diff --git a/education/math/Partial Fractions.md b/education/math/Partial Fractions.md
index 7e80d17..c741890 100644
--- a/education/math/Partial Fractions.md	
+++ b/education/math/Partial Fractions.md	
@@ -19,8 +19,13 @@ $$ \frac{ax+2a + bx+b}{(x+1)(x+2)} $$
 $$ \frac{2x+1}{(x+1)(x+2)} = \frac{ax+2a + bx+b}{(x+1)(x+2)} $$
 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
-$$ 2x+1 =  $$
+6. Next, group your $x$ values on one side, and your constants on the other side. You'll notice that you can factor $ax + bx$, giving you $x(a+b)$.
+$$ 2x+1 = x(a + b) + (2a + b) $$
+7. With the above equation, each side is in the same form. it's $x$ multiplied by a constant ($2$ on the left, and $(a+b)$ on the right, and with a constant of $1$ on the left and $2a + b$) on the right, letting you find the two equations below:
+$$ 2 = a + b $$
+$$ 1 = 2a + b $$
+
+
 ## Degree of the numerator is equal
 1. First perform polynomial division.
 2. Then find a partial fraction with the remainder