Answer:
Yes, we can find a unique price for an apple and an orange.
Step-by-step explanation:
Let x be the price of one apple and y be the price of one orange.
We have been given that a fruit stand has to decide what to charge for their produce. They need $5.30 for 1 apple and 1 orange.
We can represent this information in an equation as:
![x+y=5.30...(1)](https://tex.z-dn.net/?f=x%2By%3D5.30...%281%29)
They also need $7.30 for 1 apple and 2 oranges.
Upon substituting our given information we formed a system of equations. Let us see if this system is solvable or not.
For a unique solution
, where
and
are constant of x and y variables of 1st equation respectively.
and
are constant of x and y variables of 2nd equation respectively.
Let us check our system of equations for unique solution.
![\frac{1}{1} \neq\frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B1%7D%20%5Cneq%5Cfrac%7B1%7D%7B2%7D)
![1 \neq\frac{1}{2}](https://tex.z-dn.net/?f=1%20%5Cneq%5Cfrac%7B1%7D%7B2%7D)
We can clearly see that 1 is not equal to half, therefore, we can find a unique price for an apple and orange using our system of equations.
Upon subtracting our 1st equation from 2nd equation we will get,
![x-x+2y-y=7.30-5.30](https://tex.z-dn.net/?f=x-x%2B2y-y%3D7.30-5.30)
![y=2](https://tex.z-dn.net/?f=y%3D2)
Therefore, price of one orange is $2.
Upon substituting y=2 in equation 1 we will get,
![x+2=5.30](https://tex.z-dn.net/?f=x%2B2%3D5.30)
![x=5.30-2](https://tex.z-dn.net/?f=x%3D5.30-2)
![x=3.30](https://tex.z-dn.net/?f=x%3D3.30)
Therefore, price of one apple is $3.30.