Question

a fruit company packages its fruit into two types of boxes: large and small. this morning, the company made two deliveries. the table below shows the number of boxes in each delivery and the total weight (in kilograms).

	number of large boxes	number of small boxes	total weight (in kilograms)
second delivery	2	6	110

let ( x ) be the weight (in kilograms) of each large box.
let ( y ) be the weight (in kilograms) of each small box.

(a) write a system of equations that could be used to find the weight (in kilograms) of each type of box.
( square x + square y = square )
( square x + square y = square )

(b) how much does each type of box weigh (in kilograms)?
weight of each large box: ( square ) kilograms

Explanation:

Part (a)

Step1: Analyze first delivery

First delivery has 5 large boxes (each \( x \) kg) and 3 small boxes (each \( y \) kg), total weight 116 kg. So equation: \( 5x + 3y = 116 \).

Step2: Analyze second delivery

Second delivery has 2 large boxes and 6 small boxes, total weight 110 kg. So equation: \( 2x + 6y = 110 \).

Step1: Simplify the second equation

Divide \( 2x + 6y = 110 \) by 2: \( x + 3y = 55 \), so \( x = 55 - 3y \).

Step2: Substitute into first equation

Substitute \( x = 55 - 3y \) into \( 5x + 3y = 116 \):
\( 5(55 - 3y) + 3y = 116 \)
\( 275 - 15y + 3y = 116 \)
\( 275 - 12y = 116 \)
\( -12y = 116 - 275 \)
\( -12y = -159 \)
\( y = \frac{-159}{-12} = 13.25 \)? Wait, no, miscalculation. Wait, 275 - 116 is 159? Wait 275 - 116 = 159? No, 275 - 116 = 159? Wait 200 - 100 = 100, 75 - 16 = 59, so 100 + 59 = 159. Then \( -12y = -159 \)? Wait no, 275 - 12y = 116 → -12y = 116 - 275 = -159 → y = (-159)/(-12) = 13.25? Wait but let's check again. Wait first equation: 5x + 3y = 116, second: 2x + 6y = 110. Let's use elimination. Multiply first equation by 2: 10x + 6y = 232. Subtract second equation (2x + 6y = 110) from this: (10x + 6y) - (2x + 6y) = 232 - 110 → 8x = 122 → x = 122/8 = 15.25? Wait no, 232 - 110 is 122? 232 - 110 = 122. Then x = 122/8 = 15.25? Wait but then substitute back. Wait maybe I made a mistake in substitution. Let's do elimination properly.

First equation: \( 5x + 3y = 116 \)
Second equation: \( 2x + 6y = 110 \) → divide by 2: \( x + 3y = 55 \) → \( x = 55 - 3y \)

Substitute into first equation:
5(55 - 3y) + 3y = 116
275 - 15y + 3y = 116
275 - 12y = 116
-12y = 116 - 275
-12y = -159
y = (-159)/(-12) = 13.25? Wait 159 divided by 12: 1213=156, 159-156=3, so 13 + 3/12 = 13.25. Then x = 55 - 3(13.25) = 55 - 39.75 = 15.25. Wait but let's check with first equation: 515.25 + 313.25 = 76.25 + 39.75 = 116. Correct. Second equation: 215.25 + 613.25 = 30.5 + 79.5 = 110. Correct. So x = 15.25, y = 13.25? Wait but maybe the problem expects integer? Wait no, the numbers are as per calculation. Wait maybe I miscalculated 275 - 116. 275 - 100 = 175, 175 - 16 = 159. Yes. So:

Step3: Solve for x

From \( x = 55 - 3y \), and \( y = 13.25 \), so \( x = 55 - 3*13.25 = 55 - 39.75 = 15.25 \). Wait but maybe the problem has integer solutions? Wait let's check the table again. First delivery: 5 large, 3 small, total 116. Second: 2 large, 6 small, total 110. Let's multiply first equation by 2: 10x + 6y = 232. Second equation: 2x + 6y = 110. Subtract second from first: 8x = 122 → x = 122/8 = 15.25. Yes, that's correct. So:

Answer:

For the first equation: \( 5x + 3y = 116 \)
For the second equation: \( 2x + 6y = 110 \)

Part (b)