### What is algebra?

$$3 + 7 = \underline{\hspace{1cm}}.$$

But once you've written down "10", you've only solved that one single problem, and there are an infinite number of such problems. But, ... they all have the same pattern.

What if there was a way to solve all problems with the same pattern at once? Then the numbers in the actual problem of interest could be substituted into the solved pattern, and out would pop the solution.

For example, what if we wanted to find any solution to the equation with the pattern

$$a + 7 = b$$

With a simple algebraic rearrangement (rearrangement is mostly what algebra is), we can have a = b - 7 (we simply subtracted 7 from both sides of the equation), so that for any b, we can always find exactly the right a. That solves an infinite set of equations of the type a = b - 7. Just plug in anything for b (which we'll call the independent variable because we can make it anything we want) and a is easy to calculate - just subtract 7 from b. We'll call a the dependent variable because it depends on our choice of b.

That's algebra.

#### A quick note about variables

We once held a math haiku contest at our shool. One of the winning haikus read:

I don't not like math,

I just liked it better before

We'll be using letters to represent numbers now. Get used to it. Letters are just stand-ins for numbers, and we'll treat them just as we would numbers. As you work through your education in math and science, it will be of great value to learn to do all of these algebraic manipulations with symbols (letters) rather than numbers, plugging numbers in for variables only once all the algebra is done. This will be especially important when you come to chemistry or physics, where units (Kilograms, Newtons, Ohms &c.) will be treated just like variables.

Thanks to René DesCartes, we have the following convention for letters and the kinds of numbers they represent. This convention isn't always followed to the letter (pun intended), but it will be most of the time in the problems you encounter:

#### Letter conventions in algebra

$x, \; y, \; z$   represent unknown numbers

$a, \; b, \; c$   represent known quantities, e.g. "constants"

$i, \; j, \; k, \; l, \; m$ and $n$   usually represent known integers, usually "counters" or "index variables"

All other letters are multi-purpose.

The most important thing about algebra is that it's all about maintaining balance across the equal sign. We'll mostly be working with equations, the root of which is "equal", which means to be the same as in number or amount. An equation signifies that the expressions (mixtures of numbers and/or variables) on the left and right sides of the equal sign are the same, or represent the same thing.

In algebra, we want to isolate or "solve for" one variable. Usually that means manipulating the equation, one step at a time, so that only that variable remains on one side of the equal sign (which side doesn't matter). We can do anything we want to an equation, provided that we do the same thing on both sides of the equal sign. Think of a balancing scale like the ones shown below. Equal means that the scales are balanced. Put a 1Kg mass on both pans → balance. Add 1 gram to each side → still in balance. Subtract 1 gram from only one side→ imbalance.

Here is a scale with the same algebraic expression, a + b, on both sides.

The next set of scales shows that if we start out with equivalent expressions on either side of the equal sign, we can subtract (or add) the same thing from both sides and still have a true statement: a = a.

The same is true of addition (remember that subtraction is just addition of the negative, so this will work with subtraction, too). Likewise, we can divide each side by the same number, and division is just multiplication by the reciprocal, so it works with division, too:

When we try to do different things to the expressions on each side of an equation, or when we do something to just one side, we create an imbalance. We've created an untrue statement.

The simplest way to state this rule is that it's OK to do anything to both sides of an equation, as long as what you're doing is to the whole side of the equation.

Another important concept in algebra is that multiplication (or division) by the number 1 doesn't change anything. It's the one thing we can do to only one side of an equation. That sounds useless at first, but it's actually one of the most important techniques for solving algebraic equations. And remember, anything divided by itself, like x/x, is one: 3/3 = a/a = (2y)/(2y) = 1.

#### Essence of algebra

Maintain Balance: Anything done to one side of an equation must also be done to the other side.

### Inverse operations are crucial

In working to isolate a variable on one side of an equation, the concept of inverse operations is very important.

Addition and subtraction are inverse operations - one "undoes" the other. If we add 3 to x, then subtract 3, we end up where we started, with x.

Here's an example:

Here's another with a variable, a, added to x instead of a number. The idea is just the same:

When the variable of interest (x) is multiplied by a constant, we divide by that constant on both sides to isololate x.

