To solve for the variable isolate it on one side of the equation by moving all other constants to the other side. To do this, perform opposite operations to manipulate the equation.

\(\displaystyle \frac{5}{4}x=1200\) 

Multiply \(\displaystyle \frac{4}{5}\) on both sides.

\(\displaystyle x=960\)

Add 14 on both sides.

\(\displaystyle 8x-14+14=37+14\)

Simplify both sides.

\(\displaystyle 8x=51\)

Divide by eight on both sides.

\(\displaystyle \frac{8x}{8}=\frac{51}{8}\)

The answer is:  \(\displaystyle \frac{51}{8}\)

Divide by two on both sides.

\(\displaystyle \frac{2(2x-3)}{2}=\frac{18}{2}\)

Simplify both sides.

\(\displaystyle 2x-3=9\)

Add three on both sides.

\(\displaystyle 2x-3+3=9+3\)

Simplify both sides.

\(\displaystyle 2x=12\)

Divide by two on both sides.

\(\displaystyle \frac{2x}{2}=\frac{12}{2}\)

\(\displaystyle x=6\)

The answer is:  \(\displaystyle 6\)

Notice that the denominators on the left side can be converted to 33 by multiplying the top and bottom by a value of three.  This will save time from having to convert between steps.

\(\displaystyle (\frac{3}{3})(\frac{2}{11}x+\frac{1}{11} )= \frac{1}{33}\)

Simplify by distribution.

\(\displaystyle \frac{6}{33}x + \frac{3}{33} = \frac{1}{33}\)

Multiply by 33 on both sides in order to cancel the denominators.

\(\displaystyle 33(\frac{6}{33}x + \frac{3}{33}) = 33(\frac{1}{33})\)

The equation becomes:  \(\displaystyle 6x+3=1\)

Subtract three on both sides.

\(\displaystyle 6x=-2\)

Divide by six on both sides and reduce the fraction.

\(\displaystyle \frac{6x}{6}=\frac{-2}{6}\)

The answer is:  \(\displaystyle x=-\frac{1}{3}\)

In order to isolate the \(\displaystyle x\) variable, first square both sides of the equation to eliminate the radical.

\(\displaystyle (\sqrt{2x+3})^2=( 6)^2\)

Simplify both sides.

\(\displaystyle 2x+3 = 36\)

Subtract three from both sides.

\(\displaystyle 2x=33\)

Divide by two on both sides.

\(\displaystyle \frac{2x}{2}=\frac{33}{2}\)

The answer is:  \(\displaystyle \frac{33}{2}\)

Group the constants and variables on a separate side of the equation.

Subtract \(\displaystyle 3x\) on both sides.

\(\displaystyle 3x+9-(3x)=6x-9-(3x)\)

\(\displaystyle 9=3x-9\)

Add nine on both sides.

\(\displaystyle 9+9=3x-9+9\)

Simplify both sides.

\(\displaystyle 18=3x\)

Divide by three on both sides.

The answer is:  \(\displaystyle x=6\)

Divide by two on both sides.

\(\displaystyle \frac{2(6x+9) }{2}=\frac{ -36}{2}\)

Simplify both sides of the equation.

\(\displaystyle 6x+9=-18\)

Subtract nine from both sides.

\(\displaystyle 6x+9-9=-18-9\)

Simplify both sides.

\(\displaystyle 6x=-27\)

Divide by six on both sides.

\(\displaystyle \frac{6x}{6}=\frac{-27}{6}\)

Reduce the fractions.

\(\displaystyle x=\frac{-3\times 9}{3\times 2} = -\frac{9}{2}\)

The answer is:  \(\displaystyle -\frac{9}{2}\)

First, multiply both sides of the equation by \(\displaystyle x\) to eliminate the denominators of the terms in the left-hand side:

\(\displaystyle (x)(\frac{3}{x} + \frac{6}{x} - \frac{1}{x}) = 2x(x)\)

\(\displaystyle \Rightarrow 3+6-1= 2x^2\)

\(\displaystyle \Rightarrow\) \(\displaystyle 8=2x^2\)

Next, divide both sides by \(\displaystyle 2\) to yield a power of \(\displaystyle x\) on the right-hand side:

\(\displaystyle \frac{8}{2} = \frac{2x^2}{2}\)

\(\displaystyle \Rightarrow4=x^2\)

Take the square root of both sides to solve for the unknown variable \(\displaystyle x\):

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

\(\displaystyle \Rightarrow\pm2=x\)

or

\(\displaystyle x=2,-2\)

We now have two solutions for the unknown variable. Substitute \(\displaystyle 2\) and \(\displaystyle -2\) into the original equation to verify that both solutions yield an identity:

If \(\displaystyle x=2\), then

\(\displaystyle \frac{3}{2} + \frac{6}{2} - \frac{1}{2} = 2(2)\)

\(\displaystyle \Rightarrow \frac{8}{2} = 4\)

which is true.

If \(\displaystyle x=-2\), then

\(\displaystyle -\frac{3}{2} - \frac{6}{2} + \frac{1}{2} = 2(-2)\)

\(\displaystyle \Rightarrow -\frac{8}{2} = -4\)

which is also true.

Hence, the two solutions for the given equation are

\(\displaystyle x=2,-2\)

First, subtract \(\displaystyle 1\) from both sides of the equation:

\(\displaystyle 1-1+\frac{5}{3x-7}=2-1\)

\(\displaystyle \Rightarrow \frac{5}{3x-7}=1\)

Next, multiply both sides by \(\displaystyle 3x-7\) to eliminate the denominator of the term on the left-hand side of the equation:

\(\displaystyle \frac{5}{3x-7}(3x-7)=1(3x-7)\)

\(\displaystyle \Rightarrow5=3x-7\)

Add \(\displaystyle 7\) to both side of the equation to isolate the term containing the unknown variable on the right-hand side of the equation:

\(\displaystyle 5+7=3x-7+7\)

\(\displaystyle \Rightarrow12=3x\)

Divide both sides by \(\displaystyle 3\):

\(\displaystyle \frac{12}{3}=\frac{3x}{3}\)

\(\displaystyle \Rightarrow x=4\)

Checking the solution \(\displaystyle x=4\) by substituting it into the original equation yields:

\(\displaystyle 1+\frac{5}{3(4)-7}=2\)

\(\displaystyle \Rightarrow 1+\frac{5}{5}=2\)

\(\displaystyle \Rightarrow 1+1=2\)

which is true. Hence, the solution to this equation is \(\displaystyle x=4\).

Group like terms.  Start with the x-variable.

Add \(\displaystyle 9x\) on both sides.

\(\displaystyle 9x-3+(9x)=-9x-6+(9x)\)

\(\displaystyle 18x-3=-6\)

Add three on both sides.

\(\displaystyle 18x-3+3=-6+3\)

Simplify both sides.

\(\displaystyle 18x=-3\)

Divide by 18 on both sides.

\(\displaystyle \frac{18x}{18}=\frac{-3}{18}\)

Simplify both fractions.

\(\displaystyle x=-\frac{1}{6}\)

The answer is:  \(\displaystyle -\frac{1}{6}\)