Question

A fruit stand has to decide what to charge for their produce. They need $5.30/$5.30 for 11 apple and 11 orange. They also need $7.30/$7.30 for 11 apple and 22 oranges. We put this information into a system of linear equations.

Can we find a unique price for an apple and an orange?

Answer 1

No; the system has no solution.

Explanation:

A fruit stand has to decide what to charge for their produce. They decide to charge \$5.30$ 5, point, 30 for 111 apple and 111 orange. They also plan to charge \$14$14 dollar sign, 14 for 222 apples and 222 oranges. We put this information into a system of linear equations.

Can we find a unique price for an apple and an orange?

Answer 2

Yes, we can find a unique price for an apple and an orange.

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)`

They also need $7.30 for 1 apple and 2 oranges.

`x+2y=7.30...(2)`

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
`(a_1)/(a_2) \\eq(b_1)/(b_2)`, where
`a_1` and
`b_1` are constant of x and y variables of 1st equation respectively.
`a_2` and
`b_2` are constant of x and y variables of 2nd equation respectively.

Let us check our system of equations for unique solution.

`(1)/(1) \\eq(1)/(2)`

`1 \\eq(1)/(2)`

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`

`y=2`

Therefore, price of one orange is $2.

Upon substituting y=2 in equation 1 we will get,

`x+2=5.30`

`x=5.30-2`

`x=3.30`

Therefore, price of one apple is $3.30.