Basic Tips on Solving for XDate: 11/16/2000 at 22:25:13 From: Megan Subject: How to solve for the variable "X" The following problems have tripped me up. I can't seem to figure out how to solve them. I've tried, several times, and had my algebra teacher explain them, but I still can't get the answers. 3x - 11 = 6 2x - 7 = 3 + x 3 + 3x = 9 + 2x 3x - 3 = 6 + x Thanks, Megan Date: 11/17/2000 at 11:03:31 From: Doctor Ian Subject: Re: How to solve for the variable "X" Hi Megan, Suppose I start with something like x = 5 That's no fun, is it? There's nothing to figure out. But I can multiply both sides of the equation by 3 without changing the truth of the equation, right? 3x = 15 And there are a lot of different ways I can write 3: 3 = 1 + 2 = 7 - 4 = -5 + 8 So I can change my equation to look like this: 3x = 15 x(3) = 15 x(7 - 4) = 15 7x - 4x = 15 And since I can add or subtract anything on both sides of an equation without changing the truth of the equation, I can add 4x to both sides: 7x - 4x = 15 7x - 4x + 4x = 15 + 4x 7x = 15 + 4x And of course, I can do the same thing with any number I want: 7x = 15 + 4x 7x - 5 = 15 + 4x - 5 7x - 5 = 15 - 5 + 4x 7x - 5 = 10 + 4x Now I have something that looks just like your problems, don't I? The trick, then, is to reverse the steps that I just did, to get back to the original, simple, boring equation: 7x - 5 = 10 + 4x 7x - 5 + 5 = 10 + 4x + 5 Add 5 to both sides 7x = 15 + 4x 7x - 4x = 15 + 4x - 4x Subtract 4x from both sides 3x = 15 (3x)/3 = 15/3 Divide both sides by 3 x = 5 It's easy to forget what's really going on when you're solving an algebraic expression for a variable. You're really just trying to rewrite a complicated expression in a simpler way, without doing anything to change the meaning of the expression. It's a little like what you would do to change a complicated sentence like: "Under a mammal of feline persuasion is located, at this time, in an unspecified location, a rectangle of woven fabric." into a simpler one like: "The cat is on the mat." Algebra is basically tying knots and then untying them again. With that in mind, let's look at one of your example problems: 3 + 3x = 9 + 2x The first thing to notice is that we have numbers on both sides. We know that someone put them there by adding a number to both sides of the equation, so we can un-put them there by subtracting a number from both sides: 3 + 3x = 9 + 2x 3 + 3x - 3 = 9 + 2x - 3 (3 - 3) + 3x = (9 - 3) + 2x 3x = 6 + 2x Now we have multiples of x on both sides. We can do the same kind of thing to fix that: 3x = 6 + 2x 3x - 2x = 6 + 2x - 2x (3 - 2)x = 6 + (2 - 2)x x = 6 Can you follow the same kinds of steps to solve the rest of the problems? I hope this helps. Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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