xaktly | Algebra | ax + b = c

Algebra practice 1

Solving problems of the type   $ax + b = c$

This is one of the simplest kinds of algebra problems and a good introduction to solving for x. Here's an example:

$$4x + 5 = 7$$

To solve it, we need to work on it until it's in the form $x = c,$ where c is our solution. To do that, consider that x has two things "stuck" to it that we'll have to remove – but by the rules. The first is the added 5 and the second is the multiplied 4.

The easiest thing to get rid of is the 5 (pick the low-hanging fruit). Do that by subtracting it from both sides (because it's added on the left and we use inverse operations).

\begin{align} & 4x + 5 = 7 \\ &\underline{\phantom{00} -5 -5}\\ &\phantom{000} 4x = 7 \end{align}

Now we just need to move the 4 over by division:

$$\frac{4x}{4} = \frac{7}{4} \longrightarrow \bf x = \frac{7}{4}$$

You can practice problems of this type below. Follow the steps and enter your answer as an integer a fraction of integers, like 3/4, or a decimal number like 1.55. Do as many problems as you need to get good at these. You'll form a solid foundation for what comes next.

Next we'll add some rational fractions like $\frac{2}{5}$ to complicate things a little more. Fractions are your friends!

Inverse operations

In this section, we'll refer often to inverse operations. Inverse operations are opposite, and one can be used to undo the action of the other.

• Addition and subtraction are inverse operations.

• Multiplication and division are inverse operations. Other Algebra practice problems

There are a number of these pages you can use for algebra practice. Just pick the rough type of problem you need to work on.

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:   $ax^2 + \frac{b}{c} = \frac{d}{e}$  xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2012, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.