Applications of linear systems

First, define variables for the quantities in the text:

gallons of of premium gasoline $\widehat{=}$ $x$

gallons of of regular gasoline $\widehat{=}$ $y$

Next, you need to organize all the information. On the one hand you have gallons loaded and on the other, money spent.

Important: $x$ and $y$ are included in both data.

So:

Gallons: $x$ gallons, $y$ gallons, $5$ gallons

Expenses: $\$1.35\cdot x$ , $\$1.20\cdot y$ , $\$6.50$

In the last step you set up one equation from each set of respective data.

You get the first equation by the information that $x$ gallons and $y$ gallons add up to $5$ gallons.

You get the second equation by the information that $\$1.35\cdot x$ and $\$1.20\cdot y$ add up to $\$6.50$ .

Both equations $\mathrm{I}$ and equation $\mathrm{II}$ must be satisfied according to the problem setting. So, you can represent the problem from the setting by the following system of equations.

$\begin{array}{rrlll}\mathrm{I}& x& +& y& =5\\ \mathrm{II}& 1.35\cdot x& +& 1.2\cdot y& =6.5\end{array}$

A system of linear equations is given, which can be solved via elimination by equating coefficients

$\phantom{\mathrm{I}}\mathrm{I}x+y=5$

$\mathrm{II}1.35\cdot x+1.2\cdot y=6.5$

1. Solve both equations for y

Solve both equations for one variable. In this case, it does not matter whether you take $x$ or $y$.

2. Equate

Set the two equations $\mathrm{I}$ and $\mathrm{I}{\mathrm{I}}^{\prime }$ equal.

$\Rightarrow 5-x=\frac{65}{12}-1.125\cdot x$

3. Solve equation for x

4. Plug in x to find out y

Plug $x$ into ${\mathrm{I}}^{\prime }$ or $\mathrm{I}{\mathrm{I}}^{\prime }$.

$\phantom{\mathrm{I}}{\mathrm{I}}^{\prime }y=5-\frac{10}{3}=\frac{5}{3}$

Write down the solution set

$${\displaystyle L=\left\{(\frac{10}{3};\frac{5}{3})\right\}}$$

Dave has loaded $x=\frac{10}{3}\approx 3.3$ gallons of regular gasoline and $y=\frac{5}{3}\approx 1.7$ gallons of premium gasoline.