The most useful rules of basic algebra,
free, simple, & intuitively organized

Howdy! Here are a few very handy rules of algebra. These basic rules are useful for everything from figuring out your gas mileage to acing your next math test — or even solving equations from the far reaches of theoretical physics. Happy calculating!

Algebra Rules for Arithmetic

a(b+c)=ab+aca(b+c) = ab+ac

Multiplication can be distributed across addition.

If you're multiplying something with a sum of two or more other values, you can distribute the multiplication to each of the values, then sum the result.

```3(4+5) = 3*4 + 3*5 = 27```
a(bc)=abca \left( {b \over c} \right) = { ab \over c }

Multiplying the numerator is the same as multiplying the fraction.

A multiplication or division of the numerator of a fraction affect the fraction as a whole (and vice versa). If you need to multiply a fraction, multiplying the numerator does the job.

```6\left({1 \over 3}\right) = {(6*1) \over 3} = 2```
(ac)b=abc{\left({a \over c}\right) \over b} = {a \over bc}

Dividing the numerator equals multiplying the denominator.

If you divide the numerator by a particular number, it has the same effect on the fraction's overall value as if you multiply the denominator by that same number. A division above equals a multiplication below.

```{ \left({1 \over 5}\right) \over 2} = {1 \over 10} = {1 \over (2*5)}```
``` {a \over \left({b \over c}\right)} = {ac \over b} ```

Dividing the denominator equals multiplying the numerator.

In the same fashion as the above rule, dividing the denominator of a fraction has the same effect as multiplying the numerator. A fraction below equals a multiplication above.

```{1 \over \left({6 \over 2}\right)} = {(1*2) \over 6} = {1 \over 3}```
``` \left({a \over b}\right)+\left({c \over d}\right) = {(ad+bc) \over bd} ```

Fractions can be summed by multiplying across between numerators and denominators, and multiplying denominators for a common denominator.

If top and bottom of a fraction are both multiplied by the same number, the fraction remains unchanged. So, we can sum two fractions by first multiplying each fraction's numerator and denominator with the other fraction's denominator. Then, since both the fractions now have the same denominator (the product of the two denominators), we can combine them into one fraction, with the sum in the numerator.

```{3 \over 5}+{1 \over 3} = {(3*3) \over (5*3)} + {(1*5) \over (3*5)} = {(1*5)+(3*3) \over (3*5)} = {14 \over 15}```
``` \left({a \over b}\right)-\left({c \over d}\right) = {(ad-bc) \over bd} ```

How to turn a subtraction of two fractions into one fraction.

This is another version of rule 5, but for subtraction of two fractions, rather than addition.

```{3 \over 5}-{1 \over 3} = {(3*3) \over (5*3)} - {(1*5) \over (3*5)} = {(3*3)-(1*5) \over (3*5)} = {9-5 \over 15} = {4 \over 15}```
``` {(a-b) \over (c-d)} = {(b-a) \over (d-c)} ```

Reversing a subtraction in both the numerator and the denominator of a fraction leaves the fraction's value unchanged.

Reversing a subtraction gives the inverse result: ``5-3 = 2; 3-5 = -2``. In a fraction, if both the numerator and the denominator are inverted, the value of the fraction stays the same. So, if we reverse a subtraction in both the numerator and denominator, the value of the fraction is unchanged.

```{3-5 \over 2-1} = {-2 \over 1} = {5-3 \over 1-2} = {2 \over -1} = -2``` or ```{-3 \over -4} = 0.75 = {3 \over 4}```
``` {(a+b) \over c} = {a \over c} + {b \over c} ```

Fractions with common denominators can be combined.

Two fractions with common denominators can be added by adding the numerators and leaving the denominator unchanged. Going the other direction, we can also break apart a fraction with an addition in the numerator into two fractions (each with the common denominator).

```{(1+2) \over 4} = {3 \over 4} = {1 \over 4} + {2 \over 4}```
``` {ac+bc \over c} = a+b ```

Multiplication and division (by the same number) cancel each other out.

Division is the inverse of multiplication: if ``{a \over b} = c`` then ``b*c = a``. This means that the fraction ``{ac \over c}`` is equal to ``a``, since we are multiplying ``a`` by ``c`` and then immediately dividing it by ``c`` again, which puts us right back where we started. Since we know that ``{ac+bc \over c} = {ac \over c} + {bc \over c}`` (see rule 8), and based on the above we can see that ``{ac \over c} = a`` and ``{bc \over c} = b``, we have our result: ``a+b``.

```{(4*5)+(2*5) \over 5} = {4*5 \over 5}+{2*5 \over 5} = 4+2```
``` {\left({a \over c}\right) \over \left({b \over d}\right)} = {ad \over bc} ```

If the numerator and denominator of a fraction are both fractions, it can be converted into a fraction of two multiplications.

