To determine the price of one used book ([tex]\(u\)[/tex]) and one new book ([tex]\(n\)[/tex]), we are given two pieces of information:
1. A purchase of 10 used books and 5 new books costs [tex]$150.
2. A purchase of 15 used books and 1 new book costs $[/tex]108.
We can set up two equations to represent these purchases:
- For the first customer:
[tex]\[
10u + 5n = 150
\][/tex]
- For the second customer:
[tex]\[
15u + n = 108
\][/tex]
We now have a system of linear equations:
1. [tex]\(10u + 5n = 150\)[/tex]
2. [tex]\(15u + n = 108\)[/tex]
To find the values of [tex]\(u\)[/tex] and [tex]\(n\)[/tex], we can solve this system of equations. We will use the method of elimination or substitution. Let's detail the steps:
### Step 1: Simplify the Second Equation
Starting with the second equation:
[tex]\[ 15u + n = 108 \][/tex]
We can isolate [tex]\(n\)[/tex] to simplify substitution:
[tex]\[ n = 108 - 15u \][/tex]
### Step 2: Substitute [tex]\(n\)[/tex] into the First Equation
Now, substitute [tex]\(n\)[/tex] from the second equation into the first equation:
[tex]\[ 10u + 5(108 - 15u) = 150 \][/tex]
### Step 3: Simplify and Solve for [tex]\(u\)[/tex]
Expand and simplify the equation:
[tex]\[ 10u + 540 - 75u = 150 \][/tex]
[tex]\[ 540 - 65u = 150 \][/tex]
[tex]\[ -65u = 150 - 540 \][/tex]
[tex]\[ -65u = -390 \][/tex]
[tex]\[ u = \frac{-390}{-65} \][/tex]
[tex]\[ u = 6 \][/tex]
### Step 4: Solve for [tex]\(n\)[/tex]
Use the value of [tex]\(u\)[/tex] in the simplified second equation to find [tex]\(n\)[/tex]:
[tex]\[ n = 108 - 15(6) \][/tex]
[tex]\[ n = 108 - 90 \][/tex]
[tex]\[ n = 18 \][/tex]
Therefore, the cost of one used book is [tex]\(u = \$6\)[/tex] and the cost of one new book is [tex]\(n = \$18\)[/tex].
Thus, the correct answer is:
C. [tex]\(u = \$6\)[/tex], [tex]\(n = \$18\)[/tex]
