diff --git a/Mathematics/Prealgebra/Add_Subtract_Fractions.md b/Mathematics/Prealgebra/Add_Subtract_Fractions.md
new file mode 100644
index 0000000..8234a53
--- /dev/null
+++ b/Mathematics/Prealgebra/Add_Subtract_Fractions.md
@@ -0,0 +1,41 @@
+---
+tags:
+  - Mathematics
+  - Prealgebra
+  - fractions
+  - division
+---
+
+# Adding and subtracting fractions
+
+## Adding/ subracting fractions with common denominators
+
+For two fractions $\frac{a}{c}$ and  $\frac{b}{c}$ with a common denominator, their sum is defined as:
+
+$$
+    \frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}
+$$
+
+For example:
+
+$$
+    \frac{2}{8} + \frac{3}{8} = \frac{5}{8}
+$$
+
+The same applies to subtraction:
+
+$$
+    \frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}
+$$
+
+
+## Adding/ subracting fractions without common denominators
+
+* Find the lowest common denominator for the two fractions
+* Use this to create two equivalent fractions
+* Add/subtract
+* Reduce 
+
+Demonstration: $\frac{4}{9} + \frac{1}{6}$ :
+
+The lowest common denominator is **the smallest number divisible by both of the denominators of the fractions without remainder**.
diff --git a/Mathematics/Prealgebra/Dividing_fractions.md b/Mathematics/Prealgebra/Dividing_fractions.md
index 459bf4a..5a12a45 100644
--- a/Mathematics/Prealgebra/Dividing_fractions.md
+++ b/Mathematics/Prealgebra/Dividing_fractions.md
@@ -6,23 +6,26 @@ tags:
   - division
 ---
 
+# Dividing fractions
+
 Suppose you have the following shape:
 
-![draw.io-Page-9.drawio 1.png](../../img/draw.io-Page-9.drawio%201.png)
+![draw.io-Page-9.drawio 1.png](../../img/draw.io-Page-9.drawio.png)
 
 One part is shaded. This represents one-eighth of the original shape.
 
 ![one-eighth-a.png](../../img/one-eighth-a.png)
 
-Now imagine there are four instances of the shape and one-eighth remains shaded. How man one-eighths are there in four? 
+Now imagine there are four instances of the shape and one-eighth remains shaded. How man one-eighths are there in four?
 
-![draw.io-Page-9.drawio 2.png](../../img/draw.io-Page-9.drawio%202.png)
+![draw.io-Page-9.drawio 2.png](../../img/draw.io-Page-9.drawio.png)
 
 The shaded proportion represents $\frac{1}{8}$ of the shape. Imagine four of these shapes, how many eighths are there?
 
 This is a division statement: to find how many one-eighths there are we would calculate:
+
 $$
-4 \div \frac{1}{8} 
+4 \div \frac{1}{8}
 $$
 
 But actually it makes more sense to think of this as a multiplication. There are four shapes of eight parts meaning there are $4 \cdot 8$ parts in total, 32. One of these parts is shaded making it equal to $\frac{1}{32}$.
@@ -33,37 +36,37 @@ $$
 4 \div \frac{1}{8} = 4 \cdot 8 = 32
 $$
 
-Note that we omit the numerator but that technically the answer would be $\frac{1}{32}$. 
+Note that we omit the numerator but that technically the answer would be $\frac{1}{32}$.
 
 ### Formal specification of how to divide fractions
 
 We combine the foregoing (that it is easier to divide by fractional amounts using multiplication) with the concept of a [reciprocol](Reciprocals.md) to arrive at a definitive method for dividing two fractions.
-It boils down to: *invert and multiply*:
+It boils down to: _invert and multiply_:
 
- > 
- > If $\frac{a}{b}$ and $\frac{c}{d}$ are fractions then: $$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}$$
+If $\frac{a}{b}$ and $\frac{c}{d}$ are fractions then: $$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}$$
 
-We invert the divisor (the second factor) and change the operator from division to multiplication. 
+We invert the divisor (the second factor) and change the operator from division to multiplication.
 
 #### Demonstration
 
-*Divide $\frac{1}{2}$ by $\frac{3}{5}$*
+Divide $\frac{1}{2}$ by $\frac{3}{5}$
 
 $$
-\\begin{split}
-\\frac{1}{2} \div \frac{3}{5}  = \frac{1}{2} \cdot \frac{5}{3}  \\
+\begin{split}
+\frac{1}{2} \div \frac{3}{5}  = \frac{1}{2} \cdot \frac{5}{3}  \\
 = \frac{5}{5}
-\\end{split}
+\end{split}
 $$
 
-*Divide $\frac{-6}{x}$ by $\frac{-12}{x^2}$*
+Divide $\frac{-6}{x}$ by $\frac{-12}{x^2}$
 
 $$
-\\begin{split}
-\\frac{-6}{x} \div \frac{12}{x^2} = \frac{-6}{x} \cdot \frac{x^2}{-12} \\ = 
-\\frac{(\cancel{3} \cdot \cancel{2} )}{\cancel{x}} \cdot \frac{(\cancel{x} \cdot \cancel{x} )}{\cancel{3} \cdot \cancel{2} \cdot 2} \\ =
-\\frac{x}{2}
+\begin{split}
+\frac{-6}{x} \div \frac{12}{x^2} = \frac{-6}{x} \cdot \frac{x^2}{-12} \\ =
+\frac{(\cancel{3} \cdot \cancel{2} )}{\cancel{x}} \cdot \frac{(\cancel{x} \cdot \cancel{x} )}{\cancel{3} \cdot \cancel{2} \cdot 2} \\ =
+\frac{x}{2}
+
+\end{split}
 
-\\end{split}
 
 $$
diff --git a/Mathematics/Prealgebra/Handling_negative_fractions.md b/Mathematics/Prealgebra/Handling_negative_fractions.md
index b642a61..4fc5fbe 100644
--- a/Mathematics/Prealgebra/Handling_negative_fractions.md
+++ b/Mathematics/Prealgebra/Handling_negative_fractions.md
@@ -6,31 +6,33 @@ tags:
   - negative-numbers
 ---
 
+# Negative fractions
+
 To work with negative fractions we draw on the [Rules for operations on like and unlike terms](Rules%20for%20operations%20on%20like%20and%20unlike%20terms.md).
 
 ## Fractions with unlike terms
 
-* A fraction is just one number divided by another. $\frac{5}{10}$ is just ten divided by 5.
+- A fraction is just one number divided by another. $\frac{5}{10}$ is just ten divided by 5.
 
-* A positive integer divided by a negative or vice versa will always result in a negative. Thus $\frac{5}{-15}$ is equal to $-3$.
+- A positive integer divided by a negative or vice versa will always result in a negative. Thus $\frac{5}{-15}$ is equal to $-3$.
+
+- We can therefore express the whole fraction as a negative:
 
-* We can therefore express the whole fraction as a negative:
   $$
   -	\frac{5}{15}
-  $$ 
+  $$
 
-* Or we could apply the negative symbol to the numerator. It would stand for the same value:
+- Or we could apply the negative symbol to the numerator. It would stand for the same value:
   $$
   \\frac{-5}{15}
   $$
 
 Therefore:
 
- > 
- > Let $a,b$ be any integers. The following three fractions are [equivalent](Equivalent%20fractions.md):	$$\frac{-5}{15}, \frac{5}{-15}, - \frac{5}{15}$$
+> Let $a,b$ be any integers. The following three fractions are [equivalent](Equivalent%20fractions.md): $$\frac{-5}{15}, \frac{5}{-15}, - \frac{5}{15}$$
 
 ## Fractions with like terms
 
-* In cases where both the numerator and denominator are both negative, the value that the fraction represents will be positive overall. This is because the quotient of a negative integer divided by a negative integer will always be positive.
+- In cases where both the numerator and denominator are both negative, the value that the fraction represents will be positive overall. This is because the quotient of a negative integer divided by a negative integer will always be positive.
 
-* Thus:  $$	\frac{- 12xy^2}{ - 18xy^2} = \frac{12xy^2}{18xy^2}$$
+- Thus: $$ \frac{- 12xy^2}{ - 18xy^2} = \frac{12xy^2}{18xy^2}$$
diff --git a/Mathematics/Prealgebra/Least_common_denominator.md b/Mathematics/Prealgebra/Least_common_denominator.md
new file mode 100644
index 0000000..e69de29
diff --git a/Mathematics/Prealgebra/Reciprocals.md b/Mathematics/Prealgebra/Reciprocals.md
index 6143b62..7b54c35 100644
--- a/Mathematics/Prealgebra/Reciprocals.md
+++ b/Mathematics/Prealgebra/Reciprocals.md
@@ -7,21 +7,22 @@ tags:
   - theorems-axioms-laws
 ---
 
+# Recipricols
+
 The [Property of Multiplicative Identity](Multiplicative%20identity.md) applies to fractions as well as to whole numbers:
 
 $$
-\\frac{a}{b} \cdot 1 = \frac{a}{b}
+\frac{a}{b} \cdot 1 = \frac{a}{b}
 $$
 
 With fractions there is a related property: the **Multiplicative Inverse**.
 
- > 
- > If $\frac{a}{b}$ is any fraction, the fraction $\frac{b}{a}$ is called the *multiplicative inverse* or *reciprocol* of $\frac{a}{b}$. The product of a fraction multiplied by its reciprocol will always be 1. $$ \frac{a}{b} \cdot \frac{b}{a} = 1$$
+> If $\frac{a}{b}$ is any fraction, the fraction $\frac{b}{a}$ is called the _multiplicative inverse_ or _reciprocol_ of $\frac{a}{b}$. The product of a fraction multiplied by its reciprocol will always be 1. $$ \frac{a}{b} \cdot \frac{b}{a} = 1$$
 
 For example:
 
 $$
-\\frac{3}{4} \cdot \frac{4}{3} = \frac{12}{12} = 1
+\frac{3}{4} \cdot \frac{4}{3} = \frac{12}{12} = 1
 $$
 
 In this case $\frac{4}{3}$ is the reciprocol or multiplicative inverse of $\frac{3}{4}$.
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