Q&A for How to Solve One Step Equations

An extraneous solution refers to a result that may appear valid when substituting a certain value for X in an equation, but it often leads to a situation where the expression becomes indeterminate, such as division by zero. When working with the original (parent) function, extraneous solutions may not be immediately apparent, as simplification steps can eliminate them. However, upon revisiting the original function, plugging in an extraneous solution may render the function undefined or result in a zero-over-zero scenario. In essence, extraneous solutions are deemed irrelevant as they don't contribute valid answers to the problem, given that zero over zero or undefined values are not considered part of the solution set.

Divide both sides by 8.7 so that x is isolated. You get x = 5.17.

Let some variable (such as x) represent the quantity you're looking for. Then express in mathematical terms the conditions set forth in the problem. For example, if you're told, "three times a number plus 30 equals 45. What's the number?" you'd let x be the unknown number. Then "3x" is "three times the number," and "3x + 30" is "three times the number plus 30." Finally, "three times a number plus 30 equals 45" is expressed as "3x + 30 = 45." When you solve that equation you find that x = 5.

Take your answer and plug it into the original equation. If the equation is true with your value in it, you know your answer is correct. If not, check your work.

Add 8 to both sides of the equation: 6b = 24. Now divide both sides by 6: b = 4. See Solve Two-Step Algebraic Equations: A Straightforward Guide .