Between the last two days of the week, write an inequality to determine the minimum number of hours he needs to practice on each of the two days.

A. [tex]5 \frac{1}{3} + 2x \leq 7[/tex]
B. [tex]5 \frac{1}{3} x + 2 \leq 7[/tex]
C. [tex]5 \frac{1}{3} x + 2 \geq 7[/tex]
D. [tex]5 \frac{1}{3} + 2x \geq 7[/tex]



Answer :

To solve the inequality [tex]\(5 \frac{1}{3} + 2x \geq 7\)[/tex], let's take it step by step:

1. Convert the mixed number to an improper fraction:
[tex]\(5 \frac{1}{3}\)[/tex] can be written as [tex]\(5 + \frac{1}{3}\)[/tex].
Converting this into an improper fraction, we get:
[tex]\(5 + \frac{1}{3} = \frac{15}{3} + \frac{1}{3} = \frac{16}{3}\)[/tex].

2. Write the inequality with the improper fraction:
[tex]\(\frac{16}{3} + 2x \geq 7\)[/tex]

3. Isolate the term with [tex]\(x\)[/tex]:
To isolate [tex]\(2x\)[/tex], subtract [tex]\(\frac{16}{3}\)[/tex] from both sides of the inequality:
[tex]\(2x \geq 7 - \frac{16}{3}\)[/tex]

4. Convert 7 to a fraction with the same denominator:
[tex]\(7\)[/tex] can be written as [tex]\(\frac{21}{3}\)[/tex].
Thus, the inequality becomes:
[tex]\(2x \geq \frac{21}{3} - \frac{16}{3}\)[/tex]

5. Combine the fractions:
[tex]\(\frac{21}{3} - \frac{16}{3} = \frac{5}{3}\)[/tex]
Now the inequality is:
[tex]\(2x \geq \frac{5}{3}\)[/tex]

6. Solve for [tex]\(x\)[/tex]:
Divide both sides of the inequality by [tex]\(2\)[/tex]:
[tex]\(x \geq \frac{\frac{5}{3}}{2} = \frac{5}{6}\)[/tex]

Therefore, the solution to the inequality is:

[tex]\[ x \geq \frac{5}{6} \][/tex]

This means that the minimum number of hours he needs to practice on each of the last two days is [tex]\(\frac{5}{6}\)[/tex] hours.