Gillian purchased 25 books at the library book sale. Each hardcover book cost $[tex]$1.50$[/tex], and each paperback book cost $[tex]$0.50$[/tex]. Gillian spent a total of $[tex]$26.50$[/tex]. The book costs can be represented by the system of equations below.
[tex]
\begin{array}{c}
h+p=25 \\
1.50 h+0.50 p=26.50
\end{array}
[/tex]
How many paperback books did Gillian buy?
Answer (1)
Express h in terms of p using the first equation: h = 25 − p .
Substitute this expression into the second equation: 1.50 ( 25 − p ) + 0.50 p = 26.50 .
Simplify and solve for p : 37.5 − 1.50 p + 0.50 p = 26.50 ⇒ − p = − 11 ⇒ p = 11 .
Gillian bought 11 paperback books.
Explanation
Analyze the problem and data We are given a system of two equations with two variables, h and p , representing the number of hardcover and paperback books, respectively. Our goal is to find the value of p , which represents the number of paperback books Gillian bought. The system of equations is:
h + p = 25 1.50 h + 0.50 p = 26.50
Express h in terms of p We can solve this system of equations using substitution or elimination. Let's use the substitution method. From the first equation, we can express h in terms of p :
h = 25 − p
Substitute into the second equation Now, substitute this expression for h into the second equation:
1.50 ( 25 − p ) + 0.50 p = 26.50
Distribute Distribute the 1.50 :
37.5 − 1.50 p + 0.50 p = 26.50
Combine like terms Combine like terms:
37.5 − p = 26.50
Isolate p Subtract 37.5 from both sides:
− p = 26.50 − 37.5 − p = − 11
Solve for p Multiply both sides by − 1 :
p = 11 So, Gillian bought 11 paperback books.
Examples
Imagine you're planning a school event and need to buy snacks. You have a budget and know the prices for juice boxes and bags of chips. By setting up a system of equations similar to this problem, you can determine exactly how many of each item you can buy to stay within your budget. This method is useful for managing resources and making informed purchasing decisions in everyday situations.
