Question

A fruit stand has to decide what to charge for their produce. They need $10 for 4 apples and 4 oranges.

They also need $12 for 6 apples and 6 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

There cannot be unique price of an apple and an orange for which price of four apples and four oranges is $10 and price of 6 apples and 6 oranges is $ 12.

Solution:

Let’s first create linear equations from given information.

Let assume price of 1 apple = $x

And price of 1 orange = $y

For 4 apples and 4 oranges amount required is $10.

4x + 4y = 10

On simplification, we get

2x + 2y = 5 ------(1)

And for 6 apples and 6 oranges amount required is $12.

6x + 6y = 12

On simplification we get

x + y =2 ------(2)

So two required equations are

2x + 2y = 5 ------(1)

x + y =2 ------(2)

According to the system of linear equations ,

Two equations
`\mathrm{a}_(1) \mathrm{x}+\mathrm{b}_(1) \mathrm{y}=\mathrm{c}_(1) \text { and } \mathrm{a}_(2) \mathrm{x}+\mathrm{b}_(2) \mathrm{y}=\mathrm{c}_(2)` will have unique solution if
`\left((a_(1))/(a_(2))\right) \text { is not equal to }\left((b_(1))/(b_(2))\right)`

Let’s check our two equations for unique solution.

In our case
`\mathrm{a}_(1)=2, \mathrm{b}_(1)=2, \mathrm{a}_(2)=1, \mathrm{b}_(2)=1`

`(a_(1))/(a_(2))=(2)/(1)=2 \text { and } (b_(1))/(b_(2))=(2)/(1)=2`

Since in our case
`(a_(1))/(a_(2))=(b_(1))/(b_(2))=2`

So our equation will not have unique solution.

Now let’s check
`(c_(1))/(c_(2))`

`(c_(1))/(c_(2))=(5)/(2)`

So final condition for our two linear equation is
`(a_(1))/(a_(2)) \text { is equal to } (b_(1))/(b_(2)) \text { and not equal to } (c_(1))/(c_(2))`

According to system if linear equation above condition is for no solution means equation (1) and equation (2) do not have any solution.

Hence there cannot be unique price of an apple and an orange for which price of four apples and four oranges is $10 and price of 6 apples and 6 oranges is $ 12.