Combining rules 3 and 4, we can multiply the denominator of the bottom fraction with the numerator of the upper fraction, which gives the combined numerator, and cancels the denominator of the lower fraction; we can then multiply the denominator of the upper fraction with the numerator of the lower fraction, to give the combined denominator and cancel the denominator of the upper fraction.

```{\left({4 \over 5}\right) \over \left({2\over1}\right)} = {0.8 \over 2} = {4 \over 10} = {4*1 \over 2*5}```

Algebra Rules for Exponents

``` a^ma^n = a^{m+n} ```

The product of two powers with the same base is equal to that base raised to the sum of the two exponents.

As with many rules related to exponents, writing out the exponents as multiplications makes it obvious why the rule is true

```3^3*3^2 = (3*3*3)(3*3) = 3^5```
``` (a^n)^m = a^{nm} ```

A number raised to a power raised to a power is equal to that number raised to the product of the two exponents.

Like the previous rule, this one can be demonstrated simply by expanding the exponents out into a series of multiplications

```(4^2)^3 = (4*4)^3 = (4*4)(4*4)(4*4) = 4*4*4*4*4*4 = 4^6 = 4^{2*3}```
``` (ab)^n = a^nb^n ```

Convert a multiplication with an exponent into the product of two factors each raised to the exponent.

Thanks to the commutative property of multiplication, any series of multiplications can be rearranged without changing its value. This means that we can take a multiplication raised to a power and rearrange the resulting series of multiplications to make two exponents

```(4*5)^3 = (4*5)(4*5)(4*5) = 4*5*4*5*4*5 = 4*4*4*5*5*5 = 4^3*5^3```
``` a^{-n} = {1\over a^n} ```

The result of a negative exponent is the inverse of the same positive exponent.

It might seem odd to have a negative exponent (since you can't multiply something by itself a negative number of times). However, if we take a closer look at the rule ``a^na^m = a^{n+m}`` we can see that it implies that ``a^{-n}`` must equal ``{1 \over a^n}``, the multiplicative inverse or reciprocal of ``a^n``.

This becomes clear looking at the ``a^{n+m}`` side of the equation from rule 11. What happens if ``m`` is negative? Obviously, this will reduce the combined value of the exponent (for example, ``2^{4-2} = 2^2``). What does this mean for the left hand side of the ``a^na^m = a^{n+m}`` equation? It means that the value of, for example, ``2^4`` must be reduced to ``2^2`` when it is multiplied by ``2^{-2}``. If, as this rule states, ``a^{-n} = {1 \over a^n}``, this works out perfectly: ``2^4 * 2^{-2} = 2^4 * {1 \over 2^2} = 16 * {1 \over 4} = 4 = 2^2 = 2^{4-2}``

```2^{-2} = {1\over2^2} = {1\over 4}```
``` \left({a \over b}\right)^{-n} = \left({b\over a}\right)^n ```

A fraction raised to a negative exponent equals the inverse of the fraction raised to a positive exponent.

The reciprocal of a fraction is the fraction turned on its head: the reciprocal of ``{2 \over 3}`` is ``{3 \over 2}``. We know from the previous rule that ``a^{-n}`` is the reciprocal of ``a^n``, so we can simply convert the fraction to its reciprocal by exchanging the numerator and denominator, and then the exponent becomes positive. Positivity is such a nice thing!

```\left({1 \over 2} \right)^{-2} = {1 \over (1/2)*(1/2)} = {1 \over 1/4} = 4 = 2^2 = \left({2\over1} \right)^2```
``` \left({a \over b}\right)^{n} = {a^n\over b^n} ```

A fraction with an exponent is equal to the same fraction with the exponent on the numerator and denominator.

This looks weird at first, but the reasons behind it are pretty simple. If we were paying attention when someone told us how to multiply fractions (this is doubtful, but we'll continue anyhow) we will remember that to multiply two fractions you simply multiply the numerators with each other and multiply the denominators with each other to get the resulting fraction. This rule follows from that fact.

```\left(\frac{3}{4}\right)^2 = \left(\frac{3}{4}\right)\left(\frac{3}{4}\right) = {3*3\over4*4} = {3^2 \over 4^2}```
``` {a^n\over a^m} = a^{n-m} ```

If the top and bottom of a fraction are both exponents with the same base, the fraction is equal to the base raised to the numerator exponent minus the denominator exponent.

This one is very simple. Since division is the inverse of multiplication, multiplying a number by itself a few times and then dividing it by itself multiplied a few time is the same as just multiplying it by itself a few less times.

```{4^4\over4^2} = {4*4*4*4 \over 4*4} = {256 \over 16} = { 16 \over 1} = 16 = 4^2 = 4^{4-2} ```
``` a^0 = 1 ```

Anything raised to the power of zero is equal to 1.

This rule may seem arbitrary, but it is necessary in order to maintain consistency with other properties of exponents. Consider the rule ``a^na^m = a^{n+m}``. What happens if ``m = 0``? The right hand side of the equation will be ``a^{n+0}``, or ``a^n``. This means that in the left hand side, ``a^n`` has to be multiplied by the value of ``a^0``, but remain unchanged. The only way for this to be the case is if ``a^0 = 1``. (For some discussion of the peculiar case of ``0^0`` and why it should (probably) equal ``1``, see this article.)

```123^0 = 1 = \pi^0 = 1 = (everything)^0 = 1```

Algebra Rules for Radicals

``` a^{1 \over n} = \sqrt[n]{a} ```

The nth root of a number is the same as the number raised to one over n.

