let's sayÂ
c = calories in the cheeseburgers
f  = calories in the fries
[tex]\bf \begin{array}{llcll}
10c+5f=6800&\boxed{\times -3}\implies &-30c-15f=-20400\\\\
5c+3f=3590&\boxed{\times 5}\implies &25c+15f=17950\\
&&----------\\
&&-5c+0=-2450
\end{array}
\\\\\\
c=\cfrac{-2450}{-5}\implies \boxed{c=490}\\\\
-------------------------------\\\\
\textit{now let's plug that in the first equation}\\\\
10(490)+5f=6800\implies 4900+5f=6800\implies 5f=1900
\\\\\\
f=\cfrac{1900}{5}\implies \boxed{f=380}[/tex]
notice, using elimination, the idea for the -3 and 5 multipliers, is to get a -15f atop and 15f at the bottom, thus they "eliminate" the variable "f"