Likewise, when x is divided by a constant (and that's really just multiplied by the reciprocal of the denominator), we multiply both sides by that constant to isolate x:

Finally, when our variable is inside of some function, we will need to apply the inverse of that function to both sides of the equation. You'll learn a lot more about that later, but for now, let's work with squares and square roots – one undoes the other. If we take the square root of something squared, we just get that thing. If we square something inside of a square root, we just "undo" the square root. Here's an example:

### Practice problems — 1

Use inverse operations to isolate x in each of these equations. Do the easiest rearrangements first (usually addition & subtraction), then then do division & multiplication.

 1 $x - 3 = y + 4$ Solution Our x has a three attached to it by a subtraction operation. We'll get rid of it by adding 3 to both sides: \begin{align} x &= y + 4 + 3 \\ x &= y + 7 \end{align} 2 $x - 7 = y - 7$ Solution The -7 on both sides can be removed because it's OK to remove the same thing from both sides of the = sign. Or you can just add 7 to both sides. Either way you get $$x = y$$ 3 $x - 2 = x + y - 4$ Solution This equation has an x on both sides, so if we subtract x from both sides, x is gone ... nothing to solve for any more. $$-2 = y - 4$$ We can solve for y, though. Why not ... Add 4 to both sides to get $$y = 2$$ 4 $x + 27 = y + 32$ Solution Subtract the 27 from both sides to get x by itself. \begin{align} x &= y + 32 - 27 \\ x &= y + 5 \end{align} 5 $\frac{x}{4} + 3 = y$ Solution Our x is divided by 4 and has a 3 added to it. The easiest thing to do first is to subtract 3 from both sides. $$\frac{x}{4} = y + 3$$ Now multiply both sides by 4. Remember that the 4 will multiply (y + 3) on the right - use parentheses. \begin{align} x &= 4(y + 3) \\ x &= 4y + 12 \end{align} 6 $9x - 12 = 6y$ Solution The easiest thing to move away from the x (which also has a 9 stuck to it) is the 12. Do so by adding 12 to both sides: $$9x = 6y + 12$$ Now divide both sides by 9: \begin{align} x &= \frac{6y + 12}{9} \\ x &= \frac{6y}{9} + \frac{12}{9} \\ \\ x &= \frac{2y}{3} + \frac{4}{3} \\ \\ x &= \frac{2y + 4}{3} \end{align} The last few steps were just to simplify as much as possible.

 7 $-12x + 9 = -4$ Solution first move the 9 to the right by subtracting 9 from both sides, then divide both sides by -12 to free up the x on the left. \begin{align} -12x &= -4 - 9 \\ -12x &= -13 \\ \\ x &= \frac{13}{12} \end{align} 8 $x^2 - 3 = 6$ Solution First move the 3 over to the right by adding 3 to both sides. Then get x from x2 by taking the square root of both sides: \begin{align} x^2 &= 6 + 3 \\ x^2 &= 9 \\ \sqrt{x^2} &= ±\sqrt{9} \\ x &= ± 3 \end{align} Note that the ± has to be there because both 32 and (-3)2 are 9. So this equation has two solutions. 9 $\frac{x^2}{2} + 3 = 5$ Solution First get rid of the 3 by subtracting 3 from both sides. Then multiply both sides by 2, remembering to use ( ) on the right, and finally, take the square root of both sides. \begin{align} \frac{x^2}{2} &= 5 - 3 \\ \frac{x^2}{2} &= 2 \\ x^2 &= 4 \\ \sqrt{x^2} &= \sqrt{4} \\ x &= ±2 \end{align} The ± needs to be there because $2^2 = 4$ AND (-2)^2 = 4. 10 $\frac{x^2}{9} + 3y = 5y$ Solution First move the 3y to the right by subtracting 3y from both sides. Then multiply both sides by 9, and finally, take the square root of both sides. \begin{align} \frac{x^2}{9} &= 5y - 3y \\ \frac{x^2}{9} &= 2y \\ x^2 &= 18y \\ \sqrt{x^2} &= \sqrt{18y} \\ x &= ±\sqrt{18y} \\ x &= ± 3\sqrt{2y} \end{align} In the last step, notice that 18 = 9·3, and 9 is a perfect square. Always try to reduce roots like this if you can. 11 $-2x + 3y = 4$ Solution First subtract 3y from both sides, then divide both sides by -2. \begin{align} -2x &= 4 - 3y \\ x &= \frac{4 - 3y}{-2} \\ x &= \frac{3y - 4}{2} \; \; \text{ or} \\ x &= \frac{3}{2} y - 2 \end{align} 12 $\frac{1}{9} x + 4y = 5y$ Solution First move the 4y to the right by subtracting 4y from both sides, then multiply both sides by 9. \begin{align} \frac{x}{9} &= y \\ x &= 9y \end{align}

