Algebra Basics: Refresh Your Math Skills

algebra basicsAlgebra is a branch of math used to help you solve specific kinds of problems faster. It uses unknown values known as variables instead of using only numerical values. The difference between algebra and basic arithmetic is the equation.

You can try this online beginning algebra course to help build up your basic math skills.

Arthimetic equations are those you would likely be familiar with from very early grades: 3 + 4 =? Algebraic equations, on the other hand, involve variables that are represented by letters of the alphabet: x + 4 = 7 where 4 and 7 are constants in the problem.

These kinds of equations require you to balance both sides to find what number x represents. You might recognize the setup of that basic algebraic equation as others you might have done during basic math, but instead of using a letter of the alphabet, the equation was often listed in one of three ways:

__ + 10 = 20

? + 10 = 20

Ο + 10 = 20

The Ο in the last example represents the circle or square that might have been on your worksheets in these classes. Now that you know how algebra compares to other basic skills you have, let’s begin working on how to solve these kinds of equations.

How to Solve Simple Algebraic Equations

Let’s make up an example problem to solve:

x + 17 = 34

We’re going to solve this one by imagining we have a scale we need to balance, and we’re using beads instead of weights. Imagine the bar between the two weight pans as being equivalent to the = sign in an algebraic equation. The pan on the right side of the bar is where the answer would sit, the 34. So, you have 34 beads sitting in the weight pan to the right of the bar. You have a number of other beads in a small bag to add to the other side that has only 17 beads. So, how do you find out how many beads you need to add? You could just add one bead at a time until you finally have 34 on each side, but that would take a lot of time that you probably don’t want to waste just counting beads. Instead of doing that, you could just ask yourself this: If I take 17 away from 34, how much is left? Or, another way to put it would be 34 – 17 = ?, which is a basic arithmetic-style problem. This would leave you with an answer of 17, meaning x = 17. You would then take 17 beads and add them to the other side of the scale.

Solving this problem is written in this algebraic fashion. The bold statements are added for clarification:

Original problem: x + 17 = 34

Subtract 17 from both sides: x + 17 – 17 = 34 – 17

17 – 17 = 0, leaving x alone on one side: +17 – 17 = 34 – 17

34 – 17 = 17, giving you the solution to x: x = 17

As you can see, the answer is the same as the extended explanation above, but the algebraic fashion is clear cut and quick, and you don’t even have to include the bold statements. The whole point of algebra is to allow you to solve large problems quickly. You can take an online algebra 1 course to continue working on more problems like this.

How to Solve More Complex Equations

Here’s an example of a more complex equation:

x + 13 = 5x + 104

As you can see, this equation involves more than just addition. There is the added element of multiplication. Obviously, the simple imagined scale with its system of weights and balances won’t be enough to explain this problem. Rather than messing around with long, drawn-out explanations we can apply the steps we learned earlier to solve this problem algebraically.

Original problem: x + 13 = 5x + 104

Subtract 104 from both sides: x + 13 – 104 = 5x + 104 – 104

104 – 104 = 0, leaving 5x on one side: x + 13 – 104 = 5x + 104 – 104

13 – 104 = -91, giving you a new equation: x – 91 = 5x

Subtract x from both sides: x – x – 91 = 5x – x

x – x = 0, leaving -91 on one side: x – x – 91 = 5x – x

5x – x = 4x, giving you your final equation: -91 = 4x

Divide both sides by 4: -91/4 = 4x/4

4/4 = 1, leaving x alone on one side: -91/4 = 4x/4

-91/4 = -22.75, giving you the solution to x: -22.75 = x

It’s important to note that when it comes to algebra, the variable can be on either side of the equals sign. There is no rule stating that the x must be on the left side. As you can see, you can use a similar format to solve simple and complex algebra equations. If you feel ready, you can take this prep course to test your knowledge and move on to the next step in math.

Remember to Follow the Order of Operations

You’re probably familiar with PEMDAS. Many of you likely memorized it as “Please, excuse my dear aunt Sally” or some other variation of the same phrase. If you’ll remember, each letter in PEMDAS stands for a different operation to tell you the order you should perform operations:

  • P: parenthesis ( )
  • E: exponents x³
  • M: multiplication *
  • D: division /
  • A: addition +
  • S: subtraction –

It’s important to note that multiplication and division can be done in opposite order provided you follow the problem from left to right. The same can also be done for addition and subtraction. Here is an example of solving a problem in the exact order of PEMDAS:

7 + (6 * 5² + 3) – 12/4 * 3 + 4

Solve everything in the parenthesis first: (6 * 5² + 3)

 Follow PEMDAS even within the parenthesis, meaning you would do the exponent first (5² = 25): (6 * 25 + 3)

Do the multiplication within the parenthesis (6 * 25 = 150): (150 + 3)

There is no division so move on to addition: 150 + 3 = 153

There is no subtraction so return to the original problem with your new answer for the area in parenthesis: 7 + 153 – 12/4 * 3 + 4

Exponents were completed within the parenthesis so move on to multiplication (4 * 3 = 12): 7 + 153 – 12/12 + 4

Move on to division (12/12 = 1): 7 + 153 – 1 + 4

Do addition (7 + 153 = 160, 1 + 4 = 5): 160 – 5

Do subtraction to get your answer: 155

 Here is an example of solving the same problem but doing multiplication and division or addition and subtraction in opposite order while following the problem from left to right:

You can begin the problem with the parenthesis already solved: 7 + 153 – 12/4 * 3 + 4

Following the problem left to right, you run into division before multiplication (12/4 = 3): 7 + 153 – 3 * 3 + 4

You will now do multiplication (-3 * 3 = -9): 7 + 153 – 9 + 4

You run into addition first so do that first (7 + 153 = 160): 160 – 9 + 4

Subtraction is next when following left to right so do that first (160 – 9 = 151): 151 + 4

Do the last bit of addition to get the same answer as before: 155

If you’re interested in doing more advanced algebraic problems, you can take a course in advanced algebra. Just remember to keep PEMDAS in mind whenever solving equations. If you’re interested in just doing extra work at home, there are worksheets available that you can print out.