# Math Refresher — Basic Algebra

Algebra is all about keeping things “balanced.” You’ve probably been taught: whatever you do to one side of the equation, you must also do to the other side. Think of an equation as a balance, with the \(=\) being the balance point.

In the following discussion, \(a, b, c,\) and \(d\) are variables. That means that they can represent any number — they can *vary.*

Many times, you are told to “solve for \(x\)” (or some other variable). What this means is to put the requested variable on one side of the equation, with everything else on the other.

Here, I want to teach you a new way of thinking about basic Algebra.

## Addition and Subtraction

**Definition:**Additive (or subtractive) terms are terms in an equation that are joined together by \(+\) or \(–\) operators.

Here’s an example:

$$a + b = c + d$$

Now, subtract \(b\) from both sides

$$a = c + d – b$$

Notice that when \(b\) (an additive term) crosses the \(=,\) it changes sign (\(+\) to \(–\) and vice versa). Now, subtract \(d\) from both sides

$$a – d = c – b$$

Again, when an additive term crosses the \(=,\) it changes sign.

Now, start over with the original equation

$$a + b = c + d$$

then solve for \(d.\) That means to isolate \(d\) on one side of the \(=\) and everything else on the other side. So, to isolate \(d,\) we need to move the \(c\) to the other side. Instead of thinking “subtract \(c\) from both sides,” think “push \(c\) across the \(=\) and change its sign.” Then we have

$$a + b – c = d$$

and here we have solved for \(d.\)

**Examples.**

- Solve for \(a\)

$$a + 4 = b + 7$$

Push the \(4\) across the \(=\) and change its sign

$$a = b + 7 – 4$$

or

$$a = b + 3 .$$

- Solve for \(a\)

$$a + 3 – b = 3c + d$$

Push everything but the \(a\) term to the right side, and changing signs gives,

$$a = 3c + d – 3 + b .$$

## Multiplication and Division

**Definition:**Multiplication terms are terms in an equation that are multiplied (or divided) with other terms.

Here’s an example:

$$ab = cd$$

now, divide both sides by \(b\)

$$a = \frac{cd}{b}$$

Notice that \(b\) was in the numerator, and is now in the denominator (top goes to bottom). Then, divide both sides by \(d\)

$$\frac{a}{d} = \frac{c}{b}$$

Again, top goes to bottom.

If you need to take a multiplied term across the \(=\) sign, bottom goes to top; top goes to bottom. In other words, when you need to take a multiplied term across the \(=,\) simply move it “up” or “down,” as needed. Example: using the previous equation,

$$\frac{a}{d} = \frac{c}{b}$$

solve for \(b.\) Another way to think about what this means: numerator \(=\) numerator, and denominator \(=\) denominator. So, it’s prefectly legal to “flip” both sides, giving

$$\frac{d}{a} = \frac{b}{c}$$

Then, to isolate \(b,\) we need to move \(c\) across the \(=.\) Remember, bottom goes to top, giving

$$\frac{cd}{a} = b$$

Also notice that we could do two operations at once. Since top goes to bottom and bottom goes to top, you can swap terms — numerator on one side and denominator on the other. For example, say we now want to solve for \(a.\) Since \(a\) is in the denominator on one side and \(b\) is in the numerator on the other side, we can just swap them, and in one step, we have

$$\frac{cd}{b} = a$$

and we have solved for \(a.\)

**Examples.**

- Solve for \(a\)

$$ \frac{a}{2} = \frac{b}{4}$$

Push the \(2\) across; bottom goes to top, giving

$$a = \frac{2b}{4}$$

which simplifies to

$$a = \frac{b}{2} .$$

- Solve for \(b\)

$$\frac{2ab}{c} = d + 4$$

push the \(c\) across; bottom goes to top

$$ 2ab = c(d + 4)$$

then push the \(2a\) across; top goes to bottom, giving

$$b = \frac{c(d + 4)}{2a} .$$

## Fractions

### Adding

When adding fractions —

*do not*find a common denominator! Cross-multiply on top, direct multiply on the bottom. Like this:

$$\frac{a}{b} + \frac{c}{d} = \frac{ad + cb}{bd} \qquad {\rm done!}$$

Now reduce if possible — process done, easy and quick.

### Subtracting

Same with subtracting fractions — cross multiply with a minus sign in the middle — like this:

$$\frac{a}{b} – \frac{c}{d} = \frac{ad – cb}{bd} \qquad {\rm done!}$$

Now reduce if possible — process done, easy and quick.

### Multiplying

When multiplying fractions, it’s numerator \(\times\) numerator over denominator \(\times\) denominator.

$$\frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd}$$

### Dividing

When dividing fractions — invert and multiply — like so

$$\frac{ \ \frac{a}{\ b\ } \ }{ \ \frac{\ c\ }{d} \ } = \frac{a}{b} \times \frac{d}{c} = \frac{ad}{bc}$$

What if we only have a fraction on top? That is, a fraction divided by a single variable, like this

$$\frac{ \ \frac{a}{b} \ }{ c }$$

Remember that we can divide a lone variable by 1 to “make it a fraction” — like so: \(c = c/1.\) Now divide “fractions” as above, and we get

$$\frac{ \ \frac{a}{b} \ }{ \frac{ c }{ 1 } } = \frac{a}{b} \times \frac{1}{c} = \frac{a}{bc}$$

**Shortcut.** Think of it this way: a denominator on top moves to the bottom and is multiplied with the original denominator.

In summary,

$$\frac{ \ \frac{a}{b} \ }{ c } = \frac{a}{bc}$$

Similarly, what if we have a lone variable divide by a fraction? Following the process from above

$$\frac{ a }{ \ \frac{b}{c} \ } = \frac{a}{1} \times \frac{c}{b} = \frac{ac}{b}$$

**Shortcut.** Think of it this way: a denominator on bottom moves to the top and is multiplied with the original numerator.

In summary,

$$\frac{ a }{ \ \frac{b}{c} \ } = \frac{ac}{b}$$

**Examples.**

- Solve for \(a\)

$$ab – bc + cd – da = 4$$

What we are given has all the \(a\) terms on one side, so just push across the other terms (and change their signs)

$$ab – da = 4 + bc – cd$$

Now factor out \(a\)

$$a (b – d) = 4 + bc – cd$$

and the multiplier of \(a\) moves from top to bottom, giving the answer

$$a = \frac{4 + bc – cd}{b – d}$$

- Solve for \(a\)

$$abc + cd = ac$$

collect the \(a\) terms on one side — change their sign as they cross \(=\)

$$abc – ac = -cd$$

Now factor out the \(a\) terms

$$a (bc – c) = -cd$$

and the multiplier of \(a\) moves from top to bottom, giving the answer

$$a = \frac{-cd}{bc – c}$$

Since each of the 3 terms on the right has a \(c,\) we can simplify by cancelling a \(c\) out of each term, like so

$$a = \frac{-d}{b – 1}$$

This is correct, but it’s usually good to leave your answer in*standard form.*Here that means we want the “leading term” to be positive, so multiply the right side by \(-1/-1\) like so

$$a = \frac{d}{1 – b}$$

## Comments

Math Refresher — Basic Algebra— No Comments