# How To Solve Equations With Fractions.

Solving equations with fractions is not much different to the normal method of solving equations.

A lot of students struggle to solve for x with fractions. This is due to the fact that most have difficulty solving fractions.

Simplifying fractions was introduced as far back as grade 4 or 5 but when you get to FET phase, you are required to recall this. In fact, it is also covered in grade 8. But we why do students fear this? Well, we do not know.

So if you are like some of us, fearing this, today you can 'chill'. And we must say, the methods get better and better with time. To help you solve equations with fractions, follow this procedure.

1. Ensure your denominators are factorised.
3. Find the lowest common denominator (LCD) or lowest common multiple (LCM).
4. Multiple the expression with the LCD or LCM with the expression
5. Solve for x

Now tet us try a few examples.

1. Solve for x: $$\displaystyle \frac{x+1}{x+2} = \frac{x-3}{x+1}$$

Following the steps above.

We do not need to factorise the denominators. They are simplified.

When we restrict equations, we need to make sure that the denominators are not equal to zero.

\begin{gather} x+2=0\\ x=-2 \end{gather}

The restriction is therefore $$x \neq -2$$

The second restriction:

\begin{gather} x+1=0\\ x=-1 \end{gather}

The restriction is therefore $$x \neq -1$$

\begin{gather} \frac{x+1}{x+2} = \frac{x-3}{x+1}, x\neq-2, x \neq-1 \end{gather}

The LCM of this equation is $$(\,x+2)\,(\,x+1)\,$$. When we multiply by the LCM, we get:

\begin{gather} (\,x+1)\,(\,x+1)\,=(\,x-3)\,(\,x+2)\, \\ x^2+2x+1=x^2-x-6\\ 3x=-7\\ x= -\frac{7}{3} \end{gather}

The value is not one of the restricted. Therefore $$x = -\frac{7}{3}$$.

2. Solve for x: $$\displaystyle \frac{x+3}{x-1} + \frac{x-1}{x+2} =2$$

When restricted:

\begin{gather} \frac{x+3}{x-1} + \frac{x-1}{x+2} =2, x\neq 1, x \neq-2 \end{gather}

LCM: $$(x-1$$$$(x+2)$$

Now we have

\begin{gather} (x+3)(x+2) + (x-1)(x-1) =2(x-1)(x+2) \\ x^2+5x+6+x^2-2x+1=2(x^2+x-2)\\ 2x^2+3x+7=2x^2+2x-4\\ x=-11 \end{gather}

\begin{gather} (x+3)(x+2) + (x-1)(x-1) =\\ 2(x-1)(x+2) \\ x^2+5x+6+x^2-2x+1=2(x^2+x-2)\\ 2x^2+3x+7=2x^2+2x-4\\ x=-11 \end{gather}

3. Solve for x: $$\displaystyle \frac{3-x}{x^2+4x+3} =\frac{2}{x+1}$$

Factorise:

\begin{gather} \frac{3-x}{(x+3)(x+1)} =\frac{2}{x+1} \end{gather}

Restricting: \begin{gather} \frac{3-x}{(x+3)(x+1)} =\frac{2}{x+1}, x \neq-1, x\neq -3 \end{gather}

Restricting: \begin{gather} \frac{3-x}{(x+3)(x+1)} =\frac{2}{x},x \neq-1,\\ x\neq -3 \end{gather}

LCM= $$(x+1)(x+3)$$

Now we have

\begin{gather} 3-x =2(x+3)\\ 3-x=2x+6\\ -3=3x\\ -1=x \end{gather}

One of the restrictions was that $$x\neq -1$$. The answer we got here is $$x = -1$$. Therefore, we have no solution