As in some of the exponent properties, this rule is not an intuitive extension of the typical meaning of an exponent. Nevertheless, it fits with the all-important exponent rule ``a^na^m = a^{n+m}``.

```16^{{1\over2}+{1\over4}} = 16^{{3\over4}} = \left(16^{1\over4}\right)^3 = \left(\sqrt[4]{16}\right)^3 = 2^3 = 16^{1\over4}*16^{1\over2} = \sqrt[4]{16}*\sqrt[2]{16} = 2*4 = 8```
``` \sqrt[n]{ab} = \sqrt[n]{a}\sqrt[n]{b} ```

The root of a multiplication equals the product of the roots of its factors.

If ``x = \sqrt{a}`` and ``y = \sqrt{b}`` then ``\sqrt{ab} = \sqrt{x^2*y^2}`` If we write out the multiplication, this turns into ``\sqrt{x*x*y*y}``. Thanks to the commutative property of multiplication, we can rearrange the Xs and Ys and get ``\sqrt{x*y*x*y} = \sqrt{(x*y)(x*y)} = \sqrt{(x*y)^2} = x*y = \sqrt{a}*\sqrt{b}``

```\sqrt{4*9} = \sqrt{36} = 6 = 2*3 = \sqrt{4}*\sqrt{9}```
``` \sqrt[m]{\sqrt[n]{a}} = \sqrt[nm]{a} ```

Converting a root of a root into a single root.

Once again, by working backwards from the value of these two expressions we can see why they are equal. If ``\sqrt[m]{\sqrt[n]{a}} = x``, then we can construct ``a`` out of combinations of ``x`` and see how the whole equation works. To make things simple, we'll start with given values of ``m`` and ``n``. If ``\sqrt[2]{\sqrt[3]{a}} = x``, then ``x = \sqrt[2]{x*x}``, which also means that ``\sqrt[2]{\sqrt[3]{(x*x)*(x*x)*(x*x)}} = x``. So ``a = (x*x)*(x*x)*(x*x) = x^6 = x^{mn}``. And happily, ``\sqrt[mn]{x^{mn}} = x`` by definition, so we have ``\sqrt[m]{\sqrt[n]{a}} = x = \sqrt[mn]{x^{mn}}``. Our example is a specific case where ``m = 2`` and ``n = 3``, but since ``a`` will always be equal to ``x^{mn}``, the equation holds regardless of the values of ``m`` and ``n``

```\sqrt[2]{\sqrt[3]{729}} = \sqrt[2]{9} = 3 = \sqrt[6]{729}```
``` \sqrt[n]{a \over b} = {\sqrt[n]{a} \over \sqrt[n]{b}} ```

Convert the root of a fraction into a fraction of two roots.

To see how this works, we can use a similar trick to the rule above. If ``x = \sqrt{a} `` and ``y = \sqrt{b}``, then: ``\sqrt{{a \over b}} = \sqrt{{x*x \over y*y}} = \sqrt{{x \over y}*{x \over y}} = \sqrt{{x \over y}^2} = {x \over y} = {\sqrt{a} \over \sqrt{b}}``

```\sqrt[3]{8 \over 1} = 2 = {2\over 1} = {\sqrt[3]{8} \over \sqrt[3]{1}}```
``` \tiny\text{if n is odd:} \newline \normalsize \sqrt[n]{a^n} = a \newline \tiny\text{if n is even:} \newline \normalsize \sqrt[n]{a^n} = |a| ```

The nth root of the nth power of a number equals that number, or its absolute value.

If ``a`` is a positive number, then ``\sqrt[n]{a^n}`` will always equal ``a``. However, if ``a`` is negative, then the result will be positive if ``n`` is even, but it will be negative if ``n`` is odd. This comes from the fact that multiplying a negative number an even number of times always produces a positive result; an odd number of multiplications will produce a negative result. (This has the interesting side effect that there are no real numbers that are even-numbered roots of a negative number.)

`````` Positive number, even root/exponent: ```\sqrt[2]{3^2} = \sqrt{9} = 3``` Positive number, odd root/exponent: ```\sqrt[3]{3^3} = \sqrt[3]{27} = 3``` Negative number, even root/exponent: ```\sqrt[2]{-3^2} = \sqrt[2]{-3*-3} = \sqrt[2]{9} = 3``` Negative number, odd root/exponent: ```\sqrt[3]{-3^3} = \sqrt[3]{-3*-3*-3} = \sqrt[3]{-27} = -3```

A little bit about

Algebra rules is a project by two of the folks who run The Autodidacts.

A couple of autodidact math enthusiasts, we were looking for all the rules of basic algebra concisely presented in one place. We couldn’t find such a place, so we made

These simple rules — applied with a pinch of imagination and a dash of arithmetic — can divide, conquer, and solve just about any practical algebra problem.

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