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Difficulty: Very hardLinear Inequalities in One Variable

Let pp and qq be constants. The inequality p(x−5)<q(x+4)p(x - 5) < q(x + 4) has the solution set x>2x > 2. If p−q=−3p - q = -3, what is the value of p+qp + q?

  1. A
    1
  2. B
    39
  3. -1Answer
  4. D
    -3

Answer

-1
The correct answer is −1-1. Expanding the inequality p(x−5)<q(x+4)p(x - 5) < q(x + 4) gives px−5p<qx+4qpx - 5p < qx + 4q. Grouping the xx terms yields (p−q)x<5p+4q(p - q)x < 5p + 4q. Since p−q=−3p - q = -3, which is negative, dividing both sides by p−qp - q reverses the inequality to x>5p+4qp−qx > \frac{5p + 4q}{p - q}. Comparing this to the given solution x>2x > 2, the boundary value must satisfy 5p+4q−3=2\frac{5p + 4q}{-3} = 2, which simplifies to 5p+4q=−65p + 4q = -6. Using the system of equations p−q=−3p - q = -3 and 5p+4q=−65p + 4q = -6, we find p=−2p = -2 and q=1q = 1. Therefore, the value of p+qp + q is −2+1=−1-2 + 1 = -1.

Step-by-Step Solution

1
Expand both sides of the inequality and group the terms with xx on one side.
(p−q)x<5p+4q(p - q)x < 5p + 4q
Expanding p(x−5)<q(x+4)p(x - 5) < q(x + 4) gives px−5p<qx+4qpx - 5p < qx + 4q. Subtracting qxqx and adding 5p5p to both sides yields px−qx<5p+4qpx - qx < 5p + 4q, which factors to (p−q)x<5p+4q(p - q)x < 5p + 4q.
2
Determine the direction of the inequality when isolating xx.
x>5p+4qp−qx > \frac{5p + 4q}{p - q}
Since we are given p−q=−3p - q = -3, the quantity p−qp - q is negative. Dividing both sides of the inequality by a negative value reverses the inequality symbol from << to >>.
3
Equate the resulting boundary expression to the boundary of the given solution set x>2x > 2.
5p + 4q = -6
The boundary value of the solution set is 22, so we set 5p+4qp−q=2\frac{5p + 4q}{p - q} = 2. Substituting p−q=−3p - q = -3 gives 5p+4q−3=2\frac{5p + 4q}{-3} = 2, which simplifies to 5p+4q=−65p + 4q = -6.
4
Solve the system of linear equations for pp and qq.
p=−2p = -2 and q=1q = 1
We have the system of equations p−q=−3p - q = -3 and 5p+4q=−65p + 4q = -6. From the first equation, p=q−3p = q - 3. Substituting this into the second equation gives 5(q−3)+4q=−6  ⟹  9q−15=−6  ⟹  9q=9  ⟹  q=15(q - 3) + 4q = -6 \implies 9q - 15 = -6 \implies 9q = 9 \implies q = 1. Substituting q=1q = 1 back into p=q−3p = q - 3 gives p=−2p = -2.
5
Calculate the value of p+qp + q.
-1
Adding the values of pp and qq gives p+q=−2+1=−1p + q = -2 + 1 = -1.

Key Concept

Solving linear inequalities in one variable with variable coefficients and applying sign reversal rules.
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Topics

Linear Inequalities in One Variable

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