Question 1138848
<br>
c = # of cheese pizzas
p = # of pepperoni pizzas
s = # of supreme pizzas<br>
{{{c+p+s = 26}}}  the total number of pizzas was 26
{{{6c+9p+12s = 222}}}  the total cost was $222
{{{9p = 2(6c)}}}  the cost of the pepperoni pizzas was twice the cost of the cheese pizzas<br>
There is an endless number of different paths for solving that system of 3 equations in 3 variables.  The path I chose involved many twists and turns and was not particularly pleasant....<br>
So instead of finishing the problem that way, let's spend a little effort to set up the problem using a single equation in a single variable and see if the resulting path to the solution is easier.<br>
He spent twice as much on the $9 pepperoni pizzas as he spent on the $6 cheese pizzas:<br>
{{{9p = 2(6c)}}}
{{{9p = 12c}}}
{{{3p = 4c}}}<br>
Using this, we can let p=4x and c=3x; then 3p=4c=12x.<br>
And then with p=4x and c=3x, and with 26 pizzas in all, the number of supreme pizzas is 26-7x.<br>
Then...<br>
{{{6(3x)+9(4x)+12(26-7x) = 222}}}
{{{18x+36x+312-84x = 222}}}
{{{-30x+312 = 222}}}
{{{90 = 30x}}}
{{{x = 3}}}<br>
ANSWER:
cheese: 3x = 9
pepperoni: 4x = 12
supreme: 26-7x = 5<br>
With the two ways I chose to solve the problem, the little extra effort required to set up an equation using a single variable resulted in an equation that required far less effort to solve than the system of 3 equations in 3 variables.<br>
While it is of course important to understand how to set up a problem like this with 3 variables directly from the given information, it is also important to know that very often the overall effort required to solve a problem will be greatly reduced if you can set up the problem using a single variable.