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Answer :
We start by letting [tex]$x$[/tex] be the number of the first paperback books (weighing [tex]$\frac{2}{3}$[/tex] pounds each) and [tex]$y$[/tex] be the number of the second paperback books (weighing [tex]$\frac{3}{4}$[/tex] pounds each). The problem tells us that the total number of books is 179 and their total weight is 128 pounds. This information leads to the following system of equations:
[tex]$$
x + y = 179
$$[/tex]
[tex]$$
\frac{2}{3} x + \frac{3}{4} y = 128
$$[/tex]
Below is a step-by-step explanation that verifies each statement given in the problem:
1. The statement “The system of equations is [tex]$x+y=179$[/tex] and [tex]$\frac{2}{3}x+\frac{3}{4}y=128$[/tex]” uses the totals for books and weight correctly. Thus, this statement is true.
2. The statement “The system of equations is [tex]$x+y=128$[/tex] and [tex]$\frac{2}{3}x+\frac{3}{4}y=179$[/tex]” incorrectly swaps the total number of books with the total weight. Hence, it is false.
3. The statement “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” is not sufficient. Multiplying the weight equation by 3 gives
[tex]$$
3\left(\frac{2}{3}x + \frac{3}{4}y\right)= 2x + \frac{9}{4}y = 384,
$$[/tex]
but the [tex]$x$[/tex]-coefficient in the simple equation remains 1. This does not immediately cancel the [tex]$x$[/tex]-term upon addition or subtraction without further adjustments, so the statement is false.
4. The statement “To eliminate the [tex]$y$[/tex]-variable from the equations, you can multiply the equation with the fractions by [tex]$-4$[/tex] and multiply the other equation by [tex]$3$[/tex]” is a valid elimination strategy. Here’s how it works:
- Multiply the weight equation by [tex]$-4$[/tex]:
[tex]$$
-4\left(\frac{2}{3}x+\frac{3}{4}y\right) = -\frac{8}{3}x - 3y = -512.
$$[/tex]
- Multiply the books count equation by [tex]$3$[/tex]:
[tex]$$
3(x+y) = 3x+3y = 537.
$$[/tex]
Adding these two new equations yields:
[tex]$$
3x - \frac{8}{3}x + (3y - 3y) = 537 - 512,
$$[/tex]
which simplifies to an equation in [tex]$x$[/tex] only. Hence, this elimination method works, and the statement is true.
5. Finally, we solve the system to check the statement “There are 104 copies of one book and 24 copies of the other.”
We already have:
[tex]$$
x+y =179.
$$[/tex]
To eliminate the fractions in the weight equation, multiply it by 12:
[tex]$$
12\left(\frac{2}{3}x+\frac{3}{4}y\right)= 8x + 9y = 1536.
$$[/tex]
Substitute [tex]$x = 179 - y$[/tex] into the equation:
[tex]$$
8(179-y) + 9y = 1536.
$$[/tex]
Compute [tex]$8(179)$[/tex]:
[tex]$$
1432 - 8y + 9y = 1536.
$$[/tex]
This simplifies to:
[tex]$$
1432 + y = 1536.
$$[/tex]
Solving for [tex]$y$[/tex] gives:
[tex]$$
y = 1536 - 1432 = 104.
$$[/tex]
Then, [tex]$x = 179 - 104 = 75$[/tex].
Since we obtain 75 copies of one book and 104 copies of the other (and not 24 copies), the statement is false.
In summary, the only true statements are the first and the fourth.
[tex]$$
x + y = 179
$$[/tex]
[tex]$$
\frac{2}{3} x + \frac{3}{4} y = 128
$$[/tex]
Below is a step-by-step explanation that verifies each statement given in the problem:
1. The statement “The system of equations is [tex]$x+y=179$[/tex] and [tex]$\frac{2}{3}x+\frac{3}{4}y=128$[/tex]” uses the totals for books and weight correctly. Thus, this statement is true.
2. The statement “The system of equations is [tex]$x+y=128$[/tex] and [tex]$\frac{2}{3}x+\frac{3}{4}y=179$[/tex]” incorrectly swaps the total number of books with the total weight. Hence, it is false.
3. The statement “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” is not sufficient. Multiplying the weight equation by 3 gives
[tex]$$
3\left(\frac{2}{3}x + \frac{3}{4}y\right)= 2x + \frac{9}{4}y = 384,
$$[/tex]
but the [tex]$x$[/tex]-coefficient in the simple equation remains 1. This does not immediately cancel the [tex]$x$[/tex]-term upon addition or subtraction without further adjustments, so the statement is false.
4. The statement “To eliminate the [tex]$y$[/tex]-variable from the equations, you can multiply the equation with the fractions by [tex]$-4$[/tex] and multiply the other equation by [tex]$3$[/tex]” is a valid elimination strategy. Here’s how it works:
- Multiply the weight equation by [tex]$-4$[/tex]:
[tex]$$
-4\left(\frac{2}{3}x+\frac{3}{4}y\right) = -\frac{8}{3}x - 3y = -512.
$$[/tex]
- Multiply the books count equation by [tex]$3$[/tex]:
[tex]$$
3(x+y) = 3x+3y = 537.
$$[/tex]
Adding these two new equations yields:
[tex]$$
3x - \frac{8}{3}x + (3y - 3y) = 537 - 512,
$$[/tex]
which simplifies to an equation in [tex]$x$[/tex] only. Hence, this elimination method works, and the statement is true.
5. Finally, we solve the system to check the statement “There are 104 copies of one book and 24 copies of the other.”
We already have:
[tex]$$
x+y =179.
$$[/tex]
To eliminate the fractions in the weight equation, multiply it by 12:
[tex]$$
12\left(\frac{2}{3}x+\frac{3}{4}y\right)= 8x + 9y = 1536.
$$[/tex]
Substitute [tex]$x = 179 - y$[/tex] into the equation:
[tex]$$
8(179-y) + 9y = 1536.
$$[/tex]
Compute [tex]$8(179)$[/tex]:
[tex]$$
1432 - 8y + 9y = 1536.
$$[/tex]
This simplifies to:
[tex]$$
1432 + y = 1536.
$$[/tex]
Solving for [tex]$y$[/tex] gives:
[tex]$$
y = 1536 - 1432 = 104.
$$[/tex]
Then, [tex]$x = 179 - 104 = 75$[/tex].
Since we obtain 75 copies of one book and 104 copies of the other (and not 24 copies), the statement is false.
In summary, the only true statements are the first and the fourth.
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