Solving Equations with Algebraic Fractions

A taxi home costs twenty-four pounds, split evenly between everyone who gets in. Two more people jump in, and your share drops by two pounds. So how many of you were in that taxi to begin with? Call that number x. Your first share was twenty-four over x. Your new share was twenty-four over x plus two, and the drop was two pounds. So twenty-four over x, minus twenty-four over x plus two, equals two.

This is video eight of eight in our chapter on quadratic and simultaneous equations. If factorising a quadratic feels shaky, go back to M A one two, oh one, solving x squared plus b x plus c by factorising.

There is an x under a fraction bar there. Your usual moves slide straight off it. So here is the method, in three steps. Step one. Clear it. Kill every fraction before you solve anything. Step two. Solve it. What is left will be an equation you already know. Step three. Check the bottom. No answer may make a denominator zero. Step one has a reason. You cannot collect like terms over different denominators. So multiply both sides by a number every denominator divides into exactly. Both sides get the same treatment, so the equation stays balanced, and every fraction cancels. That number is the lowest common multiple of the denominators. Try it. Solve x over four, plus x plus two over three, equals three. Four and three both divide into twelve, so twelve is the one you want. Every term of this equation gets multiplied by twelve. That plain three on the right has no fraction near it. Does it stay three, become twelve, or become thirty-six? Take a moment and pick one. I'll wait. It becomes thirty-six. That three is a term like any other. Skip one term, and the two sides stop being equal. That is the mistake to watch for. Twelve over four leaves three x. Twelve over three leaves four lots of x plus two. So the equation reads three x, plus four lots of x plus two, equals thirty-six. Step two, and it is easy now. Expand: three x plus four x plus eight equals thirty-six. So seven x is twenty-eight, and x is four. Step three is quick here. Four and three are plain numbers, so neither could ever be zero.

Now for the harder version, where x sits in the denominator. The same three steps still work. Solve three over x plus one, equals two over x minus two. Neither of those denominators divides into the other. So what do you multiply both sides by? x plus one, x minus two, or both together? Have a think about that one. I'll wait. Both together. That product is the smallest expression both denominators divide into. On the left, x plus one cancels, leaving three lots of x minus two. On the right, x minus two cancels, leaving two lots of x plus one. People call that shortcut cross-multiplying. It only works with one fraction on each side. Step two. Three x minus six equals two x plus two, so x equals eight. Step three, and this time it matters. Eight plus one is nine. Eight minus two is six. Neither is zero, so x equals eight stands. Dividing by zero means nothing, so any value that empties a denominator has to go.

Back to the taxi, with all three steps in your hands. Step one. The denominators are x and x plus two, so multiply every term by x times x plus two. The first term leaves twenty-four lots of x plus two. The second leaves twenty-four x. And the lonely two on the right becomes two x times x plus two. That right-hand term is the one people forget. Forget it and the equation breaks. Step two. Expand: twenty-four x plus forty-eight, minus twenty-four x, equals two x squared plus four x. The x terms on the left cancel, leaving forty-eight equals two x squared plus four x. Rearrange and halve it: x squared plus two x minus twenty-four equals zero. Solving that belongs to the video on factorising x squared plus b x plus c. In brief: x plus six, times x minus four, equals zero. So x is four, or minus six. There are no minus six people in a taxi. And x was the number of people. Four people. Twenty-four split four ways is six pounds each. Split six ways it is four. There is your two-pound drop. One first move turned an equation you could not touch into one you already knew.

Two things worth knowing before one of these turns up in an exam. First, solve means give the value or values of x. A line of working with a fraction still in it is not a solved equation. Second, step three is not politeness. Here is a question where it bites. Solve x over x minus three, equals three over x minus three. Multiply both sides by x minus three, and x equals three. That looks finished. But three makes both original denominators zero. So it is not a solution, and this equation has none. So write your answer down, and then check every denominator you started with, every single time.

Your turn now, with the whole method in your hands. Clear the fractions here. Two over x, plus three over x plus one, equals two. Do not solve it. Multiply every term through, and write down what is left. Pause here and work it out. I'll wait. Multiply every term by x times x plus one. You get two lots of x plus one, plus three x, equals two x times x plus one. Tidy it up: two x squared minus three x minus two equals zero. No fraction in sight. If a fraction survived your first step, one term did not get multiplied.

Let's pull those three steps back together. Clear it. Find the lowest common multiple of every denominator, and multiply every term by it. Solve it. The fractions have gone, so what is left is something you already handle. Check the bottom. Put your answer back into the original denominators, and make sure none is zero.

That completes our chapter on quadratic and simultaneous equations. The next chapter is sequences and series.

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