Question 1181298
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A total of 20 gifts that cost either 295 or 500 euros each are being bought, and a total of 6515 euros is being spent.  You need to determine how many of each are being bought.<br>
For a student just learning basic algebra, I see three basic ways to set up this problem for solving.<br>
(1) Using two equations and elimination<br>
let x = # of gifts at 295 euros each
let y = # of gifts at 500 euros each<br>
x+y=20
295x+500y=6515<br>
With two equations in this form, usually solving by elimination is easiest.<br>
(2) Using two equations and substitution.<br>
You can start the same way as in (1), but instead of using elimination, solve the first equation to get y=20-x and substitute that into the second equation.<br>
(3) Using a single equation<br>
let x = # of gifts at 295 euros each
then 20-x = # of gifts at 500 euros each<br>
Then you only need to solve a single equation to find the answer:<br>
295(x)+500(20-x)=6515<br>
I personally would never use method (2) above, because it is exactly like method (3) but with an extra step.  For me, methods (1) and (3) are equally good.<br>
Try the different methods and find one that "works" best for you.<br>
And here is my personal favorite for a fast way to solve this kind of problem, if a formal algebraic solution is not required.<br>
20 gifts all at 295 euros each would cost 5900 euros; the actual total is 6515 euros, which is another 615 euros.
The difference in the cost between the two gifts is 500-295=205 euros.
To make the needed additional 615 euros, the number of higher priced gifts must be 615/205=3.<br>
ANSWER: 3 gifts at 500 euros each; 17 at 295 euros each.<br>
CHECK: 500(3)+295(17) = 1500+5015 = 6515<br>