diff --git a/education/math/Partial Fractions.md b/education/math/Partial Fractions.md
index 2b28616..7e80d17 100644
--- a/education/math/Partial Fractions.md	
+++ b/education/math/Partial Fractions.md	
@@ -8,16 +8,19 @@ This is the "main" method of solving, and the next two headings both focus on ge
 
 ### Example
 $$ \frac{2x+1}{(x+1)(x+2)} $$
-Given the above fraction, the denominator is already factored, so we can move onto the next step, where two fractions are made with $a$ and $b$ in the numerator, and each of the denominator components in the denominator:
+1. Given the above fraction, the denominator is already factored, so we can move onto the next step, where two fractions are made with $a$ and $b$ in the numerator, and each of the denominator components in the denominator:
 $$ \frac{a}{x+1} + \frac{b}{x+2} $$
-Next, find a common denominator so that you can add the two fractions together. In this case, it's $(x+1)(x+2)$. 
-
-Then to make the denominators equal, you're going to multiply the numerator and the denominator by the component that the denominator is missing. In this example, the denominator for $\frac{a}{x+1}$ is missing $x+2$, so you're going to multiply by $\frac{x+2}{x+2}$, and vice versa for $\frac{b}{x+2}$:
-$$ \frac{a(x+2)}{(x+1)(x+2)} + \frac{b(x+1)}{(x+2(x+1))} $$
-You can now add the two equations together and distribute $a$ and $b$, giving you:
-$$ \frac{ax+2a + bx+b}{(x+1)(x+2)}$$
-This equals the first equation
-
+2. Next, find a common denominator so that you can add the two fractions together. In this case, it's $(x+1)(x+2)$
+3. Then to make the denominators equal, you're going to multiply the numerator and the denominator by the component that the denominator is missing. In this example, the denominator for $\frac{a}{x+1}$ is missing $x+2$, so you're going to multiply by $\frac{x+2}{x+2}$, and vice versa for $\frac{b}{x+2}$:
+$$ \frac{a(x+2)}{(x+1)(x+2)} + \frac{b(x+1)}{(x+2)(x+1)} $$
+4. You can now add the two equations together and distribute $a$ and $b$, giving you:
+$$ \frac{ax+2a + bx+b}{(x+1)(x+2)} $$
+	This equals the first equation, so:
+$$ \frac{2x+1}{(x+1)(x+2)} = \frac{ax+2a + bx+b}{(x+1)(x+2)} $$
+5. Notice that the denominator on both sides is equal, meaning you can cancel them out, giving you:
+$$ 2x + 1 = ax + 2a + bx + b $$
+6. Next, group your $x$ values on one side, and your constants on the other side. You'll notice that $ax
+$$ 2x+1 =  $$
 ## Degree of the numerator is equal
 1. First perform polynomial division.
 2. Then find a partial fraction with the remainder