Solving Basic Algebra

Algebra in any form can often be a pain to deal with, especially when you are first learning the concepts. With basic algebra, once you get the hang of it, it will be a piece of cake. Your basic algebra skills will be used in everything you do later in algebra and further math classes. Much of your basic algebra skills are initially taught prior to high school.

With most math, you think of purely numbers, but algebra throws in letters to represent unknown variables. A variable is just a number that you don’t know.
The most common are x, y, and z, but depending on the problem given you can see nearly any letter (Not e. e is a certain value).

For solving a basic algebra problem, you may need to use multiplication, division, addition, and/or subtraction. Some problems may only require a couple, but some may include each of those operations.

For a problem, you can not always work in a left to right direction. You have know your order of operations. PEMDAS. Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.

Whatever is inside parentheses in the problem must be answered first, then the exponents. After those, you multiply and divide, before you finally add and subtract.

Multiplication and division are two sides of the same coin. They are inverse operations just like addition and subtraction, so if you change their order you should be fine, but you can not swap the order of addition and division for example. If you do, it will change your final answer to be incorrect.


The given problem: 3-104+11 = ?*
This is what happens when you work only from left to right:
3-104+11 = ?
4+11 = ?
-28+11 = ?
-17 = ?

When you correctly follow the order of operations:
3-10*4+11 = ?
3-40+11 = ?
-37+11 = ?
-26 = ?

See how differently things turn out if you don’t follow the order of operations?

Here’s another example.

(8-2) * 4 * 10/2 + 2^2 = X <—the given problem
(6) * 4 * 10/2 + 2^2 = X <—first the parenthesis
6 * 4 * 10/2 + 4 = X <—then exponents
24 * 10/2 + 4 = X <—then multiplication
240/2 + 4 = X
120 + 4 = X <—then divide
124 = X <—add and solve!

Now here’s where things get trickier. The unknown value is not always going to be isolated on one side. Sometimes the problem will be structured like this, 3x + 2 = 32 so you will have to watch both sides. What you do to one side, you have to do to the other.

3x + 2 = 32 <—the given problem
3x + 2 - 2 = 32 - 2 <—subtract 2 from both sides
3x = 30 <—simplify
3x/3 = 30/3 <—divide both sides by 3
x = 10 <—simplify and solve

Here’s a more complicated example.

4X - 24 = 3X/2 + 6 <—the given problem
4X - 30 = 3X/2
8X - 60 = 3X
-60 = -5X
60 = 5X
12 = X

Sometimes you will face a problem with more than one variable, which you will have to solve for one of them. You can only combine Ike terms. Terms with X Can only combine with other X’s, etc.

Solve for Y
6X + 2Y = 14 <—the given problem
2Y = 14 - 6X <— subtract 6X from each side
Y = 7 - 3X <—divide by 2 to solve for Y

Solve for X
6X + 2Y = 14 <—the given problem
6X = 14 - 2Y <— subtract 2Y from each side
X = 14/6 - 2Y/6 <—divide by 6 to isolate X
X = 7/3 - Y/3 <—simplify

This is just the very basics of algebra that you use in all later math. There’s much more to go over, but this is the basics to build from. It can never hurt to look back to remind yourself of these basics. If you need other basic algebra help, feel free to ask, or give your own tips and pointers!


This is so helpful! Thank you!

I used to really like algebra :see_no_evil:

@HomeworkHelp maybe some of you find this helpful as well?

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