A linear equation is any equation where the highest power of the unknown, which I shall call x, is 1.  To illustrate more clearly with a few examples:

x+1 = 4; 10x = 3; x = 18 – 4x are three examples of linear equations

x² + 2 = 2x and x³ = 8 are not linear equations because there are x’s that are raised to a higher power than 1.

A linear equation with 1 variable is the simplest type to solve.  There is 1 equation and 1 unknown, which means that the unknown can always be determined.  To solve such an equation, you need to rearrange the equation to have like terms on either side of the equal sign.  Put another way, you are trying to isolate x (or whatever the variable is called) on one side of the equation.

For example, if 2x = 234, to isolate x, we have to divide the entire equation by 2.  Doing this, we get x = 117.

If there are x’s and numbers on either side of the equal sign, we add and subtract values to isolate x on one side.  Suppose 2x – 17 = 18 – 3x

The first thing we could do is to add 17 to both sides to get: 2x – 17 + 17 = 18 – 3x + 17

This reduces to 2x = 35 – 3x

Now, we need to have all the x’s on one side so we add 3x to both sides to get: 2x + 3x = 35 – 3x + 3x

This reduces to 5x = 35

Dividing by 5 on both sides, we get x = 7

What I just went through was a fairly simple algebraic equation.  The questions on the GMAT will look more complicated but you are essentially doing the same thing: manipulating both sides of the equation in the same way to isolate x.  Let’s try a practice problem from Grockit.

To tackle this question, we multiply both sides by 2+3/x to get 3 = 2+3/x. .  (This is also known as cross multiplying where in general if a/b-c/d,

then ad = bc

To simplify 3 = 2+3/x,

we multiply the entire equation by x to get 3x = 2x + 3.  This leaves you with a much simpler equation that you already know how to solve.

What’s a little trickier than manipulating algebraic equations is translating a word problem into an algebraic equation.  Here’s another practice problem:

Jack and his brother are sharing a monster piece of licorice that is 28 inches long. Since Jack is older, his share is 8 inches longer than his brother’s. How long, in inches, is Jack’s brother’s piece?

The way to solve this problem is to let something be x.  Here’s what happens if we let Jack’s piece be x inches.

Jack’s piece = x inches

Jack’s brother’s piece = x – 8 inches

Total length of licorice = Jack’s piece + Jack’s brother’s piece = 28 = x + (x-8)

This means that x = 18 inches.  But remember that the question wants the length of Jack’s brother’s piece, which we have defined as x – 8.  So the correct answer is 10 inches.

Here’s what happens if we let Jack’s brother’s piece be x inches.

Jack’s brother’s piece = x inches

Jack’s piece = x + 8 inches

Total length of licorice = 28 = x + (x+8) and we determine that x = 10.  In this case, since we have already defined Jack’s brother’s piece to be x, there is no further step we need to take.

In general, here are a few things to keep in mind.

• if there is only one unknown, you only need one equation to determine the value of the unknown
• in dealing with algebraic equations, remember that anything you do to one side (be it adding, subtracting, multiplying or dividing) you need to do to this other side too.
• in dealing with word problems, define something to be x and see if you can define other things in terms of x only.  (For example, in the question about licorice, you would not want to let Jack’s piece be x inches and his brother’s be y inches)  Don’t introduce unnecessary variables if it can be expressed in terms of an existing variable.