Math Help: Intermediate Algebra

How to Succeed in Intermediate Algebra

Ask for Help

If you're struggling with intermediate algebra, you should probably seek math help right away so you don't fall behind. You may want to first seek help by asking your math teacher to meet with you before or after school. Many teachers are willing to meet with students, re-teach lessons and provide extra practice material.

Your school also might have a math lab or tutoring service. Even if they don't provide tutors, asking your teacher for help may lead to a recommendation for a good tutor. It always helps to let your teacher know you're trying to improve.

Simplifying Double Negatives

Often, the trick to solving algebra problems is to break them up into more manageable pieces. When you encounter double negative expressions, like 14 - - 3, the first step is to simplify the operations. This problem has a subtraction symbol followed by a negative number (-3). Two negatives in math work the same way double negatives work in language - they reverse the meaning of the symbols, transforming the equation into addition. If you say, 'I don't not like her,' the two negatives would change the meaning of them both, so you would essentially be saying, 'I like her.' So 14 - - 3 = 14 + 3. The answer is 17.

Mixed Equations

Sometimes you'll encounter problems that have multiple operations, like 5 - + - 4. This problem has two subtraction symbols with an addition symbol stuck in between them. With this type of problem, remember that the addition symbol won't alter the directions given by other symbols. So, if you have a problem with one subtraction symbol followed by an addition mark, like 4 - + 2, you ignore the plus sign and re-write the equation as 4 - 2. When you have 5 - + - 4, you ignore the plus sign, which leaves you with 5 - - 4. Now you can use the double negative rule to change the problem to 5 + 4. The final answer is 9.

Variables

In intermediate algebra, you'll encounter equations with variables, which are letters that stand for unknown numbers. Here's an example: 14 = 2x - 4. The first step is to isolate the variable. In this example, that means we have to add 4 to both sides of the equation, like this:

14 = 2x - 4

14 + 4 = 2x - 4 + 4

18 = 2x

Now, divide both sides by 2, like this:

18 ÷ 2 = 2x ÷ 2

9 = x

Eventually, you'll run into equations with two variables, like 6y = x + 3. To solve this equation in terms of x, you'll follow the same steps as above.

6y = x + 3

6y - 3 = x + 3 - 3

6y - 3 = x