Question

A fruit stand has to decide what to charge for their produce. They need \$5.30$5.30dollar sign, 5, point, 30 for 111 apple and 111 orange. They also need \$7.30$7.30dollar sign, 7, point, 30 for 111 apple and 222 oranges. We put this information into a system of linear equations.

Answer 1

To solve this problem, we can set up a system of linear equations to represent the charges for apples and oranges at the fruit stand. We can then solve this system using the elimination method to find the cost of an apple and an orange.

Step-by-step explanation:

To solve this problem, we can set up a system of linear equations to represent the charges for apples and oranges at the fruit stand. Let's use x to represent the cost of an apple and y to represent the cost of an orange.

From the given information, we can set up the following equations:

111x + 111y = 5.30

111x + 222y = 7.30

Next, we can solve this system of equations by using substitution or elimination. Let's use the elimination method:

Multiply the first equation by 2, and multiply the second equation by 1:

222x + 222y = 10.60

111x + 222y = 7.30

Subtract the second equation from the first:

111x = 3.30

Divide both sides by 111:

x = 0.03

Substitute this value back into either equation to solve for y:

111(0.03) + 222y = 7.30

3.33 + 222y = 7.30

222y = 3.97

y = 0.01

Therefore, the cost of an apple is $0.03 and the cost of an orange is $0.01 at the fruit stand.

Answer 2

3.30 for apple and 2.00 for an orange

Step-by-step explanation: