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Answer :
To solve this problem, we need to define and work with a system of equations based on the following information:
1. Two types of paperback books, with a total of 179 books.
2. The total weight of these books is 128 pounds.
3. The weight of each book of the first type is [tex]\(\frac{2}{3}\)[/tex] of a pound.
4. The weight of each book of the second type is [tex]\(\frac{3}{4}\)[/tex] of a pound.
Let's set up the system of equations:
- Let [tex]\( x \)[/tex] be the number of the first type of paperback books.
- Let [tex]\( y \)[/tex] be the number of the second type of paperback books.
From the information given, we can create these equations:
Equation 1: The total number of books is 179.
[tex]\[
x + y = 179
\][/tex]
Equation 2: The total weight of the books is 128 pounds.
[tex]\[
\frac{2}{3}x + \frac{3}{4}y = 128
\][/tex]
Now we'll solve this system of equations:
1. Multiply Equation 1 by 3 to help eliminate fractions in Equation 2:
[tex]\[
3(x + y = 179) \implies 3x + 3y = 537
\][/tex]
2. Solve Equation 2 for elimination of fractions:
Multiply through by 12 (the least common multiple of 3 and 4) to clear the fractions:
[tex]\[
12 \left(\frac{2}{3}x + \frac{3}{4}y = 128 \right) \implies 8x + 9y = 1536
\][/tex]
Now we have the modified system:
[tex]\[
3x + 3y = 537
\][/tex]
[tex]\[
8x + 9y = 1536
\][/tex]
3. Solve the system:
- First, let's express [tex]\( y \)[/tex] from the first equation:
[tex]\[
3x + 3y = 537 \quad \implies \quad y = 179 - x
\][/tex]
- Substitute [tex]\( y = 179 - x \)[/tex] in the second equation:
[tex]\[
8x + 9(179 - x) = 1536
\][/tex]
[tex]\[
8x + 1611 - 9x = 1536
\][/tex]
[tex]\[
-x + 1611 = 1536
\][/tex]
[tex]\[
-x = 1536 - 1611
\][/tex]
[tex]\[
-x = -75
\][/tex]
[tex]\[
x = 75
\][/tex]
4. Find [tex]\( y \)[/tex]:
Substitute [tex]\( x = 75 \)[/tex] back into [tex]\( y = 179 - x \)[/tex]:
[tex]\[
y = 179 - 75
\][/tex]
[tex]\[
y = 104
\][/tex]
In conclusion, the solution to the system of equations tells us there are 75 copies of the first type of book and 104 copies of the second type of book.
Now, let's evaluate the given statements:
- The system of equations is [tex]\( x + y = 179 \)[/tex] and [tex]\( \frac{2}{3} x + \frac{3}{4} y = 128 \)[/tex]. True.
- The system of equations is [tex]\( x + y = 128 \)[/tex] and [tex]\( \frac{2}{3} x + \frac{3}{4} y = 179 \)[/tex]. False.
- 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. False. (You should multiply both equations appropriately to eliminate fractions.)
- 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. False. (You don't eliminate variables that way here.)
- There are 104 copies of one book and 24 copies of the other. False. (Correct count is 75 and 104.)
1. Two types of paperback books, with a total of 179 books.
2. The total weight of these books is 128 pounds.
3. The weight of each book of the first type is [tex]\(\frac{2}{3}\)[/tex] of a pound.
4. The weight of each book of the second type is [tex]\(\frac{3}{4}\)[/tex] of a pound.
Let's set up the system of equations:
- Let [tex]\( x \)[/tex] be the number of the first type of paperback books.
- Let [tex]\( y \)[/tex] be the number of the second type of paperback books.
From the information given, we can create these equations:
Equation 1: The total number of books is 179.
[tex]\[
x + y = 179
\][/tex]
Equation 2: The total weight of the books is 128 pounds.
[tex]\[
\frac{2}{3}x + \frac{3}{4}y = 128
\][/tex]
Now we'll solve this system of equations:
1. Multiply Equation 1 by 3 to help eliminate fractions in Equation 2:
[tex]\[
3(x + y = 179) \implies 3x + 3y = 537
\][/tex]
2. Solve Equation 2 for elimination of fractions:
Multiply through by 12 (the least common multiple of 3 and 4) to clear the fractions:
[tex]\[
12 \left(\frac{2}{3}x + \frac{3}{4}y = 128 \right) \implies 8x + 9y = 1536
\][/tex]
Now we have the modified system:
[tex]\[
3x + 3y = 537
\][/tex]
[tex]\[
8x + 9y = 1536
\][/tex]
3. Solve the system:
- First, let's express [tex]\( y \)[/tex] from the first equation:
[tex]\[
3x + 3y = 537 \quad \implies \quad y = 179 - x
\][/tex]
- Substitute [tex]\( y = 179 - x \)[/tex] in the second equation:
[tex]\[
8x + 9(179 - x) = 1536
\][/tex]
[tex]\[
8x + 1611 - 9x = 1536
\][/tex]
[tex]\[
-x + 1611 = 1536
\][/tex]
[tex]\[
-x = 1536 - 1611
\][/tex]
[tex]\[
-x = -75
\][/tex]
[tex]\[
x = 75
\][/tex]
4. Find [tex]\( y \)[/tex]:
Substitute [tex]\( x = 75 \)[/tex] back into [tex]\( y = 179 - x \)[/tex]:
[tex]\[
y = 179 - 75
\][/tex]
[tex]\[
y = 104
\][/tex]
In conclusion, the solution to the system of equations tells us there are 75 copies of the first type of book and 104 copies of the second type of book.
Now, let's evaluate the given statements:
- The system of equations is [tex]\( x + y = 179 \)[/tex] and [tex]\( \frac{2}{3} x + \frac{3}{4} y = 128 \)[/tex]. True.
- The system of equations is [tex]\( x + y = 128 \)[/tex] and [tex]\( \frac{2}{3} x + \frac{3}{4} y = 179 \)[/tex]. False.
- 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. False. (You should multiply both equations appropriately to eliminate fractions.)
- 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. False. (You don't eliminate variables that way here.)
- There are 104 copies of one book and 24 copies of the other. False. (Correct count is 75 and 104.)
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