Q54. - Algebra 2

Question

Solve each system by using substitution or elimination.

 

 $\begin{array}{l}3p+6q=-3\\ 2p-3q=-9\end{array}$

Step-by-Step Solution

Answer

The solution of the system of equations: $\begin{array}{c}3p+6q=-3\\ 2p-3q=-9\end{array}$  is $\left(-3,1\right)$. 

Step 1 - Solve the first equation for p in terms of q

Write the equation $3p+6q=-3$ in terms of p.

 

$\begin{array}{c}3p+6q=-3\\ 3p+6q-6q=-3-6q\\ 3p=-3-6q\\ \frac{3p}{3}=\frac{-3-6q}{3}\\ p=-1-2q\end{array}$

Step 2 - Solve for q

Substitute $-1-2q$ for p into equation $2p-3q=-9$ so as to obtain the value of q.

$\begin{array}{c}2\left(-1-2q\right)-3q=-9\\ 2\left(-1\right)-2\left(2q\right)-3q=-9\\ -2-4q-3q=-9\\ -2-7q+2=-9+2\\ -7q=-7\\ q=1\end{array}$

Step 3 - Solve for p

Now substitute 1 for q into either original equation and solve for p. 

$\begin{array}{c}3p+6\left(1\right)=-3\\ 3p+6-6=-3-6\\ 3p=-9\\ p=-3\end{array}$

Therefore, the solution of the system is $\left(-3,1\right)$.