Linear Equation With a Fraction

Take a look at this equation

\frac{24}{x}=8.

It may be not look like a linear equation since it’s not in form ax + b = 0 but don’t get fooled. It actually is a linear equation but not in the basic form. Nevertheless we can rearrange the equation.

The question is: can we somehow make the fraction disappear? Wes, we can – we can multiply the fraction by it’s denominator. The result will be just the nominator. So if we have the fraction 24/x, we can multiply it by x and the result will be just the 24:

\frac{24}{x}\cdot x = 24

Of course under the condition that x ≠ 0 because we cannot divide by zero. We can use this knowledge to solve our little equation. We multiply the whole equation by x:

\begin{eqnarray}\frac{24}{x}&=&8\qquad \mbox{multiplying by <i>x</i>}\\\\24&=&8\cdot x\\\\8x&=&24\end{eqnarray}

From now it’s simple. We subtract the 24:

8x = 24
8x − 24 = 0

And after dividing by 8 we get the final result:

8x − 24 = 0
x − 3 = 0
x = 3

The solution of the original equation is x = 3.

More Complicated Fraction

What about this one?

\frac{64}{2x+10}=8

The fraction is little bit more complicated since the denominator doesn’t contain just the unknown variable. But that doesn’t matter, the approach is still the same. Get rid off the fraction by multiplying by the denominator and the show will go on. So first we multiply the equation by 2x + 10:

\begin{eqnarray}\frac{64}{2x+10}&=&8\\\\64&=&8\cdot (2x+10)\end{eqnarray}

We simplify the right part:

64 = 8 · (2x + 10)
64 = 16x + 80

And now we can simply move all parts to the left side:

64 = 16x + 80
16x + 16 = 0
x = − 1

The solution is x = − 1. We can verify it by substituting the result into the equation:

\begin{eqnarray}\frac{64}{2\cdot(-1)+10}&=&8\\\\\frac{64}{-2+10}&=&8\\\\\frac{64}{8}&=&8\\\\8&=&8\end{eqnarray}

Even More Fractions!

Okay, what about this beauty?

\frac{30}{x+2}=\frac{25}{x+1}

Nice one, right? Now we have two fractions. We must get rid off both of the fractions. The approach is still the same: multiply the equation by the denominators. We start with the first one, the x + 2 denominator:

\begin{eqnarray}\frac{30}{x+2}&=&\frac{25}{x+1}\\\\30&=&\frac{25}{x+1}\cdot(x+2)\\\\30&=&\frac{25\cdot(x+2)}{x+1}\end{eqnarray}

Now the second fraction. Still the same, we just multiply the whole equation by x + 1:

\begin{eqnarray}30&=&\frac{25\cdot(x+2)}{x+1}\\\\30\cdot(x+1)&=&25\cdot(x+2)\end{eqnarray}

See? Now it looks like any other linear equation we know. It’s quite simple to compute the result from this form.

30 · (x + 1) = 25 · (x + 2)
30x + 30 = 25x + 50
5x − 20 = 0
5x = 20
x = 4

What does it mean in terms of function graphs? We can draw graphs of the two function f(x) = 30/(x + 2) and g(x) = 25/(x + 1):

Two functions intersects

You can clearly see that the graphs intersects in the point A which has x-coordinate equals to x = 4. Coincidence? I don’t think so! The x-coordinate is exactly the solution of our equation. We’ve found the x for which f(x) = g(x). This is exactly the x-coordinate of the intersection point of these two functions.