Equations with Algebraic Fractions
Speed is distance divided by time, so the moment you ask "how fast must I go to make this trip
in exactly an hour?" the unknown lands underneath a fraction. Equations like that look nasty —
but a single move clears every fraction at once.
An equation like \dfrac{x}{3} + \dfrac{x}{4} = 7 looks like trouble.
Two different denominators, an x buried in each fraction — where do
you even start? But there is a single, almost magical move that makes the whole thing collapse
into something easy: multiply everything by a common denominator, and every
fraction vanishes at once.
Multiply each term of \dfrac{x}{3} + \dfrac{x}{4} = 7 by
12 (a number both 3 and
4 divide into) and watch:
12\cdot\frac{x}{3} + 12\cdot\frac{x}{4} = 12\cdot 7 \;\Longrightarrow\; 4x + 3x = 84
The fractions are gone. What's left, 7x = 84, is a plain
linear equation you can already solve: x = 12. That one move —
clear the fractions first — turns a scary equation into an easy one. Let's make it a reliable
method.
When an equation has fractions in it, the trick is to get rid of the
fractions first. Multiply every term on both sides by the
common denominator — the same idea you used when
adding and subtracting algebraic fractions,
but here it clears the equation in one stroke. Take:
\frac{x}{2} + \frac{x}{3} = 5
The denominators are 2 and 3, so a
common denominator is 6. Multiply every term by
6:
6 \cdot \frac{x}{2} + 6 \cdot \frac{x}{3} = 6 \cdot 5
Each fraction cancels cleanly — that is the whole point of the common denominator — leaving
a tidy equation with no fractions at all:
3x + 2x = 30
Now it is an ordinary linear equation. Collect like terms and solve as usual, just like an
equation with
unknowns on both sides:
5x = 30 \quad\Rightarrow\quad x = 6
The recipe, every time: (1) find the lowest common denominator of all the
fractions; (2) multiply every term — both sides, fractions and whole numbers
alike — by it; (3) solve the fraction-free equation that's left.
See it solved
Step through the same solution one line at a time. Watch the fractions disappear the moment
every term is multiplied by 6, then the rest is routine.
Worked example: the unknown in the denominator
Sometimes x hides underneath the line. Solve:
\frac{6}{x} = 3
The denominator to clear is x itself. Multiply both sides by
x:
x \cdot \frac{6}{x} = 3 \cdot x \;\Longrightarrow\; 6 = 3x \;\Longrightarrow\; x = 2
Always check when x was in a denominator: here
x = 2 makes the original denominator 2,
not 0, so it's valid — and \tfrac{6}{2} = 3
✓.
Worked example: an algebraic denominator
When a whole expression sits underneath, the same move works — multiply by that expression.
This is exactly cross-multiplying. Solve:
\frac{2}{x+1} = 3
Multiply both sides by (x+1) to clear it:
2 = 3(x+1) \;\Longrightarrow\; 2 = 3x + 3 \;\Longrightarrow\; 3x = -1 \;\Longrightarrow\; x = -\tfrac{1}{3}
Check the denominator isn't zero: x + 1 = -\tfrac{1}{3} + 1 = \tfrac{2}{3},
which is fine — so x = -\tfrac{1}{3} is a genuine solution. Notice
the shortcut: \dfrac{2}{x+1} = \dfrac{3}{1} cross-multiplies straight
to 2 \cdot 1 = 3 \cdot (x+1).
Clearing fractions before solving is the same move you make when
rearranging formulae
that contain fractions: multiply through to tidy up first, then carry on.
Clearing fractions is powerful, but two slips catch people out every time:
-
Multiply EVERY term — not just the fractions. In
\dfrac{x}{3} + \dfrac{x}{4} = 7, the lonely
7 on the right must be multiplied by
12 too, giving 84. Multiply only the
two fractions and forget the 7, and the equation is now
unbalanced — every later step is wrong. If you multiply one term by the common
denominator, you multiply all of them, both sides.
-
When x was in a denominator, CHECK it isn't zero.
A "solution" that makes any denominator 0 is not allowed — you
can't divide by zero, so that value was never really in the equation. For example, solving
\dfrac{1}{x-2} = \dfrac{x}{x-2} seems to give
x = 1… but if a method ever handed you
x = 2, you'd throw it out, because
x - 2 = 0 there. Always test your answer in the
original denominators.
Everything, it turns out. Clearing fractions is the move that unlocks a whole family of
real-world puzzles built on rates. The classic: one tap fills a bath in
3 hours, another in 6 hours — how long
together? In one hour they fill \tfrac{1}{3} and
\tfrac{1}{6} of the bath, so the equation is
\dfrac{t}{3} + \dfrac{t}{6} = 1. Multiply through by
6: 2t + t = 6, so
t = 2 hours. The famous "two people painting a fence together"
problem is exactly the same equation with a different story.
And it doesn't stop at baths. Physics is full of these. The thin-lens
equation for cameras and glasses is
\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}. Two resistors in
parallel combine as \dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}.
Every one of these is a rational equation — and the very first thing an engineer or physicist
does is multiply through to clear the fractions, turning it into an ordinary linear
or quadratic equation you already know how to crack.
See it explained
Sal Khan solves an equation with the unknown in a denominator by multiplying both sides to
clear the fractions.