#### Interactive algebra practice problems

A number of interactive algebra practice pages are available. They are divided into the classes of problems below. They might help you practice. It's important to get very good at these simple algebra problems.

Type 1:   $ax + by = c$

Type 2:   $\frac{a}{b}x + \frac{c}{d} = \frac{e}{f}$

Type 3:   $\frac{a}{x} + b = c$

Type 4:   $\frac{a}{x} + \frac{b}{c} = \frac{d}{e}$

Type 5:   $ax^2 + b = c$

Type 6:   $ax^2 + \frac{b}{c} = \frac{d}{e}$

### Properties of multiplication and addition

When you first learn about the commutative and distributive properties of addition and multiplication, it can seem a bit out of the blue, not useful at all. Let me assure you that having a good grasp on these properties can make algebra much easier for you.

#### Addition and multiplication are commutative

You already know that $2 + 3 = 3 + 2 = 5.$ Order doesn't matter at all when we add numbers. That's handy for many reasons. First, consider the sum

$$80 + 19 + 20 + 11 = \underline{\hspace{1cm}}$$

It's not too cumbersome just to add these numbers from left to right, but consider how this rearrangement makes things easier:

$$(80 + 20) + (19 + 11) = \underline{\hspace{1cm}}$$

Now $80 + 20 = 100.$ That's easy. And $19 + 11$ is $19 + 1 + 10 = 20 + 10 = 30,$ so our sum is 130. In that last step, I used the associative property of addition as well as the commutative property. It says that we can group parts of a sum together (in parenthesis) however we like. Although this rearrangement may seem trivial to you, consider that it might make a more complicated sum easier to figure out in your head. It will also be of great use in solving many kinds of problems in algebra. In general, we have

\begin{align} a + b &= b + a \\ a + (b + c) &= (a + b) + c \end{align}

Likewise, multiplication is commutative: $a\cdot b = b\cdot a,$ and $a\cdot (b\cdot c) = (a\cdot b)\cdot c.$

#### Multiplication is Distributive

The distributive property is one of the most important properties of number operations you'll encounter. Using it correctly will be crucial for solving algebra problems. There are a few pitfalls, mistakes often made, that you should guard against. Do the exercises here, then check out the pitfalls page for more discussion of what can go wrong. Let's look at an example using just numbers first.

$$2\cdot (3 + 4) = 2\cdot (7) = 14$$

That's pretty straightforward, but notice that the same solution can also be found like this:

$$2\cdot (3 + 4) = 2\cdot 3 + 2\cdot 4 = 6 + 8 = 14$$

In this case, the 2 multiplies the 3 and the 4, and the sum of those products is the answer. We say that the 2 "distributes" over the binomial (3 + 4).

It is very important that you remember that the distributive property can also be used in reverse. It can be used to separate a common factor from two or more terms. This is one of the kinds of operations we refer to collectively as factoring. Here are some examples:

#### Some examples

Equation Property
$xy + xz = x(y + z)$ x multiplies both y and z, so it can be moved out front.
$ax + 2x - 3x = x(a + 2 - 3) = x(a - 1)$ x multiplies all three terms, so it can be moved out front.
$c(y - 1) - 4(y - 1) = (y - 1)(c - 4)$ The binomial (y - 1) appears in both terms. It multiplies c and -4. It can be moved out front to multiply (c - 4).

