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A box containing a total of 179 copies of two different paperback books was shipped to Marci's school. The total weight of the books was 128 pounds. If the weight of each of the first paperback books was [tex]\frac{2}{3}[/tex] of a pound and the weight of each of the second paperback books was [tex]\frac{3}{4}[/tex] of a pound, which statements are true? Check all that apply.

- The system of equations is [tex]x + y = 179[/tex] and [tex]\frac{2}{3}x + \frac{3}{4}y = 128[/tex].
- The system of equations is [tex]x + y = 128[/tex] and [tex]\frac{2}{3}x + \frac{3}{4}y = 179[/tex].
- To eliminate the [tex]x[/tex]-variable from the equations, you can multiply the equation with the fractions by 3 and leave the other equation as it is.
- To eliminate the [tex]y[/tex]-variable from the equations, you can multiply the equation with the fractions by -4 and multiply the other equation by 3.
- There are 104 copies of one book and 24 copies of the other.

Answer :

To solve this problem, we need to set up and solve a system of equations based on the information provided. Let's break it down step-by-step:

1. Identify the Variables:
We have two types of paperbacks:
- Let [tex]\( x \)[/tex] be the number of the first type of paperback books, each weighing [tex]\( \frac{2}{3} \)[/tex] of a pound.
- Let [tex]\( y \)[/tex] be the number of the second type of paperback books, each weighing [tex]\( \frac{3}{4} \)[/tex] of a pound.

2. Set Up the Equations:
From the problem, we have two conditions:
- The total number of books is 179. So, the equation is:
[tex]\[
x + y = 179
\][/tex]
- The total weight of the books is 128 pounds. So, the equation is:
[tex]\[
\frac{2}{3}x + \frac{3}{4}y = 128
\][/tex]

3. Solving the System of Equations:
We have a system of linear equations:
[tex]\[
\begin{align*}
x + y &= 179 \\
\frac{2}{3}x + \frac{3}{4}y &= 128
\end{align*}
\][/tex]

To eliminate fractions, let's multiply the entire second equation by 12, which is the least common multiple of 3 and 4:
[tex]\[
4 \cdot 2x + 3 \cdot 3y = 128 \cdot 12
\][/tex]
This simplifies to:
[tex]\[
8x + 9y = 1536
\][/tex]

So now we have two equations:
[tex]\[
\begin{align*}
x + y &= 179 \\
8x + 9y &= 1536
\end{align*}
\][/tex]

Step 1: Solve the first equation for one of the variables, let's say [tex]\( y \)[/tex]:
[tex]\[
y = 179 - x
\][/tex]

Step 2: Substitute [tex]\( y \)[/tex] in the second equation:
[tex]\[
8x + 9(179 - x) = 1536
\][/tex]

Step 3: Distribute and solve for [tex]\( x \)[/tex]:
[tex]\[
8x + 9 \cdot 179 - 9x = 1536
\][/tex]
[tex]\[
-x + 1611 = 1536
\][/tex]
[tex]\[
-x = 1536 - 1611
\][/tex]
[tex]\[
-x = -75
\][/tex]
[tex]\[
x = 75
\][/tex]

Step 4: Substitute [tex]\( x = 75 \)[/tex] back into the equation for [tex]\( y \)[/tex]:
[tex]\[
y = 179 - 75
\][/tex]
[tex]\[
y = 104
\][/tex]

4. Conclusion:
There are 75 copies of one book type and 104 copies of the other. Based on the statement choices:
- "The system of equations is [tex]\( x + y = 179 \)[/tex] and [tex]\( \frac{2}{3}x + \frac{3}{4}y = 128 \)[/tex]" is true.
- "There are 104 copies of one book and 24 of the other" wasn't correctly mentioned; it's actually 104 of one and 75 of the other as we found.

These findings lead us to know which books have which values, matching these correctly with your answer choices will yield the correct understanding of the shipping details.

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