Applications of linear systems

Define some variables and set up the equations

Set up the equations such that all equal variables are standing below each other.

$x\widehat{=}$ chocolate bar, $y\widehat{=}$ ice cream

$\begin{array}{rrlll}& \mathrm{I}& 3x+2y& =& 4.80\\ & \mathrm{II}& 1x+2y& =& 3\end{array}$

As you can see, $2y$ appears twice. Therefore the addition/subtraction method is suitable here. Since the same sign is present in both equations, you must subtract the equations.

Addition method

$\begin{array}{llccllll}& & \mathrm{I}& 3x& +& 2y& =& 4.80\\ & -& \mathrm{II}& 1x& +& 2y& =& 3\\ & & & 2x& +& 0& =& 1.80\end{array}$

Solve for the first variable

$\begin{array}{rrl}2x+0& =& 1.80\\ x& =& 0.90\end{array}$

$|:2$

Plug $x$ into $\mathrm{I}$ or $\mathrm{II}$

$x$ in $\mathrm{II}$

$\begin{array}{rrl}0.90+2y& =& 3\\ 2y& =& 2.10\\ y& =& 1.05\end{array}$

Solution set:

$𝕃=\{(x|y)=(0.9|1.05)\}$

So a chocolate bar costs $\$0.90$ and an ice cream costs $\$1.05$. That means, Stephen can indeed buy an ice cream.