#### Properties of operations

 $x + y = y + x$ Commutative property of addition $xy = yx$ Commutative property of multiplication $x + (a + b) = (x + a) + b$ Associative property of addition $xab = (xa)b = x(ab)$ Associative property of multiplication $x(a + b) = xa + xb$ Distributive property of multiplication $(a + b)·(x + y) = ax + ay + bx + by$ Distributive property of multiplication (FOIL)

Another way of looking at those properties is this:

For pure multiplication or pure addition, order of operation doesn't matter, nor does how we group things.

For mixed multiplication and addition (remember that means division and subtraction, too), we have to make sure to pay attention to the distributive property

### Practice problems – 2

Use inverse operations to isolate x in each of these equations. Do the easiest rearrangements first (usually addition & subtraction), then then do division & multiplication.

 1 $(x - 3)(x + 3) = 27$ Solution \begin{align} x^2 + 3x - 3x - 9 &= 27 \\ x^2 - 9 &= 27 \\ x^2 &= 18 \\ sqrt{x^2} &= \sqrt{18} \\ x &= ±\sqrt{18} \\ x &= ±\sqrt{9 \cdot 2} \\ x &= ± 3 \sqrt{2} \end{align} 2 $x - 3x + 2 - y = 0$ Solution \begin{align} -2x &= y - 2 \\ \\ x &= -\frac{1}{2} y + 1 \end{align} 3 $5x - 4 = 21$ Solution \begin{align} 5x &= 17 \\ \\ x &= \frac{17}{5} \end{align} 4 $x(x - 1) + x = 4$ Solution \begin{align} x^2 - x + x &= 4 \\ x^2 &= 4 \\ \sqrt{x^2} &= ±\sqrt{4} \\ x &= ±2 \end{align} 5 $3x(x - 2) + 6x = 27$ Solution \begin{align} 3x^2 - 6x + 6x &= 27 \\ 3x^2 &= 27 \\ x^2 &= 9 \\ \sqrt{x^2} &= ±\sqrt{9} \\ x &= ±3 \end{align} 6 $x^2 - (x - 2)(x + 2) = 4$ Solution \begin{align} x^2 - (x^2 + 2x - 2x - 4) &= 4 \\ x^2 - (x^2 - 4) &= 4 \\ x^2 - x^2 + 4 &= 4 \\ 4 &= 4 \end{align} This result means that any value of x at all will satisfy (solve) this equation. 7 $3(x - 2) + 4(x - 2) = x$ Solution \begin{align} 3x - 6 + 4x - 8 &= x \\ 7x - 14 &= x \\ 6x &= 14 \\ x &= \frac{14}{6} \\ x &= \frac{7}{3} \end{align} 8 $x - 6 + 3(x - 2) = 12$ Solution \begin{align} x - 6 + 3x - 6 &= 12 \\ 4x - 12 &= 12 \\ 4x &= 24 \\ x &= 6 \end{align} 9 $9x^2 - 9(x - 3)^2 = 27$ Solution \begin{align} 9x^2 - 9(x^2 - 6x + 9) &= 27 \\ 9x^2 - 9x^2 + 54x - 81 &= 27 \\ 54x &= 108 \\ x &= 2 \end{align} 10 $4(x - 1) - 5(x - 1) = -x$ Solution \begin{align} 4x - 4 - 5x + 5 &= -x \\ -x + 1 &= -x \\ 1 &= 0 \end{align} This is never true, so this equation has no solutions.
 11 $-9x + 3y - z = 0$ Solution \begin{align} -9x &= z - 3y \\ x &= \frac{z - 3y}{-9} \\ x &= \frac{3y - z}{9} \\ x &= \frac{3y}{9} - \frac{z}{9} \\ x &= \frac{y}{3} - \frac{z}{9} \end{align} 12 $6x + 121 = -5x$ Solution \begin{align} 11x + 121 &= 0 \\ 11x &= -121 \\ x &= -11 \end{align} 13 $(x + 5)(x - 5) = 0$ Solution \begin{align} x^2 - 5x + 5x - 25 &= 0 \\ x^2 &= 25 \\ \sqrt{x^2} &= ±\sqrt{25} \\ x &= ±5 \end{align} 14 $(x + a)(x - a) = 0$ Solution \begin{align} x^2 -ax + ax -a^2 &= 0 \\ x^2 - a^2 &= 0 \\ x^2 &= a^2 \\ \sqrt{x^2} &= ±\sqrt{a^2} \\ x &= ±a \end{align} 15 $2(x^2 - 3x + 2) = 2(x - 1)(x + 1)$ Solution \begin{align} 2x^2 - 6x + 4 &= 2(x^2 + x - x - 1) \\ 2x^2 - 6x + 4 &= 2x^2 - 2 \\ -6x + 4 &= -2 \\ -6x &= -6 \\ x &= 1 \end{align} 16 $-4(x - 1) + 4(x - 1) = x$ Solution \begin{align} -4x + 4 + 4x -4 &= x \\ 0 &= x \end{align} 17 $7(x^2 - 3) = 7x^2 - x$ Solution \begin{align} 7x^2 - 21 &= 7x^2 - x \\ -21 &= -x \\ x &= 21 \end{align} 18 $x(x - 2) - 2x(x - 1) + x^2 = x$ Solution \begin{align} x^2 - 2x - 2x^2 + 2x + x^2 &= x \\ 2x^2 - 2x^2 -2x + 2x &= x \\ 0 &= x \end{align} 19 $x^2 = 64$ Solution \begin{align} \sqrt{x^2} &= ±\sqrt{64} \\ x &= ± 8 \end{align} 20 $-4x^2 = 64$ Solution \begin{align} -4x^2 &= 64 \\ x^2 &= -16 \\ \sqrt{x^2} &= ±\sqrt{-16} \\ x &= ±4i \end{align} You might not have any experience with the "imaginary number" i and the square roots of negative numbers. If not, don't worry about this problem. You'll get to it in time.

