# How do you solve 2p + q = 1 and 9p + 3q + 3 = 0  using substitution?

Apr 8, 2016

$\left(p , q\right) = \left(- 2 , 5\right)$
(see below for substitution methodology)

#### Explanation:

Given $2 p + q = 1$
$\rightarrow \textcolor{red}{q} = \left(\textcolor{b l u e}{1 - 2 p}\right)$

Also given $9 \textcolor{g r e e n}{p} + 3 \textcolor{b r o w n}{q} + 3 = 0$

Substituting $\left(\textcolor{b l u e}{1 - 2 p}\right)$ for $\textcolor{b r o w n}{q}$

$\textcolor{w h i t e}{\text{XXX}} 9 \textcolor{g r e e n}{p} + 3 \left(\textcolor{b l u e}{1 - 2 p}\right) + 3 = 0$

$\textcolor{w h i t e}{\text{XXX}} 3 p + 3 + 3 = 0$

$\textcolor{w h i t e}{\text{XXX}} \textcolor{c y a n}{p} = \left(\textcolor{\mathmr{and} a n \ge}{- 2}\right)$

Substituting $\left(\textcolor{\mathmr{and} a n \ge}{- 2}\right)$ for $p$ in the original equation:
$\textcolor{w h i t e}{\text{XXX}} 2 \left(\textcolor{\mathmr{and} a n \ge}{- 2}\right) + q = 1$

$\textcolor{w h i t e}{\text{XXX}} q = 5$