# How do you solve the system 2p - 5q = 14 and p + 3/2 q = 5?

Aug 21, 2017

See a solution process below:

#### Explanation:

Step 1) Solve each equation for $2 p$:

Equation 1

$2 p - 5 q = 14$

$2 p - 5 q + \textcolor{red}{5 q} = 14 + \textcolor{red}{5 q}$

$2 p - 0 = 14 + 5 q$

$2 p = 14 + 5 q$

Equation 2

$p + \frac{3}{2} q = 5$

$p + \frac{3}{2} q - \textcolor{red}{\frac{3}{2} q} = 5 - \textcolor{red}{\frac{3}{2} q}$

$p + 0 = 5 - \frac{3}{2} q$

$p = 5 - \frac{3}{2} q$

$\textcolor{red}{2} \times p = \textcolor{red}{2} \left(5 - \frac{3}{2} q\right)$

$2 p = \left(\textcolor{red}{2} \times 5\right) - \left(\textcolor{red}{2} \times \frac{3}{2} q\right)$

$2 p = 10 - 3 q$

Step 2) Substitute $10 - 3 q$ from the second equation for $2 p$ in the first equation and solve for $p$:

$2 p = 14 + 5 q$ becomes:

$10 - 3 q = 14 + 5 q$

$- \textcolor{b l u e}{14} + 10 - 3 q + \textcolor{red}{3 q} = - \textcolor{b l u e}{14} + 14 + 5 q + \textcolor{red}{3 q}$

$- 4 - 0 = 0 + \left(5 + \textcolor{red}{3}\right) q$

$- 4 = 8 q$

$- \frac{4}{\textcolor{red}{8}} = \frac{8 q}{\textcolor{red}{8}}$

$- \frac{1}{2} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{8}}} q}{\cancel{\textcolor{red}{8}}}$

$- \frac{1}{2} = q$ or $q = - \frac{1}{2}$

Step 3) Substitute $- \frac{1}{2}$ for $q$ in either of the equations from Step 1 and calculate $p$:

$2 p = 14 + 5 q$ becomes:

$2 p = 14 + \left(5 \cdot - \frac{1}{2}\right)$

$2 p = 14 - \frac{5}{2}$

$2 p = \left(\frac{2}{2} \cdot 14\right) - \frac{5}{2}$

$2 p = \frac{28}{2} - \frac{5}{2}$

$2 p = \frac{23}{2}$

$\textcolor{red}{\frac{1}{2}} \cdot 2 p = \textcolor{red}{\frac{1}{2}} \cdot \frac{23}{2}$

$1 p = \frac{23}{4}$

$p = \frac{23}{4}$

The Solution Is: $p = \frac{23}{4}$ and $q = - \frac{1}{2}$