Math equations can intimidate students because they're full of symbols and seem disconnected from the real world. But mathematics -- and in particular algebra -- is a subject that every middle and high school student will have to contend with, and along with it comes equations. They can contain multiple variables, single variables or other complex terms. Like levels of a video game, the types of equations must all be mastered before students can graduate successfully.

## Prepare

Before you even move your first variable, check to make sure you are following the directions. Most equations are straightforward. For example, you have no doubt that for the equation x + 3 = 21, they want you to solve for "x." In equations with more than one variable, such as 2x + 4y = 10, you'll have to double-check which variable you must find a value for. In addition, look for another equation that will allow you to use the substitution method to pinpoint the values.

## Gather Like Terms

Put all "like" terms together. For example, in the equation 5x + 3 = 8x - 6, move the 5x to the right side and the -6 to the left side, which leaves 3 + 6 = 8x - 5x; this simplifies into 9 = 3x. Be certain that you do the same thing to both sides of the equation, For instance, if you forget to add 6 to the right side (effectively canceling it out), you'll end up with the wrong equation, 9 = 3x - 6.

## Isolate and Solve for the Variable

Most equations for secondary school students require you to solve for only one variable. So, for the equation 5x + 7 = 10x - 3, you would move the like variables to one side and the constants to one side, leaving 10 = 5x. Then simply solve for "x." To do this, you have to isolate the "x." Because 5 is being multiplied by the variable "x," you have to divide both sides by 5. So the answer is x = 2 (10/5 = 5x/5).

## Double-check Your Answer

Once you have solved an equation, plug the final answer into the original equation. For instance, if you put 2 into the equation 5x + 7 = 10x - 3, you'll get 5*2 + 7 = 10*2 - 3. This simplifies to 10 + 7 = 20 - 3 and then 17 = 17. Because this is a true statement, you know you have the right answer. If the left and right side of the original equation are not equal, you've made a mistake somewhere along the line.

## Quadratic Equations

Quadratic equations are more difficult than the standard algebraic equation in that they have an "x" and "y" variable, as well as exponents. A typical quadratic equation may look like this: ax^2 + bx + c = 0. With this equation, you cannot simply isolate like variables and solve. Instead, you must factor to solve for "x." For example, consider x^2 + 5x + 6 = 0. You can factor it to (x + 3)(x + 2) = 0. Then it is simply a matter of solving each equation for x. In this case, the answers are x = -3 and x = -2.

## Systems of Equations

When two or more equations need to be solved at once, you usually don't have the luxury of solving for only one variable. Instead, you'll have two unknowns. The key to remember is that you have to have as many equations as you do unknowns, or you can't solve the equations simultaneously. Now, assuming you have two unknowns and two equations, you can solve one of your equations for "x," and then substitute this answer into your other equation. This will lead to an answer for your "y" variable. For example, suppose the two equations are 3x + y = 9 and 4x + 2y = 16. Pick one equation and solve for one of the variables. Solving for "y" in the first equation yields y = 9 - 3x. Now plug this "y" into the second equation: 4x + 2(9 - 3x) = 16. Simplify: 4x + 18 - 6x = 16, or 2x = 2. It's obvious that x = 1. Plug this one into the equation y = 9 - 3(1), so y = 6. So the variables "x" and "y" are 1 and 6.