diff --git a/education/math/Partial Fractions.md b/education/math/Partial Fractions.md
index 11fb782..2b28616 100644
--- a/education/math/Partial Fractions.md	
+++ b/education/math/Partial Fractions.md	
@@ -11,7 +11,12 @@ $$ \frac{2x+1}{(x+1)(x+2)} $$
 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 $(x+1)(x+2)$. 
-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
+
+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
 
 ## Degree of the numerator is equal
 1. First perform polynomial division.