Subtract six from both sides.

$\displaystyle \sqrt{3-3x}+6-6 =10-6$

$\displaystyle \sqrt{3-3x}=4$

Square both sides.

$\displaystyle (\sqrt{3-3x})^2=4^2$

$\displaystyle 3-3x=16$

Add $\displaystyle 3x$ on both sides.

$\displaystyle 3-3x+3x=16+3x$

$\displaystyle 3=16+3x$

Subtract 16 from both sides.

$\displaystyle 3-16=16+3x-16$

$\displaystyle -13 = 3x$

Divide by three on both sides.

$\displaystyle \frac{-13 }{3}=\frac{ 3x}{3}$

The answer is:  $\displaystyle -\frac{13}{3}$

Add three on both sides.

$\displaystyle -\sqrt{-9x}-3 +3=- 8+3$

$\displaystyle -\sqrt{-9x} = -5$

Divide by negative one to isolate the radical.

$\displaystyle \frac{-\sqrt{-9x}}{-1} = \frac{-5}{-1}$

The equation becomes:  $\displaystyle \sqrt{-9x} =5$

Square both sides.

$\displaystyle (\sqrt{-9x})^2 =5^2$

$\displaystyle -9x=25$

Divide by negative nine on both sides.

$\displaystyle \frac{-9x}{-9}=\frac{25}{-9}$

The answer is:  $\displaystyle -\frac{25}{9}$

Solve by first subtracting 16 on both sides.

$\displaystyle \sqrt{5x+3}+16-16= 25-16$

The equation becomes:

$\displaystyle \sqrt{5x+3} =9$

Square both sides.

$\displaystyle (\sqrt{5x+3})^2 =9^2$

$\displaystyle 5x+3=81$

Subtract three on both sides.

$\displaystyle 5x+3-3=81-3$

$\displaystyle 5x=78$

Divide by five on both sides.

$\displaystyle \frac{5x}{5}=\frac{78}{5}$

The answer is:  $\displaystyle \frac{78}{5}$

Subtract six from both sides.

$\displaystyle \sqrt[3]{3x}+6-6=8-6$

Simplify both sides.

$\displaystyle \sqrt[3]{3x} = 2$

Cube both sides to eliminate the cube root.

$\displaystyle 3x= 2^3$

$\displaystyle 3x=8$

Divide by three on both sides.

$\displaystyle \frac{3x}{3}=\frac{8}{3}$

The answer is:  $\displaystyle \frac{8}{3}$

Simplify by first subtracting 16 on both sides.

$\displaystyle \sqrt[4]{3(x+1)}+16-16=18-16$

$\displaystyle \sqrt[4]{3(x+1)}=2$

Raise both sides by the fourth power.

$\displaystyle (\sqrt[4]{3(x+1)})^4=2^4$

$\displaystyle 3(x+1)=16$

Expand the binomial.

$\displaystyle 3x+3=16$

Subtract three from both sides.

$\displaystyle 3x+3-3=16-3$

$\displaystyle 3x=13$

Divide by three on both sides.

$\displaystyle \frac{3x}{3}=\frac{13}{3}$

The answer is:  $\displaystyle \frac{13}{3}$

To eliminate the radical, we will need to cube both sides of the equation.

$\displaystyle (\sqrt[3]{9-x})^3 = 3^3$

The equation becomes:

$\displaystyle 9-x = 27$

Add $\displaystyle x$ on both sides.

$\displaystyle 9-x+(x) = 27+(x)$

$\displaystyle 9=27+x$

Subtract 27 from both sides.

$\displaystyle 9-27=27+x-27$

$\displaystyle x=-18$

This answer is valid after re-substituting back to the original equation.

The answer is:  $\displaystyle -18$

Subtract seven on both sides.

$\displaystyle \sqrt{5x-4} +7 -7= 10-7$

$\displaystyle \sqrt{5x-4} =3$

Square both sides.  This will eliminate the radical on the left side.

$\displaystyle (\sqrt{5x-4} )^2=3^2$

$\displaystyle 5x-4 = 9$

Add four on both sides.

$\displaystyle 5x-4 +4= 9+4$

$\displaystyle 5x=13$

Divide by five on both sides.

$\displaystyle \frac{5x}{5}=\frac{13}{5}$

The answer is:  $\displaystyle \frac{13}{5}$

Add three on both sides.

$\displaystyle 3\sqrt[3]{4x-3} -3+3= 1+3$

$\displaystyle 3\sqrt[3]{4x-3}=4$

Divide by three on both sides.

$\displaystyle \frac{3\sqrt[3]{4x-3}}{3}=\frac{4}{3}$

$\displaystyle \sqrt[3]{4x-3}=\frac{4}{3}$

Cube both sides to eliminate the radical.  Simplify both sides.

$\displaystyle (\sqrt[3]{4x-3})^3=(\frac{4}{3})^3$

$\displaystyle 4x-3 = \frac{64}{27}$

Add three on both sides.  This is the same as adding $\displaystyle \frac{81}{27}$ on the right side.

$\displaystyle 4x-3+3=\frac{64}{27}+\frac{81}{27}$

$\displaystyle 4x= \frac{145}{27}$

Divide by four on both sides.  This is similar to multiplying one fourth on both sides and will isolate the x-variable on the left.

$\displaystyle 4x\cdot \frac{1}{4}= \frac{145}{27}\cdot \frac{1}{4}$

$\displaystyle x=\frac{145}{108}$

The answer is:  $\displaystyle \frac{145}{108}$

In order to solve this equation, first add 28 on both sides.

$\displaystyle \sqrt{6x-3}-28+28 = 30+28$

$\displaystyle \sqrt{6x-3}=58$

Square both sides.

$\displaystyle (\sqrt{6x-3})^2=58^2$

$\displaystyle 6x-3=3364$

Add three on both sides.

$\displaystyle 6x-3+3=3364+3$

$\displaystyle 6x=3367$

Divide by six on both sides.

$\displaystyle \frac{6x}{6}=\frac{3367}{6}$

The answer is:  $\displaystyle \frac{3367}{6}$

Cube both sides.

$\displaystyle (\sqrt[3]{6x-2})^3= 4^3$

This will eliminate the radical.

$\displaystyle 6x-2= 64$

Add two on both sides.

$\displaystyle 6x-2+2= 64+2$

$\displaystyle 6x=66$

Divide by six on both sides.

$\displaystyle \frac{6x}{6}=\frac{66}{6}$

The answer is:  $\displaystyle 11$