Question

solve for c.
$2 + \frac{4}{3}c = c - \frac{1}{3}$
$c = $

Explanation:

Step1: Subtract \( c \) from both sides

To get like terms together, subtract \( c \) from each side of the equation \( 2 + \frac{4}{3}c = c - \frac{1}{3} \). This gives \( 2 + \frac{4}{3}c - c = -\frac{1}{3} \). Simplifying \( \frac{4}{3}c - c \) as \( \frac{4}{3}c - \frac{3}{3}c = \frac{1}{3}c \), so the equation becomes \( 2 + \frac{1}{3}c = -\frac{1}{3} \).

Step2: Subtract 2 from both sides

Now, subtract 2 from both sides to isolate the term with \( c \). The equation is \( \frac{1}{3}c = -\frac{1}{3} - 2 \). Convert 2 to thirds: \( 2 = \frac{6}{3} \), so \( -\frac{1}{3} - \frac{6}{3} = -\frac{7}{3} \)? Wait, no, wait. Wait, original step: Wait, in Step1, after subtracting \( c \), we have \( 2 + \frac{1}{3}c = -\frac{1}{3} \)? Wait, no, let's redo Step1 correctly. Wait, the original equation is \( 2 + \frac{4}{3}c = c - \frac{1}{3} \). Subtract \( c \) from both sides: \( 2 + \frac{4}{3}c - c = -\frac{1}{3} \). \( \frac{4}{3}c - c = \frac{4}{3}c - \frac{3}{3}c = \frac{1}{3}c \), so \( 2 + \frac{1}{3}c = -\frac{1}{3} \). Then subtract 2 from both sides: \( \frac{1}{3}c = -\frac{1}{3} - 2 \). Convert 2 to \( \frac{6}{3} \), so \( -\frac{1}{3} - \frac{6}{3} = -\frac{7}{3} \)? Wait, that can't be right. Wait, no, maybe I made a sign error. Wait, the right side after subtracting \( c \) is \( -\frac{1}{3} \), and we have \( 2 + \frac{1}{3}c = -\frac{1}{3} \). Then subtract 2: \( \frac{1}{3}c = -\frac{1}{3} - 2 \). But 2 is positive, so subtracting 2 is \( -\frac{1}{3} - \frac{6}{3} = -\frac{7}{3} \)? Wait, but that would lead to \( c = -7 \), but let's check again. Wait, maybe I messed up the direction. Wait, let's start over.

Alternative approach: Multiply both sides by 3 to eliminate denominators. The equation is \( 2 + \frac{4}{3}c = c - \frac{1}{3} \). Multiply each term by 3: \( 3*2 + 3*\frac{4}{3}c = 3*c - 3*\frac{1}{3} \). Simplify each term: \( 6 + 4c = 3c - 1 \). Now, subtract \( 3c \) from both sides: \( 6 + c = -1 \). Then subtract 6 from both sides: \( c = -1 - 6 = -7 \). Wait, that's different from my previous mistake. Wait, so where did I go wrong earlier? Ah, in Step1, when I subtracted \( c \), the right side is \( -\frac{1}{3} \), but when I did Step2, I think I messed up the sign. Wait, let's do the multiplication by 3 method, which is better. So:

Step1: Multiply all terms by 3

To eliminate the fractions, multiply each term in \( 2 + \frac{4}{3}c = c - \frac{1}{3} \) by 3. This gives \( 3*2 + 3*\frac{4}{3}c = 3*c - 3*\frac{1}{3} \). Simplifying each term: \( 6 + 4c = 3c - 1 \).

Step2: Subtract \( 3c \) from both sides

Subtract \( 3c \) from both sides to get \( 6 + c = -1 \).

Step3: Subtract 6 from both sides

Subtract 6 from both sides: \( c = -1 - 6 \), so \( c = -7 \). Wait, let's check by plugging back in. Original equation: \( 2 + \frac{4}{3}*(-7) = -7 - \frac{1}{3} \). Left side: \( 2 - \frac{28}{3} = \frac{6}{3} - \frac{28}{3} = -\frac{22}{3} \). Right side: \( -\frac{21}{3} - \frac{1}{3} = -\frac{22}{3} \). Yes, that works. So my initial Step2 had a sign error, but using multiplication by 3 is clearer.

Wait, let's redo the steps correctly with the first method:

Original equation: \( 2 + \frac{4}{3}c = c - \frac{1}{3} \)

Step1: Subtract \( c \) from both sides

\( 2 + \frac{4}{3}c - c = -\frac{1}{3} \)

Simplify \( \frac{4}{3}c - c = \frac{1}{3}c \), so:

\( 2 + \frac{1}{3}c = -\frac{1}{3} \)

Step2: Subtract 2 from both sides

\( \frac{1}{3}c = -\frac{1}{3} - 2 \)

Convert 2 to \( \frac{6}{3} \), so:

\( \frac{1}{3}c = -\frac{1}{3} - \frac{6}{3} = -\frac{7}{3} \)? Wait, no, that's wrong. Wait, no: \( -\frac{1}{3} - 2 = -\frac{1}{3} - \frac{6}{3} = -\frac{7}{3} \), but then multiplying both sides by 3: \( c = -7 \), which matches the other method. Oh, right! I thought earlier that was a mistake, but it's correct. So \( \frac{1}{3}c = -\frac{7}{3} \), then multiply both sides by 3: \( c = -7 \). Yes, that's correct.

So let's structure the steps properly:

Step1: Subtract \( c \) from both sides

\( 2 + \frac{4}{3}c - c = c - \frac{1}{3} - c \)

Simplify: \( 2 + \frac{1}{3}c = -\frac{1}{3} \)

Step2: Subtract 2 from both sides

\( \frac{1}{3}c = -\frac{1}{3} - 2 \)

Convert 2 to \( \frac{6}{3} \): \( -\frac{1}{3} - \frac{6}{3} = -\frac{7}{3} \), so \( \frac{1}{3}c = -\frac{7}{3} \)

Step3: Multiply both sides by 3

\( c = -\frac{7}{3} * 3 = -7 \)

Answer:

\( -7 \)