{"id":16700,"date":"2013-02-04T09:00:59","date_gmt":"2013-02-04T16:00:59","guid":{"rendered":"https:\/\/gmatclub.com\/blog\/?p=16700"},"modified":"2013-01-29T00:00:30","modified_gmt":"2013-01-29T07:00:30","slug":"gmat-math-can-you-divide-by-a-variable","status":"publish","type":"post","link":"https:\/\/gmatclub.com\/blog\/gmat-math-can-you-divide-by-a-variable\/","title":{"rendered":"GMAT Math: Can you divide by a variable?"},"content":{"rendered":"


\n\"\"Question #1<\/strong>: In the equation,<\/p>\n

\"\"<\/a><\/p>\n

can you divide both sides by x?<\/p>\n

Question #2<\/strong>: In the equation (x \u2013 3)(x + 5) = (2x + 1)(x + 5), can you divide both sides by (x + 5)?<\/p>\n

Question #3<\/strong>:<\/p>\n

\"\"<\/a><\/p>\n

Question #4<\/strong>:<\/p>\n

\"\"<\/a><\/p>\n

 <\/p>\n

Dividing by a variable or by an algebraic expression<\/h2>\n

The short answer is:\u00a0NO<\/strong>.<\/p>\n

You see, it's mathematically illegal to divide by zero, and if you don't know the value of the variable, then you could be breaking the law without knowing it.\u00a0 Ask any judge --- not knowing that you're breaking the law generally is not an excuse that holds very well up in court.\u00a0 In much the same way, not knowing whether you are dividing by zero, because you are dividing by an unknown, is just as bad as dividing by zero directly.<\/p>\n

What do you do instead?\u00a0 Well, there are two alternatives.\u00a0 One method is: instead of dividing by the variable, factor it out.\u00a0 For example, with Question #1:<\/p>\n

\"\"<\/a><\/p>\n

If the produce of two or three or more factors equals zero, this means\u00a0one\u00a0of the factors\u00a0must\u00a0equal zero.\u00a0 Here, either x = 0 or (x + 3) = 0, which leads to solutions of x = 0 and x = \u20133.<\/p>\n

The second method is two break the problem into two cases, one in which the variable or expression does equal zero, and one in which it doesn't.\u00a0 Treat the two cases separate.\u00a0 For example, in Question #2:<\/p>\n

Case I: let's consider the case in which (x \u2013 5) = 0.\u00a0 Well, if this equaled zero, the equation would be true, so this is a solution.\u00a0 One solution is x = 5.<\/p>\n

Case II: let's consider the case in which (x \u2013 5) \u2260 0, that is, the case in which x \u2260 5.\u00a0 Well, now we are guaranteed that\u00a0 (x \u2013 5) \u2260 0 is not equal to zero, so dividing both sides by this expression is now perfectly legal, and this leads to the simple equation x \u2013 3 = 2x + 1, which has a solution of x = \u20134.\u00a0 Thus, the overall solutions to this problem are x = 5 and x = \u20134.<\/p>\n

 <\/p>\n

Canceling a variable or expression<\/h2>\n

Similarly, the blanket answer to the cancelling question is also,\u00a0NO<\/strong>!, for the same reason.\u00a0 If there is any possibility that your variable or expression equals zero, then cancelling would be a 100% illegal activity.<\/p>\n

For Question #3 --- for all values of x other than x = 0, for the entire continuous infinity of numbers on the number line excluding that solitary value, yes, the fraction 2x\/5x would equal 2\/5.\u00a0 BUT, when x = 0, that statement is no longer true --- it is not even false --- it is profoundly meaningless.\u00a0 It would be like asking whether the number 163 has a flavor --- even posing the question implies a profound misconstruing of essential nature of what a number is.\u00a0 For this one, we would have to say --- whatever the question is asking, whatever the question is doing, we have to recognize that x = 0 is not at all a possible value; having eliminating that value, we can proceed with whatever the rest of the problem may be.<\/p>\n

Question #4 is a particularly interesting one.\u00a0 First of all, as with the previous example, we run into major difficulties when the factor-to-be-cancelled equals zero.\u00a0 As with the other questions, we can't just do a blanket cancelling with impunity.\u00a0\u00a0 As with the previous two questions, we have to consider cases.\u00a0 If (x + 2) = 0, then the expression on the left becomes 0\/0, profoundly meaningless, and any statement setting this equal to anything else would be sheer nonsense.\u00a0 If (x + 2) = 0, then nothing equals anything else in this problem, so x = \u20132 is definitely not a legitimate answer.<\/p>\n

Now, what happens in the case in which (x + 2) \u2260 0?\u00a0 Well, in this case, this factor does\u00a0not<\/em>\u00a0equal zero, so it\u00a0can<\/em>\u00a0be cancelled, which leads to:<\/p>\n

\"\"<\/a><\/p>\n

Now, we have the same expression on both sides of the equation.\u00a0 This means, these two sides would be equal for all values of x, as long as the expression is defined.\u00a0 This means the whole continuous infinity of the number line is legal, barring a couple isolated exceptions.\u00a0 One is x = \u20134, which makes the denominator equal zero --- something divided by zero cannot equal anything, because something divided by zero has already departed from the realm in which any mathematically meaningful statement is possible.\u00a0 And, of course, as we discovered above, x = \u20132 cannot be a solution either.\u00a0 Therefore, the solution consists of all real numbers, the entire continuous infinity of the real number line, except for the values x = \u20134 and x = \u20132.<\/p>\n

 <\/p>\n

Summary<\/h2>\n

Don't divide by variables or by algebraic expressions.\u00a0 Don't cancel by variables or by algebraic expressions.\u00a0 Always consider whether the factor by which you would want to divide could equal zero, and either factor it out or consider the process in separate cases.<\/p>\n

 <\/p>\n

This post was written by Mike McGarry, GMAT expert at Magoosh<\/a>, and originally posted here<\/a>.
\n<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"

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