diff --git a/education/math/Partial Fractions.md b/education/math/Partial Fractions.md
index 484f9d3..11fb782 100644
--- a/education/math/Partial Fractions.md	
+++ b/education/math/Partial Fractions.md	
@@ -1,5 +1,6 @@
 Partial fraction decomposition is when you break a polynomial fraction down into smaller fractions that add together. 
 ## Degree where the numerator is less
+This is the "main" method of solving, and the next two headings both focus on getting to this point, at which point you solve using the below steps.
 1. Factor the bottom. 
 2. Create two fractions, $\frac{a}{p1}$, and $\frac{b}{p2}$, where p1 and p2 are the polynomials you just factored out, and a/b are arbitrary variables
 3. Multiply a by p2, and b by p1., giving you: $$\frac{a*p2}{p1} + \frac{b*p1}{p2}$$
@@ -9,7 +10,8 @@ Partial fraction decomposition is when you break a polynomial fraction down into
 $$ \frac{2x+1}{(x+1)(x+2)} $$
 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:
 $$ \frac{a}{x+1} + \frac{b}{x+2} $$
-Next, find a common denominator so that you can add the two fractions together. In this case, it's $(x+1)(x+2)$.
+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
 
 ## Degree of the numerator is equal
 1. First perform polynomial division.