Find [tex]c[/tex] for the inequality [tex]\frac{2}{3}c + 7 \leq \frac{1}{3}[/tex].

A. [tex]c \leq -6[/tex]
B. [tex]c \geq -6[/tex]
C. [tex]c \leq -10[/tex]
D. [tex]c \geq -10[/tex]



Answer :

Let's find the solution for the inequality [tex]\(\frac{2}{3}c + 7 \leq \frac{1}{3}\)[/tex].

Step 1: Subtract 7 from both sides to isolate the term with [tex]\(c\)[/tex] on one side:
[tex]\[ \frac{2}{3}c + 7 - 7 \leq \frac{1}{3} - 7 \][/tex]
This simplifies to:
[tex]\[ \frac{2}{3}c \leq \frac{1}{3} - 7 \][/tex]

Step 2: Convert the numbers on the right side to have a common denominator for subtraction:
[tex]\[ \frac{1}{3} - 7 = \frac{1}{3} - \frac{21}{3} = \frac{1 - 21}{3} = \frac{-20}{3} \][/tex]
So the inequality becomes:
[tex]\[ \frac{2}{3}c \leq \frac{-20}{3} \][/tex]

Step 3: To solve for [tex]\(c\)[/tex], divide both sides by [tex]\(\frac{2}{3}\)[/tex]. Dividing by a fraction is equivalent to multiplying by its reciprocal:
[tex]\[ \frac{2}{3}c \leq \frac{-20}{3} \][/tex]
Multiplying both sides by [tex]\(\frac{3}{2}\)[/tex]:
[tex]\[ c \leq \frac{-20 \cdot 3}{3 \cdot 2} = \frac{-60}{6} = -10 \][/tex]

Therefore, the solution to the inequality is:
[tex]\[ c \leq -10 \][/tex]

Selecting the correct multiple choice answer, we have:
[tex]\[ c \leq -10 \][/tex]

The correct answer is:
[tex]\[ \boxed{c \leq -10} \][/tex]