Simplifying fractions was introduced as far back as grade 4 or 5. When you get to the FET phase (grade 10 -12), 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.
- Ensure your denominators are factorised.
- Restrict your equation.
- Find the lowest common denominator (LCD) or lowest common multiple (LCM).
- Multiply the expression with the LCD or LCM with the expression
- Solve for x
- Check your answer if it is not the restricted value.
Now tet us try a few examples.
Solve for \(\normalsize x\): \( \displaystyle \frac{x+1}{x+2} = \frac{x-3}{x+1} \)
Following the steps above.
We do not need to factorise the denominators. It is simplified.
When we restrict equations, we need to find the value that will make the denominator to beero.
$$ \normalsize \begin{align} x+2 &=0\\ x &=-2 \end{align} $$The restriction is therefore \(\normalsize x \neq -2\)
The second restriction:
$$ \normalsize \begin{align} x+1=0\\ x=-1 \end{align} $$The restriction is therefore \(\normalsize x \neq -1\). The equation with restrictions becomes:
$${\frac{x+1}{x+2} = \frac{x-3}{x+1}, x\neq-2, x \neq-1}$$LCM is \(\normalsize (x+2)(x+1)\). We will then multiply both sides of the equation by the LCM.
$$ \normalsize \begin{align*} (\,x+1)\,(\,x+1)\, &=(\,x-3)\,(\,x+2)\, \\ x^2+2x+1&=x^2-x-6\\ 3x&=-7\\ x &= -\frac{7}{3} \end{align*} $$The value is not one of the restricted. Therefore \(x = -\frac{7}{3}\).
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
$$ \normalsize \begin{align*} (x+3)(x+2) + (x-1)(x-1) &= 2(x-1)(x+2) \\ x^2+5x+6+x^2-2x+1&= 2(x^2+x-2)\\ 2x^2+3x+7 &= 2x^2+2x-4\\ x &= -11 \end{align*} $$Solve for x: \( \displaystyle \frac{3-x}{x^2+4x+3} =\frac{2}{x+1}\)
Factorise:
$${\normalsize \frac{3-x}{(x+3)(x+1)} =\frac{2}{x+1}}$$Restricting the equation:
$${\normalsize\frac{3-x}{(x+3)(x+1)} =\frac{2}{x+1}, x \neq-1, x\neq -3}$$LCM= \(\normalsize (x+1)(x+3)\)
Multiplying both sides of the equation by the LCM, we get:
$$ \normalsize \begin{align*} 3-x &=2(x+3)\\ 3-x&=2x+6\\ -3&=3x\\ -1&=x \end{align*} $$One of the restrictions was that \(x\neq -1\). The answer we got here is \(x = -1\). Therefore, we have no solution