If the weight of the math textbook is [tex]$2 \frac{1}{5}$[/tex] pounds, which shows the correct equation and value of [tex]$x$[/tex], the weight of Marta's book for language arts?

A. [tex]4x + \frac{4}{5} = 2 \frac{1}{5}; \, x = \frac{7}{20}[/tex] of a pound
B. [tex]4x - \frac{4}{5} = 2 \frac{1}{5}; \, x = \frac{3}{4}[/tex] of a pound
C. [tex]4x + \frac{4}{5} = 2 \frac{1}{5}; \, x = \frac{3}{4}[/tex] of a pound
D. [tex]4x - \frac{4}{5} = 2 \frac{1}{5}; \, x = \frac{7}{20}[/tex] of a pound



Answer :

To determine the correct equation and value of [tex]\( x \)[/tex] that represent the weight of Marta's book for language arts, let's analyze the given equations and their solutions step-by-step.

First, convert the mixed fraction [tex]\( 2 \frac{1}{5} \)[/tex] into an improper fraction and a decimal:
- [tex]\( 2 \frac{1}{5} \)[/tex] can be converted this way:
[tex]\[ 2 + \frac{1}{5} = \frac{10}{5} + \frac{1}{5} = \frac{11}{5} = 2.2 \][/tex]

Now, review the given equations (we'll identify which one leads to the correct weight):
1. Equation 1:
[tex]\[ 4x + \frac{4}{5} = 2 \frac{1}{5} \][/tex]
We are given that [tex]\( x = \frac{7}{20} \)[/tex]. Substitute [tex]\( x = \frac{7}{20} \)[/tex] into the equation:
[tex]\[ 4 \left( \frac{7}{20} \right) + \frac{4}{5} = 2.2 \][/tex]
Simplifying each term:
[tex]\[ 4 \cdot \frac{7}{20} = \frac{28}{20} = 1.4 \][/tex]
[tex]\[ 1.4 + \frac{4}{5} = 1.4 + 0.8 = 2.2 \][/tex]
Therefore, the left side indeed equals the weight of [tex]\( 2.2 \)[/tex] pounds.

2. Equation 2:
[tex]\[ 4x - \frac{4}{5} = 2 \frac{1}{5} \][/tex]
We are given that [tex]\( x = \frac{3}{4} \)[/tex]. Substitute [tex]\( x = \frac{3}{4} \)[/tex] into the equation:
[tex]\[ 4 \left( \frac{3}{4} \right) - \frac{4}{5} = 2.2 \][/tex]
Simplifying each term:
[tex]\[ 4 \cdot \frac{3}{4} = 3 \][/tex]
[tex]\[ 3 - \frac{4}{5} = 3 - 0.8 = 2.2 \][/tex]
Therefore, the left side equals the weight of [tex]\( 2.2 \)[/tex] pounds.

So, we have verified that both equations [tex]\((4x + \frac{4}{5} = 2 \frac{1}{5}\)[/tex] and [tex]\(4x - \frac{4}{5} = 2 \frac{1}{5})\)[/tex] satisfy their respective solutions:
- [tex]\( x = \frac{7}{20} \)[/tex] for Equation 1
- [tex]\( x = \frac{3}{4} \)[/tex] for Equation 2

However, based on our given multiple-choice options:

- [ ] [tex]$4 x+\frac{4}{5}=2 \frac{1}{5} ; x=\frac{7}{20}$[/tex] of a pound
- [ ] [tex]$4 x-\frac{4}{5}=2 \frac{1}{5} ; x=\frac{3}{4}$[/tex] of a pound
- [ ] [tex]$4 x+\frac{4}{5}=2 \frac{1}{5} ; x=\frac{3}{4}$[/tex] of a pound
- [ ] [tex]$4 x-\frac{4}{5}=2 \frac{1}{5} ; x=\frac{7}{20}$[/tex] of a pound

The correct options based on our verification are:
- [tex]$4 x+\frac{4}{5}=2 \frac{1}{5} ; x=\frac{7}{20}$[/tex] of a pound
- [tex]$4 x-\frac{4}{5}=2 \frac{1}{5} ; x=\frac{3}{4}$[/tex] of a pound

Thus, there are two correct answers based on the values and equations provided in the question.

Since you requested only one correct answer (without indicating multiple selection), we conclude the list of the valid option(s):
- The equations and solutions of "[tex]$4 x+\frac{4}{5}=2 \frac{1}{5} ; x=\frac{7}{20}$[/tex] of a pound" is one true answer.
- Similarly, the equation of "[tex]$4 x-\frac{4}{5}=2 \frac{1}{5} ; x=\frac{3}{4}$[/tex] of a pound" is also true.

In conclusion, based on the given options, both [tex]$4 x+\frac{4}{5}=2 \frac{1}{5} ; x=\frac{7}{20}$[/tex] of a pound and [tex]$4 x-\frac{4}{5}=2 \frac{1}{5} ; x=\frac{3}{4}$[/tex] of a pound are valid, but depending on choosing one:

The first correct choice would be: [tex]$4 x+\frac{4}{5}=2 \frac{1}{5} ; x=\frac{7}{20}$[/tex] of a pound.