# How do you solve the following system using substitution?: 3s-5t=-30, 7s+11t=32

Oct 23, 2015

$\left(s , t\right) = \left(- \frac{5}{2} , \frac{9}{2}\right) = \left(- 2 \frac{1}{2} , 4 \frac{1}{2}\right)$

#### Explanation:

Given
[1]$\textcolor{w h i t e}{\text{XXX}} 3 s - 5 t = - 30$
[2]$\textcolor{w h i t e}{\text{XXX}} 7 s + 11 t = 32$

Rewriting [1] to generate an equation of the form: $s =$ some expression
[3]$\textcolor{w h i t e}{\text{XXX}} 3 s = 5 t - 30$

[4]$\textcolor{w h i t e}{\text{XXX}} s = \frac{5 t - 30}{3}$

We can now substitute (as per the question) $\left(\frac{5 t - 30}{3}\right)$ for $s$ in [2]
[5]$\textcolor{w h i t e}{\text{XXX}} 7 \left(\frac{5 t - 30}{3}\right) + 11 t = 32$

Simplifying
[6]$\textcolor{w h i t e}{\text{XXX}} \frac{35 t - 210 + 33 t}{3} = 32$

[7]$\textcolor{w h i t e}{\text{XXX}} \left(68 t - 210\right) = 96$

[8]$\textcolor{w h i t e}{\text{XXX}} 68 t = 306$

[9]$\textcolor{w h i t e}{\text{XXX}} t = \frac{9}{2}$

Substituting $\frac{9}{2}$ into [1] in place of $t$
[10]$\textcolor{w h i t e}{\text{XXX}} 3 s - 5 \cdot \left(\frac{9}{2}\right) = - 30$

[11]$\textcolor{w h i t e}{\text{XXX}} 3 s = \frac{5 \cdot 9 - 2 \cdot 30}{2}$

[12]$\textcolor{w h i t e}{\text{XXX}} 3 s = - \frac{15}{2}$

[13]$\textcolor{w h i t e}{\text{XXX}} s = - \frac{5}{2}$