In range A, you have changed the signs of both modules, in range B you have done it only for 2x+5 and in third case you haven't changed the signs. What's the logic behind this one??
I got your point. I'm not "changing" the signs, I'm expanding the absolute values in each range.
In range A, when x<\frac{2}{3}: |3x-2|=-3x+2 and |2x-5|=-2x+5, so we get -3x+2\leq-2x+5.
In range B, when \frac{2}{3}\leq{x}\leq\frac{5}{2}: |3x-2|=3x-2 and |2x-5|=-2x+5, so we get 3x-2\leq-2x+5.
In range C, when x>\frac{5}{2}: |3x-2|=3x-2 and |2x-5|=2x-5, so we get 3x-2\leq2x-5.
In range A, you have changed the signs of both modules, in range B you have done it only for 2x+5 and in third case you haven't changed the signs. What's the logic behind this one??
I got your point. I'm not "changing" the signs, I'm expanding the absolute values in each range.
In range A, when x<\frac{2}{3}: |3x-2|=-3x+2 and |2x-5|=-2x+5, so we get -3x+2\leq-2x+5.
In range B, when \frac{2}{3}\leq{x}\leq\frac{5}{2}: |3x-2|=3x-2 and |2x-5|=-2x+5, so we get 3x-2\leq-2x+5.
In range C, when x>\frac{5}{2}: |3x-2|=3x-2 and |2x-5|=2x-5, so we get 3x-2\leq2x-5.
My slow brain is just not getting it!!! How both LHS and RHS get negated for the range x<2/3 and only the RHS gets negated for the range B and nothing for range C. What if there were 4 or 5 ranges; what is the deciding factor as to what gets negated.
Could you please help me understand?
Absolute value properties: When x\leq{0} then |x|=-x, or more generally when some \ expression\leq{0} then |some \ expression|\leq{-(some \ expression)}. For example: |-5|=5=-(-5);
When x\geq{0} then |x|=x, or more generally when some \ expression\geq{0} then |some \ expression|\leq{some \ expression}. For example: |5|=5;
In range A, when x<\frac{2}{3}: 3x-2<0 so |3x-2|=-(3x-2) and 2x-5<0 so |2x-5|=-(2x-5), and we get -3x+2\leq-2x+5.
In range B, when \frac{2}{3}\leq{x}\leq\frac{5}{2}: 3x-2>0 so |3x-2|=3x-2 and 2x-5<0 so |2x-5|=-(2x-5), so we get 3x-2\leq-2x+5.
One way to solve is to square both the terms of course , but what is other way of solving it.
First you should determine the check points (key points): \frac{2}{3} and \frac{5}{2}. Hence we'll have three ranges to check:
A. x<\frac{2}{3} --> -3x+2\leq-2x+5 --> -3\leq{x}, as x<\frac{2}{3}, then -3\leq{x}<\frac{2}{3};
B. \frac{2}{3}\leq{x}\leq\frac{5}{2} --> 3x-2\leq-2x+5 --> -x\leq\frac{7}{5}, as \frac{2}{3}\leq{x}\leq\frac{5}{2} , then \frac{2}{3}\leq{x}\leq\frac{7}{5};
C. x>\frac{5}{2} --> 3x-2\leq2x-5 --> x\leq{-3}, as x>\frac{5}{7}, then in this range we have no solution;
Ranges from A and B give us the solution as: -3\leq{x}\leq\frac{7}{5}.
Question : What was normal way of doing it back in school? I am wondering how I used to solve them ?
Working out questions from your post and other guys notes on the site in the mean time.
Thanks a lot Bunuel.
Hussain15
Re: Inequalities - Challenging and tricky One [#permalink]
Posted: Sat Jan 16, 2010 9:49 am
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Re: Inequalities - Challenging and tricky One [#permalink]
Posted: Sat Jan 16, 2010 11:39 am
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Status: The last round Joined: Thu Jun 18, 2009 Posts: 1354
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In range A, you have changed the signs of both modules, in range B you have done it only for 2x+5 and in third case you haven't changed the signs. What's the logic behind this one??
One way to solve is to square both the terms of course , but what is other way of solving it.
First you should determine the check points (key points): \frac{2}{3} and \frac{5}{2}. Hence we'll have three ranges to check:
A. x<\frac{2}{3} --> -3x+2\leq-2x+5 --> -3\leq{x}, as x<\frac{2}{3}, then -3\leq{x}<\frac{2}{3};
B. \frac{2}{3}\leq{x}\leq\frac{5}{2} --> 3x-2\leq-2x+5 --> -x\leq\frac{7}{5}, as \frac{2}{3}\leq{x}\leq\frac{5}{2} , then \frac{2}{3}\leq{x}\leq\frac{7}{5};
C. x>\frac{5}{2} --> 3x-2\leq2x-5 --> x\leq{-3}, as x>\frac{5}{7}, then in this range we have no solution;
Ranges from A and B give us the solution as: -3\leq{x}\leq\frac{7}{5}.
that's a lovely and uncomplicated way....
fluke
Re: Inequalities - Challenging and tricky One [#permalink]
Posted: Fri Mar 11, 2011 12:42 pm
Math Forum Moderator
Joined: Mon Dec 20, 2010 Posts: 2242 Followers: 86
In range A, you have changed the signs of both modules, in range B you have done it only for 2x+5 and in third case you haven't changed the signs. What's the logic behind this one??
I got your point. I'm not "changing" the signs, I'm expanding the absolute values in each range.
In range A, when x<\frac{2}{3}: |3x-2|=-3x+2 and |2x-5|=-2x+5, so we get -3x+2\leq-2x+5.
In range B, when \frac{2}{3}\leq{x}\leq\frac{5}{2}: |3x-2|=3x-2 and |2x-5|=-2x+5, so we get 3x-2\leq-2x+5.
In range C, when x>\frac{5}{2}: |3x-2|=3x-2 and |2x-5|=2x-5, so we get 3x-2\leq2x-5.
My slow brain is just not getting it!!! How both LHS and RHS get negated for the range x<2/3 and only the RHS gets negated for the range B and nothing for range C. What if there were 4 or 5 ranges; what is the deciding factor as to what gets negated.
Could you please help me understand?
_________________ ~fluke
thesfactor
Re: Inequalities - Challenging and tricky One [#permalink]