Expand the following expressions, then collect all terms on the left side (zero on the right). Write the terms, in order of decreasing exponent, from left to right.

 21 $(x - 4)(x + 2) = x$ Solution \begin{align} x^2 + 2x - 4x - 8 &= x \\ x^2 - 2x -8 - x &= 0 \\ x^2 - 3x - 8 &= 0 \end{align} 22 $-(x - 3)^2 = 12$ Solution \begin{align} -(x^2 - 3x - 3x + 9) &= 12 \\ -x^2 + 6x + 6x - 9 -12 &= 0 \\ -x^2 + 12x - 21 &= 0 \\ \end{align} 23 $-(x - 1)(x + 4) - 8 = 0$ Solution \begin{align} -(x^2 + 4x - x - 4) - 8 &= 0 \\ -x^2 - 3x + 4 - 8 &= 0 \\ -x^2 - 3x - 4 &= 0 \end{align} 24 $-9x^2 + 2x = 4x - 15$ Solution \begin{align} -9x^2 + 2x - 4x + 15 &= 0 \\ -9x^2 - 2x + 15 &= 0 \end{align} 25 $2(x - 5)^2 = 5$ Solution \begin{align} 2(x^2 - 5x - 5x + 25) &= 5 \\ 2x^2 - 20x + 50 - 5 &= 0 \\ 2x^2 - 20x + 45 &= 0 \end{align}

 26 $-4(-4x - 4) = 44$ Solution \begin{align} 16x + 16 &= 44 \\ 16x + 16 - 44 &= 0 \\ 16x - 28 &= 0 \end{align} 27 $6x(2x - 5)(x + 1) = 0$ Solution \begin{align} 6x(2x^2 + 2x - 5x - 5) &= 0 \\ 6x(2x^2 - 7x - 5) &= 0 \\ 12x^3 - 42x^2 - 30x &= 0 \end{align} 28 $-3(8x^2 - 4x - 9) = 27$ Solution \begin{align} -24x^2 + 12x + 27 &= 27 \\ -24x^2 + 12x &= 0 \end{align} 29 $-3(8x - 4y + 5z) = 15$ Solution \begin{align} -24x + 12y -15z &= 15 \\ -24x + 12y - 15z - 15 &= 0 \end{align} 30 $4(x - 9 + s) = 31$ Solution \begin{align} 4x - 36 + 4s &= 31 \\ 4x + 4s -36 - 31 &= 0 \\ 4x + 4s - 67 &= 0 \end{align}

### Order of operations: PEMDAS

When a whole sequence of mathematical operations has to be performed, we need to have some convention (a way we all decide it works) for the order in which they are performed, or the precedence of one operation over another. Here are a couple of examples to illustrate the point:

2 + 3·x = ______ ... Does this mean (2 + 3 ) · x, or 2 + (3·x) ?

