Steve bought some apples at a cost of $.60 each and some oranges at a cost ... ...
Steve bought some apples at a cost of $.60 each and some oranges at a cost of $.50 each. If he paid a total of $4.10 for a total of 8 apples and oranges, how many apples did Steve buy?
(A) 1
(B) ...
(C) ...
(D) ...
(E) ...
*This question is included in
Nova Math - Problem Set X: Word Problems
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Steve bought some apples at a cost of $.60 each and some oranges at a cost ... ...
Posted: 08/11/2016 04:56
Can you please give detail on "solving this system"
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Posted: 08/11/2016 17:53
Hi Russ,
There are two basic ways of solving a system of two equations in two unknowns: Substitution and Elimination.
Substitution:
First solve the equation x + y = 8 for y: y = 8 - x. Next, substitute 8 - x for y in the other equation:
.60x + .50y = 4.10
.60x + .50(8 - x) = 4.10
Notice that we now have an equation in ONLY one variable, x, so we can solve it in the usual way.
Multiplying this equation by 100 to eliminate the decimals (not necessary, but it makes it easier to solve the equation) yields
60x + 50(8 - x) = 410
Distributing the 50 yields
60x + 400 - 50x = 410
10x + 400 = 410
10x = 10
x = 1
Elimination:
Multiply the equation .60x + .50y = 4.10 by 100 to eliminate the decimals:
60x + 50y = 410
Now, multiply the equation x + y = 8 by -50:
-50x - 50y = -400
So, we now have the following system of two equations:
60x + 50y = 410
-50x - 50y = -400
Notice that if we add these equations then the y variable will be eliminated:
10x + 0 = 10
10x = 10
x = 1
