Applications of linear systems

First, label the quantities with appropriate variables:

number of hamburgers $\widehat{=}$ $x$

number of cheeseburgers $\widehat{=}$ $y$

Next, you need to organize all the information. On the one hand you have burger quantities and on the other hand you have costs.

Important: $x$ and $y$ are included in both pieces of information.

So:

quantities: $x$ hamburgers, $y$ cheeseburgers, total number: $12$ burgers.

money: $\$0.99\cdot x$ , $\$1.19\cdot y$ , $\$12.68$

In the last step, you set up an equation from the respective data.

You get the first equation by the information that $x$ hamburgers and $y$ cheeseburgers add up to $12$ burgers.

You get the second equation by the information that $\$0.99\cdot x$ and $\$1.19\cdot y$ add up to $\$12.68$ .

If you want to show that $x$ and $y$ solve a system of linear equations, you only need to substitute the variables into the system. In the end, each equation must yield a true statement.

So plug $x=8$ and $y=4$ into the equations and check:

$\begin{array}{rrll}\mathrm{I}& x+y& =& 12\\ \mathrm{II}& 0.99\cdot x+1.19\cdot y& =& 12.68\end{array}$

Plug in the values.

$\begin{array}{rrll}\mathrm{I}& 8+4& =& 12\\ \mathrm{II}& 0.99\cdot 8+1.19\cdot 4& =& 12.68\end{array}$

Calculate. If the same number is on the left and on the right, the statement is true.

$\begin{array}{rrll}\mathrm{I}& 12& =& 12\\ \mathrm{II}& 12.68& =& 12.68\end{array}$

You see immediately that $x=8$ hamburgers and $y=4$ cheeseburgers provide a true statement.

A linear system of equations can have either none, exactly one, or infinitely many solutions.

The equations are linear and not identical. In addition, there are only $2$ equations. Thus there is at most one intersection point. A maximum of one intersection point means that there is either no intersection point or one intersection point. From b) you already know that the former is not true, because $x=8$ and $y=4$ is a solution. Therefore, the system of equations has exactly one solution.