The following answer assumes that you are trying to solve a linear equation with one variable such as 31 = x + 4 + 2x

1) Combine any like terms that are on the same side. 31 = 3x + 4.

2) Use the Addition Property of Equality to "move" the variables to one side and the numbers to the other. In this case, add a -4 to both sides of the equation to get 31 -4 = 3x + 4 - 4 which yields 27 = 3x.

3) Use the Multiplication Property Of Equality to "separate" the variable from its coefficient. To do this, multiply both sides by the multiplicative inverse. In this case, multiply both sides by 1/3 to get 27(1/3) = 3x(1/3) which hields 9 = x.

4) Check Your Answer - Check the mathematical calculations that you completed to solve the equation. Put the answer back into the equation to see if "it works."

If your equation is part of a word problem, make sure that you have set up the word problem correctly. Also, make sure that your answer makes sense. If you are solving for something "real" such as bags of cookies, etc., make sure your answer is not a negative number.

You should know that there are many forms of equations, each form having a particular way of solving.

Considering as example the linear equation `ax + b = 0` , you should solve it by isolating the term that contains x to the left side such that:

`ax = -b => x = -b/a`

Considering as example an equation that contains square root and the unknown is under the square root, you should solve the equation by raising to the square both sides of equal sign such that:

`sqrt(ax+b) = c => (sqrt(ax+b))^2 = c^2 => ax + b = c^2 => ax = c^2 - b => x = (c^2 - b)/a`

Considering as example a quadratic equation `ax^2 + bx + c = 0` , you should use quadratic formula, `x_(1,2) = (-b+-sqrt(b^2-4ac))/(2a)` , to find solutions.

The exponential and logarithmic equations may be solved using the properties of functions, exponential and logarithmic identities and substitutions that help for an exponential or logarithmic equation to be solved using algebraic methods.

Trigonometric equations may be solved using trigonometric identities and substitutions.

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