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Difficulty: Very hardLinear Equations in One Variable
In the equation below, pp and qq are constants.
2p(6x−9)−3q(4x+2)=−2x−9\frac{2}{p}(6x - 9) - \frac{3}{q}(4x + 2) = -2x - 9
If the equation has infinitely many solutions for xx, what is the value of p+qp + q?
  1. A
    1
  2. 5Answer
  3. C
    9
  4. D
    -5

Answer

The value of p+qp+q is 55.
For the equation to have infinitely many solutions for xx, the coefficients of xx on both sides must be equal, and the constant terms on both sides must also be equal. Expanding the left side of the equation gives (12p−12q)x−(18p+6q)=−2x−9\left(\frac{12}{p} - \frac{12}{q}\right)x - \left(\frac{18}{p} + \frac{6}{q}\right) = -2x - 9. Equating the coefficients of xx yields 12p−12q=−2\frac{12}{p} - \frac{12}{q} = -2, which simplifies to 6p−6q=−1\frac{6}{p} - \frac{6}{q} = -1. Equating the constant terms yields −18p−6q=−9-\frac{18}{p} - \frac{6}{q} = -9, which simplifies to 18p+6q=9\frac{18}{p} + \frac{6}{q} = 9. Adding these two simplified equations eliminates the term with qq, giving 24p=8\frac{24}{p} = 8, which means p=3p = 3. Substituting p=3p = 3 back into the first equation yields 2−6q=−12 - \frac{6}{q} = -1, which simplifies to 6q=3\frac{6}{q} = 3, meaning q=2q = 2. Thus, the value of p+qp + q is 3+2=53 + 2 = 5.

Step-by-Step Solution

1
Expand and group the terms on the left side of the equation.
(12p−12q)x−(18p+6q)=−2x−9\left(\frac{12}{p} - \frac{12}{q}\right)x - \left(\frac{18}{p} + \frac{6}{q}\right) = -2x - 9
To analyze the linear equation, we must group the coefficients of the variable xx and the constant terms.
2
Set up a system of equations for pp and qq using the condition for infinitely many solutions.
12p−12q=−2\frac{12}{p} - \frac{12}{q} = -2 and −18p−6q=−9-\frac{18}{p} - \frac{6}{q} = -9
A linear equation in the form Ax+B=Cx+DAx + B = Cx + D has infinitely many solutions if and only if A=CA = C and B=DB = D.
3
Simplify the system of equations.
6p−6q=−1\frac{6}{p} - \frac{6}{q} = -1 (Equation 1) and 18p+6q=9\frac{18}{p} + \frac{6}{q} = 9 (Equation 2)
Dividing Equation 1 by 22 and Equation 2 by −1-1 simplifies the coefficients, making the system easier to solve.
4
Solve for pp by adding Equation 1 and Equation 2.
24p=8  ⟹  p=3\frac{24}{p} = 8 \implies p = 3
Adding the two equations eliminates the term containing qq, allowing us to solve directly for pp.
5
Substitute p=3p = 3 back into Equation 1 to solve for qq.
63−6q=−1  ⟹  2−6q=−1  ⟹  6q=3  ⟹  q=2\frac{6}{3} - \frac{6}{q} = -1 \implies 2 - \frac{6}{q} = -1 \implies \frac{6}{q} = 3 \implies q = 2
Using the value of pp allows us to isolate and solve for qq.
6
Calculate the sum of pp and qq.
p+q=3+2=5p + q = 3 + 2 = 5
The question asks for the value of p+qp + q.

Key Concept

Conditions for a linear equation in one variable to have infinitely many solutions, and solving systems of literal equations.
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Topics

Linear Equations in One Variable

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