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Zorluk: Çok zorSystems of Linear Equations

For the constants pp and qq, the given system of linear equations in xx and yy has infinitely many solutions:

px−3y=q4x+(p−7)y=12\begin{aligned} px - 3y &= q \\ 4x + (p - 7)y &= 12 \end{aligned}

Which of the following is a possible value of qq?

  1. A
    3
  2. B
    6
  3. 9Cevap
  4. D
    16

Cevap

9
For the system of linear equations to have infinitely many solutions, the equations must be equivalent. This means the ratio of the coefficients of xx, the coefficients of yy, and the constant terms must all be equal: p4=−3p−7=q12\frac{p}{4} = \frac{-3}{p-7} = \frac{q}{12}. Cross-multiplying the first two ratios gives p(p−7)=−12p(p-7) = -12, which simplifies to p2−7p+12=0p^2 - 7p + 12 = 0. Factoring this quadratic equation gives (p−3)(p−4)=0(p-3)(p-4) = 0, which yields p=3p = 3 or p=4p = 4. Using the relation between the first and third ratios, we have q12=p4\frac{q}{12} = \frac{p}{4}, which simplifies to q=3pq = 3p. Substituting the possible values of pp gives q=3(3)=9q = 3(3) = 9 or q=3(4)=12q = 3(4) = 12. Since 9 is one of the options, it is a possible value of qq.

Adım Adım Çözüm

1
Set up the condition for infinitely many solutions by equating the ratios of the coefficients and the constants.
The coefficients and constants must satisfy the proportion: p4=−3p−7=q12\frac{p}{4} = \frac{-3}{p - 7} = \frac{q}{12}
For a system of two linear equations to have infinitely many solutions, the two equations must represent the exact same line, meaning all corresponding coefficients and constant terms must be proportional.
2
Solve the equation formed by the first two ratios to find the possible values of the parameter pp.
p4=−3p−7⇒p(p−7)=−12⇒p2−7p+12=0\frac{p}{4} = \frac{-3}{p - 7} \Rightarrow p(p - 7) = -12 \Rightarrow p^2 - 7p + 12 = 0. Factoring the quadratic yields (p−3)(p−4)=0(p - 3)(p - 4) = 0, which gives p=3p = 3 or p=4p = 4.
This step determines the values of pp for which the two lines are parallel (i.e., they have equal slopes).
3
Find the corresponding values of qq using the relation between the first and third ratios.
p4=q12⇒q=3p\frac{p}{4} = \frac{q}{12} \Rightarrow q = 3p. Substituting the values of pp:
- If p=3p = 3, then q=3(3)=9q = 3(3) = 9.
- If p=4p = 4, then q=3(4)=12q = 3(4) = 12.
This step ensures that the parallel lines are coincident by equating their y-intercepts.
4
Compare the possible values of qq with the given options.
The value 9 is a possible value of qq and matches one of the choices.
To identify which of the two mathematically valid solutions for qq is listed in the multiple-choice options.

Anahtar Kavram

Conditions for infinitely many solutions in a system of linear equations

Alternatif Yöntem

Instead of using ratios directly, we can write both equations in slope-intercept form: y=mx+by = mx + b. For the first equation, px−3y=q⇒y=p3x−q3px - 3y = q \Rightarrow y = \frac{p}{3}x - \frac{q}{3}. For the second equation, 4x+(p−7)y=12⇒y=−4p−7x+12p−74x + (p-7)y = 12 \Rightarrow y = -\frac{4}{p-7}x + \frac{12}{p-7} (assuming p≠7p \neq 7). For the system to have infinitely many solutions, the two lines must have the same slope and the same y-intercept. Equating the slopes gives p3=−4p−7\frac{p}{3} = -\frac{4}{p-7}, which simplifies to p2−7p+12=0p^2 - 7p + 12 = 0, yielding p=3p = 3 or p=4p = 4. Equating the y-intercepts gives −q3=12p−7⇒q=−36p−7-\frac{q}{3} = \frac{12}{p-7} \Rightarrow q = -\frac{36}{p-7}. Substituting p=3p=3 gives q=−36−4=9q = -\frac{36}{-4} = 9, and substituting p=4p=4 gives q=−36−3=12q = -\frac{36}{-3} = 12.
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Systems of Linear Equations

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