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

Answer:

No; the system has no solution.

Step-by-step 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

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:

$x+y=5.30...(1)$

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 $\frac{a_1}{a_2} \neq\frac{b_1}{b_2}$, 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.

$\frac{1}{1} \neq\frac{1}{2}$

$1 \neq\frac{1}{2}$

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.