Question

A fruit stand has to decide what to charge for their produce. They need $5.30 for 1 apple and 1 orange. They also need $7.30 for 1 apple and 2 oranges. We put this information into a system of linear equations.

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

- Yes; they should charge $3.30 for an apple and $2.00 for an orange.

- Yes; they should charge $2.00 for an apple and $3.30 for an orange.

- No; the system has many solutions.

- No; the system has no solution.

Answer 1

Yes; they should charge $3.30 for an apple and $2.00 for an orange.

Explanation:

We can answer this question by creating some simultaneous equations.

Let
`x` = an apple and let
`y` = an orange.

With the information given, we can say that the price for an apple and an orange is equal to $5.30 while the price for an apple and 2 oranges is equal to $7.30.

Therefore:

`x+y = 5.30` (1st equation)

`x + 2y = 7.30` (2nd equation)

There are a couple of ways to solve these equations but I'll show you the substitution method. This is when we substitute the value of one of the equation into the other one.

So let's re-arrange one of them.

`x+y=5.30` can be re-arranged into
`x= 5.30-y`

Now we know the value of
`x`, we can replace the
`x` in the second equation.

Therefore
`5.30-y + 2y=7.30`. We can simplify this to give us
`5.30 + y = 7.30` and re-arrange to solve for y:
`y = 2`.

Now we know the value of y, let's put it into the first equation:

`x+2 =5.30`

and solving for
`x` we get:
`x = 3.30`.

Therefore the price of an apple (
`x`) is $3.30 and the price of an orange (
`y`) is $2.

From the list of answers, there is only one correct option which is:

- Yes; they should charge $3.30 for an apple and $2.00 for an orange.