Answer :
Let the number of copies of the first paperback be [tex]$x$[/tex] and the number of copies of the second paperback be [tex]$y$[/tex]. Then we have:
[tex]$$
x + y = 179 \quad \text{(total number of books)}
$$[/tex]
and
[tex]$$
\frac{2}{3}x + \frac{3}{4}y = 128 \quad \text{(total weight in pounds)}.
$$[/tex]
Below is a step-by-step explanation addressing the statements:
1. The system of equations is
[tex]$$
x + y = 179 \quad \text{and} \quad \frac{2}{3}x + \frac{3}{4}y = 128.
$$[/tex]
This setup correctly represents that there are 179 books in total and their combined weight is 128 pounds. Therefore, this statement is true.
2. The alternative system given as
[tex]$$
x + y = 128 \quad \text{and} \quad \frac{2}{3}x + \frac{3}{4}y = 179
$$[/tex]
incorrectly assigns the total weight as 179 and the total number as 128. Thus, this statement is false.
3. One idea for eliminating the [tex]$x$[/tex]-variable is to multiply the equation with the fractions by 3. Multiplying
[tex]$$
\frac{2}{3}x + \frac{3}{4}y = 128 \quad \text{by } 3 \text{ gives } 2x + \frac{9}{4}y = 384.
$$[/tex]
However, the other equation, [tex]$x + y = 179$[/tex], would remain unchanged (i.e., [tex]$x$[/tex] has coefficient 1) and the coefficients of [tex]$x$[/tex] would not match for cancellation. Hence, this approach does not eliminate [tex]$x$[/tex], so this statement is false.
4. To eliminate the [tex]$y$[/tex]-variable, one effective method is to use multipliers. Multiply the equation
[tex]$$
x + y = 179
$$[/tex]
by 3 to obtain:
[tex]$$
3x + 3y = 537.
$$[/tex]
Then, multiply the weight equation
[tex]$$
\frac{2}{3}x + \frac{3}{4}y = 128
$$[/tex]
by [tex]$-4$[/tex] to get:
[tex]$$
-\frac{8}{3}x - 3y = -512.
$$[/tex]
Adding these two equations results in:
[tex]$$
\left(3x - \frac{8}{3}x\right) + (3y - 3y) = 537 - 512.
$$[/tex]
Notice that in the first term, [tex]$3x = \frac{9}{3}x$[/tex] so:
[tex]$$
\left(\frac{9}{3}x - \frac{8}{3}x\right) = \frac{1}{3}x.
$$[/tex]
This gives:
[tex]$$
\frac{1}{3}x = 25,
$$[/tex]
so
[tex]$$
x = 75.
$$[/tex]
Since the [tex]$y$[/tex]-terms cancel out exactly, this method correctly eliminates [tex]$y$[/tex], and the statement is true.
5. Now, using the first equation [tex]$x + y = 179$[/tex] and substituting [tex]$x = 75$[/tex], we have:
[tex]$$
75 + y = 179 \quad \Longrightarrow \quad y = 179 - 75 = 104.
$$[/tex]
The numbers of copies of the two books are [tex]$75$[/tex] and [tex]$104$[/tex]. The statement claiming "There are 104 copies of one book and 24 copies of the other" is incorrect since [tex]$24$[/tex] is not one of the computed values.
In summary, the true statements are:
- The system of equations is
[tex]$$
x+y=179 \quad \text{and} \quad \frac{2}{3}x+\frac{3}{4}y=128.
$$[/tex]
- To eliminate the [tex]$y$[/tex]-variable, you can multiply the equation with the fractions by [tex]$-4$[/tex] and multiply the other equation by [tex]$3$[/tex].
The final answer is that statements 1 and 4 are true, the elimination multipliers are [tex]$3$[/tex] and [tex]$-4$[/tex], and the number of copies for the two books are [tex]$75$[/tex] and [tex]$104$[/tex], respectively.
[tex]$$
x + y = 179 \quad \text{(total number of books)}
$$[/tex]
and
[tex]$$
\frac{2}{3}x + \frac{3}{4}y = 128 \quad \text{(total weight in pounds)}.
$$[/tex]
Below is a step-by-step explanation addressing the statements:
1. The system of equations is
[tex]$$
x + y = 179 \quad \text{and} \quad \frac{2}{3}x + \frac{3}{4}y = 128.
$$[/tex]
This setup correctly represents that there are 179 books in total and their combined weight is 128 pounds. Therefore, this statement is true.
2. The alternative system given as
[tex]$$
x + y = 128 \quad \text{and} \quad \frac{2}{3}x + \frac{3}{4}y = 179
$$[/tex]
incorrectly assigns the total weight as 179 and the total number as 128. Thus, this statement is false.
3. One idea for eliminating the [tex]$x$[/tex]-variable is to multiply the equation with the fractions by 3. Multiplying
[tex]$$
\frac{2}{3}x + \frac{3}{4}y = 128 \quad \text{by } 3 \text{ gives } 2x + \frac{9}{4}y = 384.
$$[/tex]
However, the other equation, [tex]$x + y = 179$[/tex], would remain unchanged (i.e., [tex]$x$[/tex] has coefficient 1) and the coefficients of [tex]$x$[/tex] would not match for cancellation. Hence, this approach does not eliminate [tex]$x$[/tex], so this statement is false.
4. To eliminate the [tex]$y$[/tex]-variable, one effective method is to use multipliers. Multiply the equation
[tex]$$
x + y = 179
$$[/tex]
by 3 to obtain:
[tex]$$
3x + 3y = 537.
$$[/tex]
Then, multiply the weight equation
[tex]$$
\frac{2}{3}x + \frac{3}{4}y = 128
$$[/tex]
by [tex]$-4$[/tex] to get:
[tex]$$
-\frac{8}{3}x - 3y = -512.
$$[/tex]
Adding these two equations results in:
[tex]$$
\left(3x - \frac{8}{3}x\right) + (3y - 3y) = 537 - 512.
$$[/tex]
Notice that in the first term, [tex]$3x = \frac{9}{3}x$[/tex] so:
[tex]$$
\left(\frac{9}{3}x - \frac{8}{3}x\right) = \frac{1}{3}x.
$$[/tex]
This gives:
[tex]$$
\frac{1}{3}x = 25,
$$[/tex]
so
[tex]$$
x = 75.
$$[/tex]
Since the [tex]$y$[/tex]-terms cancel out exactly, this method correctly eliminates [tex]$y$[/tex], and the statement is true.
5. Now, using the first equation [tex]$x + y = 179$[/tex] and substituting [tex]$x = 75$[/tex], we have:
[tex]$$
75 + y = 179 \quad \Longrightarrow \quad y = 179 - 75 = 104.
$$[/tex]
The numbers of copies of the two books are [tex]$75$[/tex] and [tex]$104$[/tex]. The statement claiming "There are 104 copies of one book and 24 copies of the other" is incorrect since [tex]$24$[/tex] is not one of the computed values.
In summary, the true statements are:
- The system of equations is
[tex]$$
x+y=179 \quad \text{and} \quad \frac{2}{3}x+\frac{3}{4}y=128.
$$[/tex]
- To eliminate the [tex]$y$[/tex]-variable, you can multiply the equation with the fractions by [tex]$-4$[/tex] and multiply the other equation by [tex]$3$[/tex].
The final answer is that statements 1 and 4 are true, the elimination multipliers are [tex]$3$[/tex] and [tex]$-4$[/tex], and the number of copies for the two books are [tex]$75$[/tex] and [tex]$104$[/tex], respectively.
