Question

use two equations in two variables to solve the problem. suppose you are a wedding photographer and you sell the following packages. package 1: one 10 × 14 and ten 8 × 10 color photos for $256.50 package 2: one 10 × 14 and five 8 × 10 color photos for $141.50 a newlywed couple buys package 1 and decides that they want one more 10 × 14 and one more 8 × 10 photograph. at the same prices, what should you charge them for each additional print? 10 × 14 $ 8 × 10 $ © andrei contiu/shutterstock.com

Explanation:

Step1: Define Variables

Let \( x \) be the price of a \( 10 \times 14 \) photo and \( y \) be the price of an \( 8 \times 10 \) photo.

Step2: Set Up Equations

From Package 1: \( x + 10y = 256.50 \)
From Package 2: \( x + 5y = 141.50 \)

Step3: Subtract Equations

Subtract the second equation from the first:
\( (x + 10y) - (x + 5y) = 256.50 - 141.50 \)
\( 5y = 115 \)

Step4: Solve for \( y \)

\( y = \frac{115}{5} = 23 \)

Step5: Solve for \( x \)

Substitute \( y = 23 \) into \( x + 5y = 141.50 \):
\( x + 5(23) = 141.50 \)
\( x + 115 = 141.50 \)
\( x = 141.50 - 115 = 26.50 \)

Step6: Calculate New Cost

The couple wants one more \( 10 \times 14 \) and one more \( 8 \times 10 \) than Package 1. So the new number of \( 10 \times 14 \) is \( 2 \), and \( 8 \times 10 \) is \( 11 \)? Wait, no—wait, Package 1 is 1 \( 10 \times 14 \) and 10 \( 8 \times 10 \). They want one more of each, so 2 \( 10 \times 14 \) and 11 \( 8 \times 10 \)? Wait, no, re-reading: "one more 10×14 and one more 8×10 photograph" than Package 1. So Package 1 has 1 \( 10 \times 14 \) and 10 \( 8 \times 10 \). So new is \( 1 + 1 = 2 \) \( 10 \times 14 \) and \( 10 + 1 = 11 \) \( 8 \times 10 \)? Wait, no, maybe I misread. Wait, the problem says: "A newlywed couple buys Package 1 and decides that they want one more 10×14 and one more 8×10 photograph." So Package 1 is 1 \( 10 \times 14 \) and 10 \( 8 \times 10 \). So adding one more of each: 2 \( 10 \times 14 \) and 11 \( 8 \times 10 \)? Wait, no, 1 (original) + 1 (more) = 2 for \( 10 \times 14 \), and 10 (original) + 1 (more) = 11 for \( 8 \times 10 \)? Wait, but let's check the equations again. Wait, maybe the initial equations: Package 1: 1 \( 10 \times 14 \) (x) and 10 \( 8 \times 10 \) (y) for $256.50: LXI24. Package 2: 1 LXI25 (x) and 5 LXI26 (y) for $141.50: \( x + 5y = 141.50 \). Then we found x = 26.50, y = 23. Now, the couple has Package 1 (1x + 10y) and wants to add 1x and 1y, so total is (1x + 10y) + (1x + 1y) = 2x + 11y? Wait, no: 1x + 10y (Package 1) + 1x + 1y (additional) = 2x + 11y. Wait, but let's compute that: 226.50 + 1123 = 53 + 253 = 306? Wait, no, maybe I made a mistake. Wait, no—wait, maybe the problem is: after buying Package 1, they want one more of each, so the total cost is Package 1 cost plus x + y. Because Package 1 is $256.50, and they want to add one x and one y. So total cost is 256.50 + x + y. Since x = 26.50 and y = 23, then 256.50 + 26.50 + 23 = 256.50 + 49.50 = 306. Wait, that makes sense. Because Package 1 is 1x + 10y. Adding 1x and 1y gives (1x + 10y) + (1x + 1y) = 2x + 11y? No, 1x + 10y + 1x + 1y = 2x + 11y. But 2x is 53, 11y is 253, 53 + 253 = 306. Alternatively, Package 1 cost is 256.50, and adding one x (26.50) and one y (23) gives 256.50 + 26.50 + 23 = 306. So that's the total cost.

Answer:

The cost for the additional package (one more of each) added to Package 1 is $306.00 (if we consider the total after adding one of each to Package 1). Wait, but the question says: "what should you charge them for each additional print?" Wait, no, re-reading the question: "what should you charge them for each additional print?" Wait, maybe I misread. Wait, the problem: "A newlywed couple buys Package 1 and decides that they want one more 10×14 and one more 8×10 photograph. At the same prices, what should you charge them for each additional print?" Wait, "each additional print"—so the price of each additional 10×14 is x = $26.50, and each additional 8×10 is y = $23.00. Wait, that makes sense. Because we found x (price of 10×14) is $26.50, y (price of 8×10) is $23.00. So the cost for one additional 10×14 is $26.50, and one additional 8×10 is $23.00.

Wait, let's re-express the problem. The key is to find the price of each size (x for 10×14, y for 8×10) using the two packages, then that's the price for each additional print of that size.

So from the two equations:

1. \( x + 10y = 256.50 \)
2. \( x + 5y = 141.50 \)

Subtract equation 2 from equation 1:

\( (x + 10y) - (x + 5y) = 256.50 - 141.50 \)

\( 5y = 115 \)

\( y = 23 \)

Then substitute y = 23 into equation 2:

\( x + 5(23) = 141.50 \)

\( x + 115 = 141.50 \)

\( x = 26.50 \)

So the price for a 10×14 (each additional) is $26.50, and for an 8×10 (each additional) is $23.00.