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Difficulty: HardLinear Equations in Two Variables

A linear relationship exists between the variables pp and qq. When the value of pp increases by 66, the value of qq decreases by 44. If q=15q = 15 when p=2p = 2, which of the following equations correctly expresses pp in terms of qq?

  1. p=−32q+492p = -\frac{3}{2}q + \frac{49}{2}Answer
  2. B
    p=−23q+493p = -\frac{2}{3}q + \frac{49}{3}
  3. C
    p=32q−412p = \frac{3}{2}q - \frac{41}{2}
  4. D
    p=−23q+12p = -\frac{2}{3}q + 12

Answer

p=−32q+492p = -\frac{3}{2}q + \frac{49}{2}
The correct equation is found by first calculating the rate of change of qq with respect to pp, which is −46=−23\frac{-4}{6} = -\frac{2}{3}. Using the point-slope form with (p,q)=(2,15)(p, q) = (2, 15), we set up the equation q−15=−23(p−2)q - 15 = -\frac{2}{3}(p - 2). To express pp in terms of qq, we isolate pp by multiplying both sides by −32-\frac{3}{2} to get −32q+452=p−2-\frac{3}{2}q + \frac{45}{2} = p - 2, and then adding 22 to both sides, which yields p=−32q+492p = -\frac{3}{2}q + \frac{49}{2}.

Step-by-Step Solution

1
Calculate the rate of change of qq relative to pp using the given changes.
The rate of change is −46=−23-\frac{4}{6} = -\frac{2}{3}.
Since qq decreases by 44 when pp increases by 66, the slope mm in terms of Δq/Δp\Delta q / \Delta p is −23-\frac{2}{3}.
2
Set up the point-slope form of the linear equation using the point (p,q)=(2,15)(p, q) = (2, 15).
q−15=−23(p−2)q - 15 = -\frac{2}{3}(p - 2)
The point-slope form is q−q1=m(p−p1)q - q_1 = m(p - p_1), where (p1,q1)=(2,15)(p_1, q_1) = (2, 15) and m=−23m = -\frac{2}{3}.
3
Solve the equation for pp by first clearing the coefficient of the pp-term.
−32(q−15)=p−2  ⟹  −32q+452=p−2-\frac{3}{2}(q - 15) = p - 2 \implies -\frac{3}{2}q + \frac{45}{2} = p - 2
Multiplying both sides by −32-\frac{3}{2} simplifies isolation of the variable pp.
4
Complete the isolation of pp by adding 22 to both sides.
p=−32q+452+2  ⟹  p=−32q+492p = -\frac{3}{2}q + \frac{45}{2} + 2 \implies p = -\frac{3}{2}q + \frac{49}{2}
Adding 22 (which is 42\frac{4}{2}) to 452\frac{45}{2} isolates pp on one side of the equation.

Key Concept

Linear relationships and isolating variables in two-variable linear equations.
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Topics

Linear Equations in Two Variables

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