Question 1191329
<br>
The equations we can form from the given information are<br>
3h+2p=16  the cost of 3 hot dogs and 2 pies is $16
4h+3p=22  the cost of 4 hot dogs and 3 pies is $22<br>
With the two equations in that form, a typical algebraic solution would use elimination.  Multiply both equations by some numbers so that the coefficients of one of the variables are opposites; then adding the two resulting equations eliminates that variable.  Then the rest of the path to the solution is easy.<br>
In this example, we can multiply the first equation by 3 and the second by -2; that makes the coefficients of p in the two equation 6 and -6.<br>
9h+6p=48
-8h-6p=-44
h=4<br>
Plug that value for h into either original equation to solve for p.<br>
3(4)+2p=16
12+2p=16
2p=4
p=2<br>
ANSWER: hot dogs cost $4 each; pies cost $2 each.<br>
The particular numbers in this example make it easy to solve the problem informally using logical reasoning.<br>
4 hot dogs and 3 pies cost $22
3 hot dogs and 2 pies cost $16<br>
The difference between the two purchases is 1 hot dog and 1 pie, for a cost of $6.  Continue subtracting 1 hot dog and 1 pie for $6 until you are left with only 1 hot dog:<br>
3 hot dogs and 2 pies cost $16
2 hot dogs and 1 pie cost $10
1 hot dog and 0 pies cost $4<br>
So each hot dog costs $4; then, since 1 hot dog and 1 pie cost $6, each pie costs $2.<br>