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Math Help - Two Precalculus Problems

  1. #1
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    Two Precalculus Problems

    1.)  3x/2 (2x-17/3)=3/2 . Note: To clarify, the "/" mark indicates fractions. Everything is a fraction except for 2x. I'm not certain how to solve this one, since I don't know exactly how to distribute fractions.

    2.) Given the following two equations, solve for q:

    Equation 1: 2p-q=50 and Equation 2: p=(100+20q)/q

    When combined, these equations should result in a quadratic equation. What I did first was substitute equation 2 for "p" in equation 1. However, I'm not certain if I solved the rest correctly from there. The quadratic equation that I found was -x^2 + 40x+150=0. However, the resulting quantity was 43.452, which doesn't seem right because it is not a whole number. Please help me find the correct solution for q.

    Thanks in advance. Your assistance is much appreciated.
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  2. #2
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    Hello, currypuff!

    1)\;\;\frac{3x}{2}\left(2x-\frac{17}{3}\right) \:=\:\frac{3}{2}

    I don't know exactly how to distribute fractions. . . . . You don't?

    Multiply both sides by 2: . 3x\left(2x-\frac{17}{3}\right) \:=\:3

    Distribute: . (3x)(2x) - (3x)\left(\frac{17}{3}\right) \;=\;3 \quad\Rightarrow\quad 6x^2 - 17x \:=\:3 \quad\Rightarrow\quad 6x^2 - 17x - 3 \:=\:0

    Factor: . (x - 3)(6x + 1) \:=\:0 . . . . . and so on




    2) Solve for q\!:\;\;\begin{array}{cccc}2p-q&=&50 & {\color{blue}[1]} \\  \\ p &=& \dfrac{100+20q}{q} & {\color{blue}[2]}\end{array}

    Substitute [2] into [1]: . 2\left(\frac{100+2q}{q}\right) - q \;=\;50

    Multiply by q\!:\;\;2(100 + 2q) - q^2 \;=\;50q \quad\Rightarrow\quad q^2 + 46q - 200 \:=\:0

    Factor: . (q - 4)(q + 50) \:=\:0 . . . . . Got it?

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