy-coefficients, −3−2=32\frac{-3}{-2} = \frac{3}{2}. Solving the equation kk−5=32\frac{k}{k-5} = \frac{3}{2} by cross-multiplying gives 2k=3(k−5)2k = 3(k-5), which simplifies to 2k=3k−152k = 3k - 15. Subtracting 3k3k from both sides gives −k=−15-k = -15, so k=15k = 15. Since the ratio of the constants is 1210=1.2\frac{12}{10} = 1.2, which is not equal to 1.51.5, this value of kk ensures the lines are parallel and distinct.

Adım Adım Çözüm

1
Set up the condition for a system of linear equations to have no solution.
The coefficients of xx and yy must be proportional, but not equal to the ratio of the constant terms: A1A2=B1B2≠C1C2\frac{A_1}{A_2} = \frac{B_1}{B_2} \neq \frac{C_1}{C_2}.
If the coefficients are proportional, the lines have the same slope. If the constant terms are not in the same proportion, the lines have different intercepts, meaning they are parallel and distinct, and thus never intersect.
2
Substitute the coefficients from the given equations into the proportion.
kk−5=−3−2\frac{k}{k-5} = \frac{-3}{-2} and verify that −3−2=1.5≠1210=1.2\frac{-3}{-2} = 1.5 \neq \frac{12}{10} = 1.2.
This establishes the relationship between the coefficients of xx and yy required to make the lines parallel, while ensuring they are not identical lines.
3
Solve the proportion for kk.
kk−5=32⇒2k=3(k−5)⇒2k=3k−15⇒k=15\frac{k}{k-5} = \frac{3}{2} \Rightarrow 2k = 3(k-5) \Rightarrow 2k = 3k - 15 \Rightarrow k = 15.
Cross-multiplying and isolating the variable kk yields the value of the constant.

Anahtar Kavram

For a system of linear equations to have no solution, the lines represented by the equations must be parallel and distinct, meaning their slopes are equal but their y-intercepts are different.
Tahmini Süre:1m 30s
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Konular

Systems of Linear Equations

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