Algebra

<math>x+3=7</math> <math>x=?</math>

This article/section deals with mathematical concepts appropriate for a student in early high school.

Algebra is a major branch of mathematics that analyzes the relationships between quantities or items. In higher math the principal fields of algebra are linear algebra, which focuses on matrices, and group theory.

Elementary Algebra

While algebra is a very wide-ranging and advanced topic in theoretical mathematics, the word also refers to a very important topic in middle-school and high-school mathematics. This topic is the handling of arithmetical expressions and operations involving quantities that aren't known numbers, but are symbols like x, y, or z. These symbols are generally referred to as "variables" or "unknowns".

Whereas, in elementary arithmetic a typical question might be:

What is the value of 3+4?

in elementary algebra a typical question might be:

If x+4=7, what is the value of x?

Of course this example is so simple as to be practically pointless—why would someone ask a question this way?—the use of algebra opens up sophisticated ways of solving sophisticated problems.

Before continuing, let's see how we actually solved that. We subtracted 4 from each side of the equation. On the left side, subtracting 4 canceled the "+4", so we got "x+4-4", which is, of course, just x. On the right side, we got 7-4, which is 3. We used the principle that

You can apply mathematical operations to something, even if it contains unknowns like "x".

The goal was to get an equation that has just "x" on the left side, with some number on the right. In the original problem, x was on the left side, but hidden with some other mathematical operations. Our job was to strip off those operations, one at a time. Here's a more complicated example:

<math>\left(\frac{\sqrt{x+31}}{3} + 1\right) \times 6 = 18\,</math>

We undo the multiplication, by dividing both sides by 6:

<math>\frac{\sqrt{x+31}}{3} + 1 = 3\,</math>

We undo the addition:

<math>\frac{\sqrt{x+31}}{3} = 2\,</math>

We undo the division:

<math>\sqrt{x+31} = 6\,</math>

We undo the square root:

<math>x+31 = 36\,</math>

We undo the addition:

<math>x = 5\,</math>

As one gets more skillful at these kinds of manipulations, one can handle more complex problems. A typical case involves an equation in which the unknown appears in more than one place. After stripping away as many operations as we can, we might be left with something like:

<math>3 \times x + 4 \times x = 56\,</math>

At this point we use the distributive law, along with the principle that

The various laws of mathematics (commutative, associative, distributive) work even when the items appearing in the expressions are unknowns.

The distributive law tells us that

<math>3 \times x + 4 \times x = (3+4) \times x\,</math>

so we get

<math>(3+4) \times x = 56\,</math>

and hence x=8. You can plug x=8 into <math>3 \times x + 4 \times x\,</math> and see that the result is 24+32, or 56.

Polynomial Manipulations

As one's proficiency increases, one can handle increasingly complicated equations. The next step involves things like this:

<math>\frac{x^2 - 8 x + 15}{x-3} = 7\,</math>

This can be solved if one knows that

<math>(x-3) \times (x-5) = x^2 - 8 x + 15\,</math>

so

<math>\frac{x^2 - 8 x + 15}{x-3} = x-5\,</math>

from which one gets x=12.

This was just a case of the principle that you can perform arithmetic operations even when the expression contain unknowns. In this case we were actually applying the distributive law 3 times.

<math>(x-3) \times (x-5) = (x) \times (x-5) + (-3) \times (x-5) = (x) \times (x) + (x) \times (-5) + (-3) \times (x) + (-3) \times (-5)\,</math>

When we multiply the sum of two things times the sum of two things, we get the sum of four things. This is an extremely common operation, and there is a mnemonic for it—"FOIL"—which stands for "first, outside, inside, last". We multiply the first item in each parenthesized expression; that's x times x. We multiply the "outside" items; that's x times -5. The "inside" items are -3 times x, and the "last" items are -3 times -5.

This same principle applies when there are more than two items in each factor; it just doesn't have a useful mnemonic word. Each term in the first factor (first parenthesized expression) get multiplied by each term in the second factor, and all the products are added up.