Linear equations with the unknown in the denominator

A linear equation that contains some fractions with a variable in the denominator is solved in a similar way to an ordinary linear equation. In the first step we rearrange the equation into the form ax + b = 0, and then we solve it as usual.

When the denominator contains only x

The simplest type of such an equation is one where the denominator contains the variable x and nothing else. An example is the equation:

$$\frac{10}{x}=5$$

It is obvious that the solution is x = 2. But how do we solve it formally, in general? We don’t like having a variable in the denominator, because we can’t sensibly work with it. The question is how to get rid of x in the denominator.

Imagine a fraction of the form

$$\frac{a}{b}$$

What do you have to multiply a fraction by to get rid of the fraction? So that the expression is no longer a quotient but a product? What do you have to multiply one third (that is, 1/3) by to get a whole number? You have to multiply it by the denominator of the fraction (or a multiple of it). If you multiply a fraction by its denominator, you get the fraction

$$\frac{a\cdot b}{b}$$

in which you can now cancel the original denominator b, leaving just the value a. Going back to the example with a third: what do you get when you multiply one third by three? You get three thirds, that is, the number one. What do you get when you multiply one sixth by six? Six sixths, that is, one. What do you get when you multiply three sevenths by seven? You get 3 · 7 = 21 twenty-one sevenths, which is three.

So we simply multiply the equation by the variable x, which is the value of the denominator. Are we allowed to do that? The answer is yes. The value of x could in principle be zero, since we assume that x is a real number, but the restrictions on the equation tell us that x ≠ 0, because x is in the denominator of a fraction and we cannot divide by zero. So this is an equivalent step.

All right, so we multiply the equation by the unknown x. What do we get?

$$\begin{array}{rcl}\frac{10}{x}& =& 5\\ x\cdot \frac{10}{x}& =& x\cdot 5\\ \frac{10x}{x}& =& 5x\end{array}$$

What can we do now? In the fraction we can cancel x. From the fraction 10x/x only 10 remains: 10 = 5x. At this point we can solve it as an ordinary linear equation. The solution is x = 2.

Does this make any intuitive sense? Yes, it does. Imagine an equation of the form:

$$\frac{x}{2}=10$$

What does this linear equation tell us? That half of x equals ten. But if we take two halves of x, they must equal twenty. And two halves of x make one whole x.

A more complicated expression in the denominator

The denominator can contain a more complicated expression than just x. Solve the equation:

$$\frac{5}{x+2}=10$$

How do we solve this equation? How do we get rid of the fraction? In the previous section we multiplied the equation by the denominator, which got rid of the fraction. Here we proceed in exactly the same way. If we multiply the equation by the expression x + 2, then x + 2 ends up in the numerator of the fraction and we can cancel it nicely. Let’s multiply:

$$\begin{array}{rcl}\frac{5}{x+2}& =& 10(\cdot (x+2))\\ (x+2)\cdot \frac{5}{x+2}& =& (x+2)\cdot 10\\ \frac{(x+2)\cdot 5}{x+2}& =& 10\cdot (x+2)(\text{cancel }(x+2))\\ 5& =& 10\cdot (x+2)\end{array}$$

Now we have got rid of the fraction, and all that remains is to expand the brackets and work out the result.

$$\begin{array}{rcl}-10x-15& =& 0\\ x& =& -\frac{3}{2}\end{array}$$

More fractions

The situation can get considerably more complicated if the equation has several fractions and several expressions in the denominators. For example:

$$\frac{5}{x+2}=\frac{4}{x+3}$$

How do we solve this equation? There are two fractions with different denominators, and both contain the variable x. The solution is to multiply the equation by both denominators. We can do this either one at a time or both at once. If we do it one at a time, the working goes like this. We start by multiplying by (x + 2).

$$\begin{array}{rcl}\frac{5}{x+2}& =& \frac{4}{x+3}(\cdot (x+2))\\ \frac{5(x+2)}{x+2}& =& \frac{4(x+2)}{x+3}\\ 5& =& \frac{4(x+2)}{x+3}\end{array}$$

Now we multiply the equation by the second denominator, (x + 3):

$$\begin{array}{rcl}5& =& \frac{4(x+2)}{x+3}(\cdot (x+3))\\ 5(x+3)& =& \frac{4(x+2)(x+3)}{x+3}\\ 5(x+3)& =& 4(x+2)\end{array}$$

Now we just expand the brackets and work out the result:

$$\begin{array}{rcl}5x+15& =& 4x+8\\ x& =& -7\end{array}$$

The second method is to multiply the equation by both denominators at the same time. That is, at the start we multiply the equation by the expression (x + 2)(x + 3). We get:

$$\begin{array}{rcl}\frac{5}{x+2}& =& \frac{4}{x+3}(\cdot (x+2)(x+3))\\ \frac{5(x+2)(x+3)}{x+2}& =& \frac{4(x+2)(x+3)}{x+3}\\ 5(x+3)& =& 4(x+2)\end{array}$$

We can do this because multiplying the equation by an expression a and then by an expression b is the same as multiplying it by a · b.