Question

Six times Jason's collection of books and one-third of Nathan's collection add up to 134 books. One-third of Jason's collection and Nathan's entire collection add up to 31 books.
The number of books in Jason's collection is blank, and the number of books in Nathan's collection is blank. so what is the answer???

Answer 1

okay so lets set some variables j = jasons collection of books and N = nathan's collection of books, now we can write an equation to be something like this
6j + \frac{1}{3}n = 134 and then \frac{1}{3}j+n = 31

Answer 2

wait the latex didnt work one sec

Answer 3

6j + (1/3)n = 134 for the first sentence and then (1/3)j + n = 31 for the second

Answer 4

ok

Answer 5

so then you need to solve for n and j

Answer 6

ok

Answer 7

thx

Answer 8

just a more easy to understand visual description of what we have:

(\(x\) is jason and \(y\) is nathan.)

Six times Jason's collection of books and one-third of Nathan's collection add up to 134 books.
\[ 6x+\frac{1}{3}y=134 \]

One-third of Jason's collection and Nathan's entire collection add up to 31 books.
\[ \frac{1}{3}x+y=31 \]

this gives us a system which looks like:
\[ \begin{cases} 6x+\frac{1}{3}y=134 \\ \frac{1}{3}x+y=31 \end{cases} \]
as you yourself wished, lets work with x.

to get the \( \frac{1}{3}x\) into \(6x\) you have to multiply by 18 right?
\[ \begin{cases} 6x+\frac{1}{3}y=134 \\ \frac{1}{3}x+y=31 \ \ \ \ \ \ \ \ (\times 18) \end{cases} \]
\[ \ \ \ \ \ \ \ \ \ \ \Updownarrow \]
\[ \begin{cases} 6x+\frac{1}{3}y=134 \\ 6x+18y=558 \end{cases} \]
\[ \ \ \ \ \ \ \ \ \ \ \Updownarrow \]
\[ \begin{cases} 6x+\frac{1}{3}y=134 \\ x=93-3y \end{cases} \]
\[ \ \ \ \ \ \ \ \ \ \ \Updownarrow \]
\[ \begin{cases} 558-\frac{53y}{3}=134 \\ x=93-3y \end{cases} \]
\[ \ \ \ \ \ \ \ \ \ \ \Updownarrow \]
\[ \begin{cases} (y=)24 \\ x=93-3y \end{cases} \]
\[ \ \ \ \ \ \ \ \ \ \ \Updownarrow \]
\[ \begin{cases} (y=)24 \\ x=93-3\times24 \end{cases} \]
\[ \ \ \ \ \ \ \ \ \ \ \Updownarrow \]
\[ \begin{cases} (y=)24 \\ (x=)21 \end{cases} \]
yeah