Linear Equations


term 3x;   1;     -5x;     -5;   3(2x-1);    \frac{2x-5}{3}
expression  3x+1;     \frac{2x-5}{3}-5x
equation  3x+1=22;    \frac{2x-5}{3}=9
relation  y=3x+1

A term is a number or an unknown, or a product (multiplication) of numbers/unknowns. A term might also be the product of expressions, such as (x-3)(x+2).

An expression is the addition (or subtraction) of one or more terms.

An equation puts a condition on an expression. For example 3x+1=22. Although we call x an unknown, in an equation we can usually figure out what it is. That’s the process of solving an equation.

A relation is when we introduce a second unknown. For example, y=3x+1. Now, should we give the x a value, then the y will have a value too and vice versa. Suppose we say that x=10. That means that y=31. If you draw a graph of a linear expression, you get a straight line, which is where the name ‘linear’ comes from. In relations, unknowns are generally referred to as variables.

Identifying a Solution to a Linear Equation by Substitution

As an introduction to algebra, an equation might look like this:

    \[3 \times \Box + 5 = 26\]

The idea is that you’re supposed to find the number that goes in the box to make the calculation correct.

The big idea of algebra is to use a letter to represent the uknown number until you figure it out (in an equation) or give it a value (in a relation). That is:

    \[3 \times x + 5 = 26\]

Which we shorten to:


In the following applet, you are given three possible solutions. Which number makes the calculation correct? Figure it out by replacing the x with one of the numbers given until you find one that fits.

Skill 4 Solving linear equal a constant

These can often be solved by common sense. Two (or more) algebraic steps will get there too. Try the ‘common sense’ approach first – ‘what number makes sense instead of the x‘?

Skill 5 Solving Linear equal Linear Equation

It can be helpful to follow some predetermined steps to solve these.

  • Expand any brackets, simplify each side if necessary;
  • Add or subtract the x term from one side so that there is an x term on one side only;
  • Continue as before, in skill 4.

There could be a better approach than these three steps – it really depends on the equation you’re solving. This is why it is good to keep thinking about the objective (figuring out what x works) and not get too hung up on one method.

When there are fractions

Albegraic problems with fractions show up in the solutions to problems in topics 2, 4, 5, 6 and 7. Invest some time here, and keep coming back until this is something you’ve mastered!

Example Solve with Fractions


You got this?

1.  Try the Test Out applet

2. This one looks a little different..


Check your solution yourself before clicking here (Khan Academy video) to see one way to work it out.

3. Here’s a set of 10: More (mostly) Linear Equations and their solutons: More (mostly) linear solutions

Unit 1 Menu

Next: Project with Connect 4 Counters