# Solving Equations I: Linear Equations

In a linear equation, each term is either a constant or the product of a constant and a single variable of degree 1. It can have one or more variables. Here are some linear equations:

$x+1=0$
$2x+4=6$
$-3x+1=x+4$
$1+5x=0$
$2-3x=3+2x$
$\frac{1}{2}x+\frac{3}{4}=\frac{1}{3}$
$\frac{x-1}{3}+\frac{1}{4}=\frac{5}{2}$
$\frac{x}{c}+b=a$ where $c\neq 0$ is a constant
$\sqrt{3}x-\sqrt{2}=\sqrt{5}$

However, the following equations are not linear:

$x^2+1=0$
$\frac{1}{x-1}+4=3$
$\sqrt{x}-2=5$
$|x|+3=1$

In order to solve a linear equation, the unknown variable must be isolated. This means to get any term with the variable in it on one side of the equation. There are two methods to isolate variables. The first method, which is very simple, is to do the same to both sides of the equal sign. Let's see an example:
Example 1: $2x+4=6$
We need to get rid of $4$ in the left hand side of the equal sign. We accomplish this by subtracting 4 on both sides.
$2x+4-4=6-4$
$2x=2$
The next step is to divide both sides by $2$:
$\frac{2x}{2}=\frac{2}{2}$
$x=1$
Therefore, the answer is $1$.
We can check the solution by substituting $1$ in the original equation for $x$.
$2x+4=2 \times 1+4=6$
Since the left side of the equation equals the right side of the equation after the substitution, the answer that we found is correct.

Lets see another examples:
Example 2:
$1+5x=0$
This an easy one! We subtract both sides of the equation by $1$.
$1+5x-1=-1$
This simplifies to
$5x=-1$
Now, we should divide both sides by $5$ to get rid of the coefficient of the variable,
$x=\frac{-1}{5}=-\frac{1}{5}$

Another method to isolate the variable is to move like terms to one side of the equation. When you move a term to the other side of a equal sign the symbol changes to the opposite sign. If it is positive it turns negative, if it is negative it turns positive. To cancel out the coefficient of the variable, you should divide all terms by the coefficient. Let's how this works by an example:

Example 3: $-3x+1=x+4$
Lets move all terms with $x$ to left side and other terms to right side:
Move $+1$ to the right side of the equal sign and change the sign from $+$ to $–$
$-3x=x+4-1$
Move $x$ to the right side of the equal sign and change the sign from $+$ to $–$
$-3x-x=4-1$
This simplifies to
$-4x=3$
Cancel out $-4$ by dividing both sides by $-4$. In other words, move it to the left side of the equal sign and flip it over.
$x=\frac{3}{-4}=-\frac{3}{4}$
Then the answer would be $x=-\frac{3}{4}$.

Example 4:
$2-3x=3+2x$
Move $2$ to the right side and change the sign from $+$ to $–$
$-3x=3+2x-2$
Move $+2x$ to the left side and change the sign from $+$ to $–$
$-3x-2x=3-2$
Simplify
$-5x=1$
Divide both sides by $-5$,
$x=\frac{1}{-5}=-\frac{1}{5}$
The answer is $x=-\frac{1}{5}$.

Example 5:
$\frac{1}{2}x+\frac{3}{4}=\frac{1}{3}$
Move $+\frac{3}{4}$ to the right side and change the sign from $+$ to $-$
$\frac{1}{2}x=\frac{1}{3}-\frac{3}{4}$

Simplify
$\frac{1}{2}x=\frac{4}{12}-\frac{9}{12}=\frac{4-9}{12}=-\frac{5}{12}$
Multiply both sides by $2$,
$\frac{2}{2}x=-\frac{5\times 2}{12}$
$x=-\frac{5}{6}$
The answer is $x=-\frac{5}{6}$.

Example 6:
$\frac{x-1}{3}+\frac{1}{4}=\frac{5}{2}$
Move $+\frac{1}{4}$ to the right side and change the sign from $+$ to $-$
$\frac{x-1}{3}=\frac{5}{2}-\frac{1}{4}$
Simplify
$\frac{x-1}{3}=\frac{5 \times 2}{2 \times 2}-\frac{1}{4}=\frac{10}{4}-\frac{1}{4}=\frac{10-1}{4}=\frac{9}{4}$
Multiply both sides by $3$,
$\frac{3\times (x-1)}{3}=\frac{9\times 3}{4}$
$x-1=\frac{27}{4}$
Move $-1$ to the right side and change the sign from $-$ to $+$
$x=\frac{27}{4}+1$
Simplify
$x=\frac{27}{4}+\frac{4}{4}=\frac{27+4}{4}=\frac{31}{4}$
The answer is $x=\frac{31}{4}$.

Example 7:
$\frac{x}{c}+b=a$ where $c\neq 0$ is a constant
Move $+b$ to the right side and change the sign from $+$ to $-$
$\frac{x}{c}=a-b$
Multiply both sides by $c$,
$\frac{c \times x}{c}=c \times (a-b)$
Simplify
$x=c \times (a-b)= ca-cb$
The answer is $x=ca-cb$.

Example 8:
$\sqrt{3}x-\sqrt{2}=\sqrt{5}$
Move $-\sqrt{2}$ to the right side and change the sign from $-$ to $+$
$\sqrt{3}x=\sqrt{5}+\sqrt{2}$
Divide both sides by $\sqrt{3}$,
$\frac{\sqrt{3}x}{\sqrt{3}}=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{3}}$
Simplify
$x=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{3}}$

To get the simplest form, we rationalizing the denominator by multiplying both numerator and denominator by $\sqrt{3}$:
$x=\frac{\sqrt{3} \times (\sqrt{5}+\sqrt{2})}{\sqrt{3} \times \sqrt{3}}=\frac{\sqrt{15}+\sqrt{6}}{3}=\frac{\sqrt{15}}{3}+\frac{\sqrt{6}}{3}$

The answer is $x=\frac{\sqrt{15}}{3}+\frac{\sqrt{6}}{3}$.