# How do you solve this system of equations: 7p + 8q = 11 and 3p + 4q = 3?

Nov 26, 2017

$p = 5 \setminus \quad , \setminus \quad q = - 3$

#### Explanation:

We can use elimination to solve this.

Let’s try to eliminate $q$, to solve for $p$ first:

$7 p + 8 q = 11$

$3 p + 4 q = 3$

Multiply both sides of the bottom equation by $- 2$:

$- 2 \left(3 p + 4 q\right) = \left(3\right) - 2$

$\setminus \rightarrow - 6 p - 8 q = - 6$

Now, the $q$ terms in both equations are opposites of each other, so they will cancel out and give us the value of $p$.

To do that, we need to add both equations together:

$\left(- 6 p - 8 q = - 6\right) \setminus \quad + \setminus \quad \left(7 p + 8 q = 1\right)$

$\setminus \rightarrow p = 5$

Knowing that, we can plug the value of $p$ into one of the equations to find the value of $q$:

$7 p + 8 q = 11$

$\setminus \rightarrow 7 \left(5\right) + 8 q = 11$

$\setminus \rightarrow 35 + 8 q = 11$

$\setminus \rightarrow 8 q = - 24$

$\setminus \rightarrow q = - 3$

Now that we have the values of both variables, we can plug them into one of the equations to check our work:

$3 p + 4 q = 3$

$3 \left(5\right) + 4 \left(- 3\right) = 3$

$15 - 12 = 3$

$3 = 3$

So it’s right.