Verifying the Solution to an Inequality ... by Substituting? by Checking Boundaries?Date: 04/28/2010 at 22:02:02 From: J.C. Subject: I'm confused with inequalities. 4 - 2/3b < 3 - 1/3b verify? O.K. I am totally confused. I have to solve this inequality, show my work, and then verify it: 4 - (2/3)b < 3 - (1/3)b I think I somewhat understand how to solve it ... but I have no idea how to verify it. Here's what I did: 4 - (2/3)b < 3 - (1/3)b Multiplied the fractions (2/3)b and (1/3)b by their lowest common denominator (3) below 3(2/3b) = (6/3)b Simplified 6/3b = 2b 3(1/3b) = (3/3)b Simplified = 3(1/3b) = 1b 4 - 2b < 3 - 1b Substituted those simplified fractions 2b + 1b < 3 - 4 Transposed, by switching the sides of the expression 1b and the quantity 4, and then changing their signs 3b < -1 3b / 3 < -1 / 3 Divided both sides by 3 b < -1/3 Normally with equations, to verify the answer I just replace the variable with what I found out the variable to be. For example, if I had to solve for x in some equation, and found out x = -4, I would replace all of the x variables in that equation with -4. But with this inequality, I got "b" as b < -1/3 So now I don't know if I have to replace "b" with something less than -1/3. Date: 04/28/2010 at 23:14:40 From: Doctor Peterson Subject: Re: I'm confused with inequalities. 4 - 2/3b < 3 - 1/3b verify? Hi, J.C. First, I want to look carefully at your work to see that it looks right; I'll write just the main steps: 4 - (2/3)b < 3 - (1/3)b 4 - 2b < 3 - 1b This is wrong: you multiplied only one term on each side by 3, which is illegal. You have to multiply each ENTIRE SIDE (every term) by 3. 2b + 1b < 3 - 4 Here you lost a negative sign; -2b is staying where it was, so its sign doesn't change. 3b < -1 b < -1/3 So your work is wrong. The solution is probably wrong, too. But I want to show you how to check; presumably, checking will also reveal the error. How do you check a solution to an inequality? Ideally, you have to check that EVERY number in the solution set (in this case, every number less than -1/3) makes the original inequality true, and that EVERY number NOT in the solution set (-1/3 and every number greater than that) makes the original inequality false. You can't do that, of course. But we know that a solution to a linear inequality will be an interval, so we just have to check that we have the right interval. To do that, we can check the BOUNDARY -- the -1/3 that you determined -- and then check which SIDE of it is correct. So I would first take b = -1/3, and see if that is on the boundary of the solution, which means that it should make the two sides of the original inequality EQUAL. Let's do that: 4 - (2/3)b =? 3 - (1/3)b let b = -1/3 4 - (2/3)(-1/3) =? 3 - (1/3)(-1/3) 4 - (-2/9) =? 3 - (-1/9) 4 2/9 =? 3 1/9 They are not equal, so this is not the correct boundary. Nevertheless, b = -1/3 is not in your claimed solution set (since only numbers LESS than this are in it), and it turns out not to be a solution (since the left side is not less than the right). So the problem is not that it's not a solution, only that it is not the correct boundary to the solution set. I'm going to continue, just to show more clearly that the solution is wrong. If it HAD worked out, my next step would have been to choose some point IN the claimed solution set, and see if it satisfies the inequality. Let's take an easy number; try b = -1, which is less than -1/3: 4 - (2/3)b <? 3 - (1/3)b let b = -1 4 - (2/3)(-1) <? 3 - (1/3)(-1) 4 - (-2/3) <? 3 - (-1/3) 4 2/3 <? 3 1/3 This value is supposed to be a solution, but it isn't. In this case, we didn't have to do this, because we already knew the solution was wrong; but if we had had the right boundary, this might have shown that we had the wrong direction for the inequality. Now correct the errors in your solution, and do the check again. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 04/28/2010 at 23:54:48 From: J.C. Subject: Thank you (I'm confused with inequalities. 4 - 2/3b < 3 - 1/3b verify?) Thank you so much. I finally understand it now. You and everybody else at Dr. Math are just the BEST! I greatly appreciate that you took time to answer my question. Once again, thank you! :D |
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