# How do you solve 2root3(10-3x)=root3(2-x)?

Jul 9, 2017

See a solution process below:

#### Explanation:

First, cube each side of the equation to eliminate the radicals while keeping the equation balanced:

${\left(2 \sqrt[3]{10 - 3 x}\right)}^{3} = {\left(\sqrt[3]{2 - x}\right)}^{3}$

${2}^{3} \left(10 - 3 x\right) = 2 - x$

$8 \left(10 - 3 x\right) = 2 - x$

Next, eliminate the parenthesis on the left side of the equation by multiplying each term within the parenthesis by the term outside the parenthesis:

$\textcolor{red}{8} \left(10 - 3 x\right) = 2 - x$

$\left(\textcolor{red}{8} \times 10\right) - \left(\textcolor{red}{8} \times 3 x\right) = 2 - x$

$80 - 24 x = 2 - x$

Then, add $\textcolor{red}{24 x}$ and subtract $\textcolor{b l u e}{2}$ from each side of the equation to isolate the $x$ term while keeping the equation balanced:

$- \textcolor{b l u e}{2} + 80 - 24 x + \textcolor{red}{24 x} = - \textcolor{b l u e}{2} + 2 - x + \textcolor{red}{24 x}$

$78 - 0 = 0 - 1 x + \textcolor{red}{24 x}$

$78 = \left(- 1 + \textcolor{red}{24}\right) x$

$78 = 23 x$

Now, divide each side of the equation by $\textcolor{red}{23}$ to solve for $x$ while keeping the equation balanced:

$\frac{78}{\textcolor{red}{23}} = \frac{23 x}{\textcolor{red}{23}}$

$\frac{78}{23} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{23}}} x}{\cancel{\textcolor{red}{23}}}$

$\frac{78}{23} = x$

$x = \frac{78}{23}$

Jul 9, 2017

$x = \frac{78}{23}$

#### Explanation:

$2 \sqrt[3]{10 - 3 x} = \sqrt[3]{2 - x}$

Cube both sides

$\therefore {\left(2 \sqrt[3]{10 - 3 x}\right)}^{3} = {\left(\sqrt[3]{2 - x}\right)}^{3}$

$\therefore \sqrt[3]{a - b} \cdot 2 \sqrt[3]{a - b} \cdot \sqrt[3]{a - b} = a - b$

$\therefore {2}^{3} \left(10 - 3 x\right) = 2 - x$

$\therefore 8 \left(10 - 3 x\right) = 2 - x$

$\therefore 80 - 24 x = 2 - x$

$\therefore - 24 x + x = 2 - 80$

$- 23 x = - 78$

Multiply both sides by $- 1$

$\therefore 23 x = 78$

:.color(magenta)(x=78/23 orcolor(magenta)(3.391304348

~~~~~~~~~~~~~~~~~~~~~~~~~~~

color(magenta)(check:

$\therefore 2 \sqrt[3]{10 - 3 \left(\textcolor{m a \ge n t a}{3.391304348}\right)} = \sqrt[3]{2 - \left(\textcolor{m a \ge n t a}{3.391304348}\right)}$

:.2root3(10-10.17391304))=root3(2-(3.391304348))

:.2root3(-0.17391304))=root3(-1.391304348)

$\therefore 2 \times - 0.558183998 = - 1.116368004$

:.color(magenta)(-1.116367997=-1.116368004