5 - 3/7 = ______ ... Does this mean (5 - 3)/7 or 5 - (3/7) ?

We have settled on an order of operations that can be remembered with the acronym PEMDAS, for "parenthesis, exponents, multiplication, division, addition and subtraction."

#### P E M D A S

PEMDAS is the correct order of operations that all computers are designed to follow, and you should, too.

1. Evaluate expressions inside Parenthesis first.
2. Perform all Exponent operations.
3. Perform Multiplications and Divisions
4. Finally, perform all Additions and Subtractions.

Now our examples make a little more sense. Your calculator should use the same operation preference, so make sure to keep it in mind when typing in a long expression. Use parenthesis liberally to make sure you get the answer you want from your calculator.

Example What to do
$2 + 3·x = 2 + (3x)$ Do the multiplication first, then the addition
$5 - 3/7 = 5 - (3/7)$ Do the division before the subtraction.
$x \cdot (2x)^2 = x \cdot (4x^2) = 4x^3$ The parenthesis contain 2x (done); do the exponent, giving (2x)2 = 4x2 [remember that (2x)2 means (2x)·(2x) = 2·x·2·x = 2·2·x·x] , then multiply by x.
$2 + (3-5)^2 + 3 · 3 = 2 + (-2)^2 + 9 \\ = 2 + 4 + 9 = 15$ Evaluate what's inside the parenthesis first, then square that. Evaluate the 3·3, then add the resulting three terms.

### Practice problems – 3

Simplify the following expressions as much as you can (use PEMDAS):

 1 $\frac{42}{7} + \frac{8}{4} - 6 \times 3 - 2$ Solution \begin{align} &= 6 + 2 - 18 - 2 \\ &= 8 - 20 \\ &= -12 \end{align} 2 $\left( \frac{3 + (-26)}{3} \right) + 12 \times \frac{1}{2}$ Solution \begin{align} &= \left( \frac{-23}{3} \right) + \frac{12}{2} \\ &= (-7.667) + 6 \\ &= -1.667 \end{align} 3 $5 + 13 \div (6 - 2) \times 3$ Solution \begin{align} &= 5 + \frac{13}{4} \times 3 \\ &= 5 + 3.25(3) \\ &= 5 + 9.75 \\ &= 14.75 \end{align} 4 $30 - 12 \div (6 - 2) + 6 \times 2^3$ Solution \begin{align} &= 30 - \frac{12}{4} + 6(8) \\ &= 30 - 3 + 48 \\ &= 27 + 48 \\ &= 75 \end{align} 5 $(7 - 1) \div 2 + 5 \times (9 - 4)^2$ Solution \begin{align} &= \frac{6}{2} + 5(5^2) \\ &= 3 + 5(25) \\ &= 3 + 125 \\ &= 128 \end{align}

 6 $(2x + 3x^2)^2 - 1$ Solution $$= 4x^2 + 12x^3 + 9x^4 - 1$$ 7 $\left( \frac{2x}{4x^2} \right)^2 - \frac{1}{x}$ Solution \begin{align} &= \left( \frac{1}{2x} \right)^2 - \frac{1}{x} \\ &= \frac{1}{4x^2} - \frac{1}{x} \\ &= \frac{1}{4x^2} - \frac{4x}{4x^2} \\ &= \frac{1 + 4x}{4x^2} \end{align} 8 $10 - 1 \div 3$ Solution \begin{align} &= 10 - \frac{1}{3} \\ &= 9 \frac{2}{3} \end{align} 9 $(10 - 1) \div 3$ Solution $$\frac{9}{3} = 3$$ 10 $(6 + x) \times 3 - 7$ Solution \begin{align} &= 18 + 3x - 7 \\ &= 11 + 3x \end{align}

### Video examples

#### Example 1

Here are six basic problems using the most basic algebra steps.

#### Example 2

Here are six problems on a slightly higher level, including the square of the variable (x2), so square roots are necessary.