# How do you solve 2root3(7b-1)-4=0?

Jun 27, 2017

See a solution process below:

#### Explanation:

First, add $\textcolor{red}{4}$ to each side of the equation to isolate the radical term while keeping the equation balanced:

$2 \sqrt[3]{7 b - 1} - 4 + \textcolor{red}{4} = 0 + \textcolor{red}{4}$

$2 \sqrt[3]{7 b - 1} - 0 = 4$

$2 \sqrt[3]{7 b - 1} = 4$

Next, divide each side of the equation by $\textcolor{red}{2}$ to isolate the radical while keeping the equation balanced:

$\frac{2 \sqrt[3]{7 b - 1}}{\textcolor{red}{2}} = \frac{4}{\textcolor{red}{2}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \sqrt[3]{7 b - 1}}{\cancel{\textcolor{red}{2}}} = 2$

$\sqrt[3]{7 b - 1} = 2$

Then, cube each side of the equation to eliminate the radical while keeping the equation balanced:

${\left(\sqrt[3]{7 b - 1}\right)}^{\textcolor{red}{3}} = {2}^{\textcolor{red}{3}}$

$7 b - 1 = 8$

Next, add $\textcolor{red}{1}$ to each side of the equation to isolate the $b$ term while keeping the equation balanced:

$7 b - 1 + \textcolor{red}{1} = 8 + \textcolor{red}{1}$

$7 b - 0 = 9$

$7 b = 9$

Now, divide each side of the equation by $\textcolor{red}{7}$ to solve for $b$ while keeping the equation balanced:

$\frac{7 b}{\textcolor{red}{7}} = \frac{9}{\textcolor{red}{7}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}} b}{\cancel{\textcolor{red}{7}}} = \frac{9}{7}$

$b = \frac{9}{7}$