# Math Help - Two Precalculus Problems

1. ## 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.

2. 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?