Hello, Muppers3262!
Maria buys three kinds of fruit for 40 pesos, 10 pesos, and 1 peso each.
She pays 259 pesos for 100 pieces of fruit.
How many pieces of the cheapest kind of fruit did she buy?
The systems of equations has three variables and only two equations.
The key is that the answers must be positive integers.
Let \(\displaystyle A,\,B,\,C\) be the number of each kind of fruit that Maria bought.
. . Then: \(\displaystyle \;A\,+\,B\,+\,C\:=\:259\;\)
[1]
\(\displaystyle A\) pieces at \(\displaystyle 40\) pesos each \(\displaystyle =\:40A\) pesos.
\(\displaystyle B\) pieces at \(\displaystyle 10\) pesos each \(\displaystyle =\:10B\) pesos.
\(\displaystyle C\) pieces at \(\displaystyle 1\) peso each \(\displaystyle =\:C\) pesos.
. . The total cost is: \(\displaystyle \;40A\,+\,10B\,+\,C\:=\:259\;\)
[2]
We have:
. [1]\(\displaystyle \;\;\;A\,+\;\;B\,+\,C\:=\:100\)
. . . . . . . . . [2]\(\displaystyle \;40A\,+\,10B\,+\,C\:=\:259\)
Subtract
[1] from
[2]: \(\displaystyle \;39A\.+\,3B\:=\:159\;\;\Rightarrow\;\;13A\,+\,3B\:=\:53\)
There are only three choices for \(\displaystyle A:\;1,\,2,\,3\).
. . (If \(\displaystyle A\,\geq\,4,\:B\) will be negative.)
If \(\displaystyle A\,=\,1\), then \(\displaystyle B\,=\,\frac{40}{3}\) ??
If \(\displaystyle A\,=\,3\), then \(\displaystyle B\,=\,\frac{14}{3}\) ??
If \(\displaystyle A\,=\,2\), then \(\displaystyle B\,=\,9\) . . .
There!
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Check
\(\displaystyle \;\,2\,@\,40\:=\:80\)
\(\displaystyle \;\:9\,@\,10\:=\:90\)
\(\displaystyle \:89\,@\,\;1\:=\:89\)
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. . . . . . .----
\(\displaystyle 100\;\;\;\;\;\;\;259\)
