$\displaystyle 5y+27-5z=30$

First subtract 27 from both sides of the equation

$\displaystyle 5y-5z=3$

Add 5z to both sides of the equation

$\displaystyle 5y=3+5z$

Lastly, divide both sides by 5 to get the y by itself

$\displaystyle y=\frac{3+5z}{5}$

To isolate the x-variable, we can multiply both sides by the least common denominator.

The least common denominator is $\displaystyle 6x$.  This will eliminate the fractions.

$\displaystyle 6x(\frac{2}{3x}+4)=\frac{1}{2x} \cdot 6x$

$\displaystyle 4+24x = 3$

Subtract 4 on both sides.

$\displaystyle 4+24x -4= 3-4$

$\displaystyle 24x=-1$

Divide by 24 on both sides.

$\displaystyle \frac{24x}{24}=\frac{-1}{24}$

The answer is:  $\displaystyle -\frac{1}{24}$

Find the least common denominator of both sides of the equation, and multiply it on both sides.  

The LCD is 60.

$\displaystyle 60(\frac{1}{3}x-\frac{1}{4}x) = \frac{1}{5}\times 60$

$\displaystyle 20x-15x = 12$

Combine like-terms on the left.

$\displaystyle 5x=12$

Divide by 5 on both sides.

$\displaystyle \frac{5x}{5}=\frac{12}{5}$

The answer is:  $\displaystyle \frac{12}{5}$

Distribute the eight through both terms of the binomial.

$\displaystyle 8x-24 = 16-x$

Add $\displaystyle x$ on both sides.

$\displaystyle 8x-24 +x= 16-x+x$

$\displaystyle 9x-24 = 16$

Add 24 on both sides.

$\displaystyle 9x-24+24 = 16+24$

$\displaystyle 9x=40$

Divide by 9 on both sides.

$\displaystyle \frac{9x}{9}=\frac{40}{9}$

The answer is:  $\displaystyle \frac{40}{9}$

Subtract $\displaystyle 7x$ from both sides.

$\displaystyle -2+7x-7x = 8x-6-7x$

$\displaystyle -2=x-6$

Add 6 on both sides.

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

$\displaystyle x=4$

The answer is:  $\displaystyle 4$

First, we want to get everything inside the square roots, so we distribute the $\displaystyle 2$:

$\displaystyle \sqrt{x^2 + 8} = \sqrt{4}\sqrt{2x-1}$

$\displaystyle \sqrt{x^2 + 8} = \sqrt{4(2x-1)}$

$\displaystyle \sqrt{x^2 + 8} = \sqrt{8x-4}$

Now we can clear our the square roots by squaring each side:

$\displaystyle (\sqrt{x^2 + 8})^{2} = (\sqrt{8x-4})^2$

$\displaystyle x^2 + 8 = 8x-4$

Now we can simplify by moving everything to one side of the equation:

$\displaystyle x^2 -8x+ 12 = 0$

Factoring will give us:

$\displaystyle (x-6)(x-2)= 0$

So our answers are:

$\displaystyle x=2, 6$

Begin by gathering all the constants to one side of the equation:

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

Now multiply by $\displaystyle -1$:

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

And finally, square each side:

$\displaystyle x=25$

This might look all fine and dandy, but let's check our solution by plugging it in to the original equation:

$\displaystyle 6-\sqrt{25}=11$

$\displaystyle 6-5=11$

$\displaystyle 1\neq 11$

So our solution is invalid, and the problem doesn't have a solution.

Add two on both sides.

$\displaystyle 3x-2+2 = 50+2$

$\displaystyle 3x = 52$

Divide by three on both sides.

$\displaystyle \frac{3x }{3}=\frac{ 52}{3}$

The answer is:  $\displaystyle \frac{52}{3}$

To isolate the x-variable, multiply both sides by the coefficient of the x-variable.

$\displaystyle \frac{2}{3}x \cdot \frac{3}{2} = \frac{5}{2}\cdot\frac{3}{2}$

The answer is:  $\displaystyle \frac{15}{4}$

Multiply both sides of the equation by $\displaystyle x^{2}-2x$ to eliminate the fraction:

$\displaystyle \frac{ x^{2}-4x}{x^{2}-2x} = 2$

$\displaystyle \frac{ x^{2}-4x}{x^{2}-2x}\cdot (x^{2}-2x) = 2 \cdot (x^{2}-2x)$

$\displaystyle x^{2}-4x = 2 x^{2}-4x$

Subtract $\displaystyle x^{2}-4x$ from both sides:

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

$\displaystyle 2 x^{2} - x^{2} -4x + 4x = 0$

$\displaystyle x^{2} = 0$

The only possible solution is $\displaystyle x = 0$, However, if this is substituted in the original equation, the expression at left is undefined, as seen here:

$\displaystyle \frac{ x^{2}-4x}{x^{2}-2x} = 2$

$\displaystyle \frac{ 0^{2}-4 \cdot 0}{0^{2}-2 \cdot 0} = 2$

$\displaystyle \frac{ 0}{ 0} = 2$

An expression with a denominator of 0 has an undefined value, so this statement is false. The equation has no solution.