Linear equations with an unknown in the denominator

A linear function that contains some fractions and has some variables in the denominator is solved similarly to a classical linear equation. In the first step, we convert the equation into its basic form and then solve it classically.

When only the denominator is x

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

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

It is obvious that the result of the equation will be x = 2. But how do we solve it formally, in general? We don't like the variable in the denominator, we need to get rid of it because it can't be reasonably manipulated. The question is how do we get rid of x in the denominator.

Imagine a fraction of the form

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

What expression do you need to multiply the fraction by to get rid of the fraction? So that the expression is no longer in the form of a fraction, but in the form of a multiple? By what do you have to multiply one-third (i.e., 1/3) to get some integer? You have to multiply it by the denominator of the fraction (or its multiple). If you multiply a fraction by its denominator, you get a fraction

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

in which you can already truncate the original denominator b to get only the value a. Going back to the one-third example, if you multiply one-third by three, what do you get? You get three thirds, i.e. the number one. What do you get when you multiply one sixth by six? Six sixths, i.e. 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 whole equation by the variable x, which is the value of the denominator. Can we do that? The answer is that we can. Although the value of x could take on zero because we assume that x is from the set of real numbers, the terms of the equation tell us that x≠0, because x is in the denominator of the fraction and we can't divide by zero. So this is an equivalent treatment of the equations.

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

$$\begin{eqnarray} \frac{10}{x}&=&5\\ x\cdot\frac{10}{x}&=&x\cdot5\\ \frac{10x}{x}&=&5x \end{eqnarray}$$

What can we do now? We can truncate x in the fraction. This leaves only 10 from the fraction 10x/x: 10 = 5x. And at this point we can solve it as a classical linear equation. The result 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. But two halves of x are equal to one whole x.

The more complex expression in the denominator

There can be a more complex expression in the denominator 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 to get rid of the fraction. In this case, we will do the same thing. If we multiply the equation by the expression x + 2, then the expression x + 2 will be in the numerator of the fraction and we can then truncate it nicely. We're going to multiply:

$$\begin{eqnarray} \frac{5}{x+2}&=&10\quad/\cdot(x+2)\\ (x+2)\cdot\frac{5}{x+2}&=&(x+2)\cdot10\\ \frac{(x+2)\cdot5}{x+2}&=&10\cdot(x+2)\quad/\mbox{ Shorten } (x+2)\\ 5&=&10\cdot(x+2) \end{eqnarray}$$

Now we've got rid of the fraction, and all that's left to do is multiply the parentheses and calculate the result.

$$\begin{eqnarray} -10x-15&=&0\\ x&=&-\frac32 \end{eqnarray}$$

More fractions

Things can get very complicated if you have multiple fractions in the equation and multiple terms in the denominator. Example:

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

What is the best way to solve this equation? We have two fractions with different denominators, both with the variable x. The solution is to multiply the equation by both denominators. We can do this either sequentially or simultaneously. If we do it sequentially, the calculation will go like this. We start by multiplying (x + 2).

$$\begin{eqnarray} \frac{5}{x+2}&=&\frac{4}{x+3}\quad/\cdot(x+2)\\ \frac{5(x+2)}{x+2}&=&\frac{4(x+2)}{x+3}\\ 5&=&\frac{4(x+2)}{x+3} \end{eqnarray}$$

And at this point, we'll multiply the equation by the second denominator, (x + 3):

$$\begin{eqnarray} 5&=&\frac{4(x+2)}{x+3}\quad/\cdot(x+3)\\ 5(x+3)&=&\frac{4(x+2)(x+3)}{x+3}\\ 5(x+3)&=&4(x+2) \end{eqnarray}$$

Now we just multiply the parentheses and calculate the result:

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

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

$$\begin{eqnarray} \frac{5}{x+2}&=&\frac{4}{x+3}\quad/\cdot(x+2)(x+3)\\ \frac{5(x+2)(x+3)}{x+2}&=&\frac{4(x+2)(x+3)}{x+3}\\ 5(x+3)&=&4(x+2) \end{eqnarray}$$

We can afford to do this because if we multiply the equation by the expression a and then multiply it by the expression b, it is the same as if we multiplied it by the expression a · b.