Algebrarules.com

The most useful rules of basic algebra,
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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

1
``` a(b+c) = ab+ac ```

This is the distributive property of multiplication. If you're multiplying something with a sum of two or more other terms, you can distribute your multiplication to each of the terms. This can be helpful for simplifying problems.

```3(4+5) = 27 = 3*4 + 3*5```
2
``` a\left({b \over c}\right) = {ab \over c} ```

Any multiplication or division of the numerator of a fraction applies to the fraction as a whole, and vice versa: if you need to multiply a fraction, multiply the numerator and your aim is achieved

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

Because of the reciprocal relationship of a fraction's numerator and 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

```\left({1 \over 5}\right)2 = {1 \over 10} = {1 \over (2*5)}```
4
``` {a \over ({b \over c})} = {ac \over b} ```

In the same fashion as the above rule, dividing the denominator of a fraction has the same effect as multiplying the numerator

```{1 \over ({6 \over 2})} = {1 \over 3} = {(1*2) \over 6}```
5
``` \left({a \over b}\right)+\left({c \over d}\right) = {(ad+bc) \over bd} ```

This follows from the fact that you can find a common denominator between two fractions by multiplying the each fraction by the other's denominator, like so:

```{3 \over 5}+{1 \over 3} = {(3*3) \over (5*3)} + {(1*5) \over (3*5)} = {9 \over 15} + {5 \over 15} = {14 \over 15} ``` or, written as in the above rule: ``` {3 \over 5}+{1 \over 3} = {(3*3) \over (5*3)} + {(1*5) \over (3*5)} = {(1*5)+(3*3) \over (3*5)}```
6
``` \left({a \over b}\right)-\left({c \over d}\right) = {(ad-bc) \over bd} ```

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

``````
7
``` {(a-b) \over (c-d)} = {(b-a) \over (d-c)} ```

Reversing the order of a subtraction operation creates the inverse result: 5-3 = 2; 3-5 = -2. A consequence of this is that for a fraction in which both the numerator and denominator contain a subtraction (ie, they each contain a negative term), if both those subtractions are reversed, the value of the fraction remains the same. This is because if the sign of both the numerator and denominator of a fraction is changed, the fraction's value remains the same.

```{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} ``` or ``` {-3 \over -4} = 0.75 = {3 \over 4}```
8
``` {(a+b) \over c} = {a \over c} + {b \over c} ```

This is simply a result of the fact that two fractions with common denominators can be added by adding the numerators and leaving the denominator unchanged. It is intuitively simple: 1 third plus 2 thirds is 3 thirds. The reverse is also true: 3 thirds minus 2 thirds equals 1 third.

```{(1+2) \over 4} = {3 \over 4} = {1 \over 4} + {2 \over 4}```
9
``` {ab+ac \over a} = b+c, a \neq 0 ```

Division can be thought of as the inverse of multiplication: if ``{a \over b} = c`` then ``b*c = a``. This means that the fraction ``ca/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 to where we started. This rule is an extension of that fact. Since we know that ``{ab+ac \over a} = {ab \over a} + {ac \over a}`` (see rule 8), and based on the above we can see that ``{ab \over a} = b`` and ``{ac \over a} = c``, from that we have our result: ``b+c``. There's a caveat here, of course: this won't work if ``a`` is zero, since division by zero is VERBOTEN!

```{(5*4)+(5*2) \over 5} = {(20 + 10) \over 5} = {30 \over 5} = 6 = 4+2```
10
``` {({a \over b})\over({c \over d})} = {ad \over bc} ```

This is a combination of rule 3 and rule 4.

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

Algebra Rules for Exponents

11
``` a^ma^n = a^{m+n} ```

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```
12
``` (a^n)^m = a^{nm} ```

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}```
13
``` (ab)^n = a^nb^n ```

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```
14
``` a^{-n} = {1\over a^n} ```

It might seem nonsensical 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}`` is the multiplicative inverse or reciprocal of ``a^n``: ``{1 \over a^n}``

```2^4*2^{-2} = 16*{1\over2^2} = 16*{1\over 4} = 4 = 2^2 = 2^{4 + -2}```
15
``` \left({a \over b}\right)^{-n} = \left({b\over a}\right)^n ```

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```
16
``` \left({a \over b}\right)^{n} = {a^n\over b^n} ```

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}```
17
``` {a^n\over a^m} = a^{n-m} ```

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} ```
18
``` a^0 = 1, a \neq 0 ```

This one I can't explain. However, it makes the other rules work in the case of an exponent of zero, so there it is.

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

Algebra Rules for Radicals

19
``` \sqrt[n]{a} = a^{1 \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 = 8 16^{1\over4}*16^{1\over2} = \sqrt[4]{16}*\sqrt[2]{16} = 2*4 = 8```
20
``` \sqrt[n]{ab} = \sqrt[n]{a}\sqrt[n]{b} ```

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}```
21
``` \sqrt[m]{\sqrt[n]{a}} = \sqrt[nm]{a} ```

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}```
22
``` \sqrt[n]{a \over b} = {\sqrt[n]{a} \over \sqrt[n]{b}} ```

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}}```
23
``` \sqrt[n]{a^n} = a, ```if n is odd;``` \sqrt[n]{a^n} = |a|, ```if n is even``` ```

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[2]{-27} = -3```

A little bit about algebrarules.com


Algebra rules is a collaboration between Automathic & Photosynthesis.


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


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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