# If -1/3x+7=4, what is 1/3x+3?

Mar 10, 2018

See a solution process below:

#### Explanation:

Step 1) Solve the equation for $x$:

$- \frac{1}{3} x + 7 = 4$

$- \frac{1}{3} x + 7 - \textcolor{red}{7} = 4 - \textcolor{red}{7}$

$- \frac{1}{3} x + 0 = - 3$

$- \frac{1}{3} x = - 3$

$\textcolor{red}{- 3} \times - \frac{1}{3} x = \textcolor{red}{- 3} \times - 3$

$\frac{\textcolor{red}{- 3}}{-} 3 x = 9$

$1 x = 9$

$x = 9$

Step 2) Substitute $\textcolor{red}{9}$ for $\textcolor{red}{x}$ in the expression and then evaluate the expression:

$\frac{1}{3} \textcolor{red}{x} + 3$ becomes:

$\left(\frac{1}{3} \times \textcolor{red}{9}\right) + 3 \implies$

$\frac{\textcolor{red}{9}}{3} + 3 \implies$

$3 + 3 \implies$

$6$

Mar 10, 2018

$x = 9$

$\frac{1}{3} x + 3 = 6$

#### Explanation:

Solve the first equation for $x$.

$- \frac{1}{3} x + 7 = 4$

Simplify $- \frac{1}{3} x$ to $- \frac{x}{3}$.

$- \frac{x}{3} + 7 = 4$

Subtract $7$ from both sides.

$- \frac{x}{3} = 4 - 7$

Simplify.

$- \frac{x}{3} = - 3$

Multiply both sides by $3$.

${\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}}^{1} \times \left(- \frac{x}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} ^ 1\right) = - 3 \times 3$

Simplify.

$- x = - 9$

Multiply both sides by $- 1$.

$x = 9$

Now substitute $9$ for $x$ in the second expression.

$\frac{1}{3} x + 3$

Simplify $\frac{1}{3} x$ to $\frac{x}{3}$.

${\textcolor{red}{\cancel{\textcolor{b l a c k}{9}}}}^{3} / {\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}}^{1} + 3$

Simplify.

$3 + 3 = 6$