Question 951824
Try using systems of linear equations.
You have two variables to be named/defined:
{{{x}}}= price (in $) for a hamburger, and
{{{y}}}= price (in $) for a pizza.
(You could also break with the tradition of using {{{x}}} and {{{y}}} ,
and use {{{H}}} for the price of a hamburger and {{{P}}} for the price of a pizza).
"Ramona bought 5 hamburgers and 3 pizzas for $167.75" translates as
{{{5x+3y=167.75}}} .
"If she had bought 4 hamburgers and 4 pizzas, then she would have paid $193" translates as
{{{4x+4y=193}}} .
Your system of linear equations is
{{{system(5x+3y=167.75,4x+4y=193)}}} .
There are many ways to go about solving it.
You could solve it "by substitution" or "by elimination".
One way to solve it "by substitution", would be
to solve {{{4x+4y=193}}} for {{{y}}} to get an equation that looks like {{{y=some}}}{{{expression}}} ;
substitute the expression you find for {{{y}}} in the other equation:
solve the resulting equation for {{{x}}} ,
and plug the {{{x}}} value found into the {{{y=some}}}{{{expression}}} equation,
to find the value for {{{y}}}.
It would go like this:
{{{4x+4y=193}}}-->{{{4y=193-4x}}}-->{{{(1/4)(4y)=(1/4)(193-4x)}}}-->{{{y=(1/4)(193)-(1/4)(4x)}}}-->{{{y=48.25-x}}}
{{{5x+3y=167.75}}}-->{{{5x+3(48.25-x)=167.75}}}-->{{{5x+3*48.25-3x=167.75}}}-->{{{5x+144.75-3x=167.75}}}-->{{{2x+144.75=167.75}}}-->{{{2x=167.75-144.75}}}-->{{{2x=23}}}-->{{{x=23/2}}}-->{{{highlight(x=11.50)}}}
{{{y=48.25-x}}}-->{{{y=48.25-11.50}}}-->{{{highlight(y=36.75)}}}