What is the solution set for |3x - 2|

Question

What is the solution set for  |3x-2|\leq|2x-5| ?

A.  x \leq -3 .

B.  -3 \leq x \leq \frac{7}{5} .

C.  -3 < x < \frac{7}{5} .

D.  -3 \leq x .

E.  x \geq \frac{7}{5} .

One way to solve is to square both the terms of course , but what is other way of solving it.

Bunuel · Accepted Answer

First, determine the check points (aka key points or transition points): \frac{2}{3} and \frac{5}{2} . Hence, we'll have three ranges to check: A. x<\frac{2}{3} For this range, 3x - 2 < 0 and 2x - 5 < 0 , so we get: -(3x-2)\leq-(2x-5) ; -3\leq{x} Since we consider x < \frac{2}{3} range, we'd get: -3\leq{x}<\frac{2}{3} . B. \frac{2}{3}\leq{x}\leq\frac{5}{2} For this range, 3x - 2 \geq 0 and 2x-5\leq 0 , so we get: 3x-2\leq-(2x-5) x\leq\frac{7}{5} Since we consider \frac{2}{3}\leq{x}\leq\frac{5}{2} range, we'd get: \frac{2}{3}\leq{x}\leq\frac{7}{5} . C. x>\frac{5}{2} For this range, 3x - 2 > 0 and 2x-5>0 , so we get: 3x-2\leq2x-5 x\leq{-3} Since we consider x>\frac{5}{2} range, there is no solution in this range. The solution from ranges A and B is: -3 \leq x \leq \frac{7}{5} .

Bunuel · Answer

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 .

GMATMadeeasy · Answer

Thank you , it helps greatly.

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 · Answer

@ Bunuel:

I didn't pick your strategy.

My way of doing is simple but it is giving different answer.

|3x-2| <= |2x-5|
We will consider two scenarios
3x-2 <= 2x-5 & -(3x-2) <= 2x-5
x<= -3 & x >= 7/5

So the range will be x<= -3 & x >= 7/5

Kindly let me know where did I go wrong???

Bunuel · Answer

If you plug the numbers from the ranges you got, you'll see that the inequality doesn't hold true.

As for the solution: we have two absolute values  |3x-2|  and  |2x-5| .  |3x-2|  changes sign at  \frac{2}{3}  and  |2x-5|  changes sign at  \frac{5}{2} .

---(I)---\frac{2}{3}---(II)---\frac{5}{2}---(III)---

We got three ranges as above. We should expand given inequality in these ranges and see what we'll get.

Hope it's clear.

Hussain15 · Answer

Thanks for the reply Bunue!

But there is one more issue.

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??

prashantbacchewar · Answer

Thanks Bunuel you posted excellent approach to deal with inequalities. I think you should write somthing on inequalities as well.
Kudos to you

puneet3085 · Answer

that's a lovely and uncomplicated way....

fluke · Answer

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?

Bunuel · Answer

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 . For more check: math-absolute-value-modulus-86462.html Hope it's clear.

thesfactor · Answer

Great solution bunuel, kudos!

Amogh · Answer

Bunuel,
when you say as -3<= X and as x<  \frac{2}{3}  then -3<=x<= \frac{2}{3}  does this mean that the range of x that satisfies the condition x< \frac{2}{3}  is all those points on the number line which satisfy both inequalities.

Also you say ranges from A and B give us solution as -3<= x<=  \frac{7}{5}  . Can you please explain this to me. Does this mean that the final solution to this problem would be all those ranges that have a solution to their respective conditions i.e the conditions x< \frac{2}{3}  ,  \frac{2}{3} <=x< \frac{5}{2}  and x>= \frac{5}{2} .

Also if for example ranges from A and B were -5<=x<2 and 4<=x<10, then would the final solution be both these ranges or do they have to have an overlap?
 
I know this is a supremely dumb and elementary question, but unfortunately inequalities and modulus happen to be my weakest areas. I cannot express in words my gratitude to you for having put up the GMAT Math Book. That thing is my Math Bible. You've made math so easy to understand.

Best wishes and many many thanks !!!
Amogh.

Bunuel · Answer

We consider three ranges:
A.  x<\frac{2}{3} ;
B.  \frac{2}{3}\leq{x}\leq\frac{5}{2} ;
C.  x>\frac{5}{2} .

In range A we get:  -3\leq{x}<\frac{2}{3} ;
In range B we get:  \frac{2}{3}\leq{x}\leq\frac{7}{5} ;
In range C there is no solution.

So, the given inequality holds true for  -3\leq{x}<\frac{2}{3}  and  \frac{2}{3}\leq{x}\leq\frac{7}{5} . Now, we can combine these two ranges and write:  -3\leq{x}\leq\frac{7}{5} .

Next, the solution set is  -3\leq{x}\leq\frac{7}{5} , means that any x from  -3\leq{x}\leq\frac{7}{5}  will satisfy  |3x-2|\leq|2x-5| .

If for some question we had -5<=x<2 from one range and 4<=x<10 from another, then the solution will be both ranges (no need for overlap).

Hope it's clear.

Ousmane · Answer

Hello Bunuel. Does this approach always work?: |3x-2| <= |2x-5| ===> (3x-2)^2<= (2x-5)^2 ===> 5x^2+ 8x -21 <= 0 solution of the ineq (-3, 7/5) Brother Karamazov

Bunuel · Answer

Do you mean squaring? If both parts of an inequality are non-negative (as in our case), then you can safely raise both parts to an even power (for example square). A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality). For example: 2<4 --> we can square both sides and write: 2^2<4^2 ; 0\leq{x}<{y} --> we can square both sides and write: x^2-2 if we square we'll get 1>4 which is not right. So if given that x>y then we can not square both sides and write x^2>y^2 if we are not certain that both x and y are non-negative. B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality). For example: -2<-1 --> we can raise both sides to third power and write: -2^3=-8<-1=-1^3 or -5<1 --> -5^2=-125<1=1^3 ; x we can raise both sides to third power and write: x^3

testcracker · Answer

excellent ..! very didactic post ...

KarishmaB · Answer

Solving the question using number line (as requested in a pm)

|3x-2|\leq|2x-5|

3 * |x-\frac{2}{3}|\leq 2*|x-\frac{5}{2}|

3 * |x-\frac{4}{6}|\leq 2*|x-\frac{15}{6}|

--------------------------- (4/6) ----------------------------(15/6)----------------

Assuming just "equal to" for now. 
Three times the distance from 4/6 is equal to two times the distance from 15/6

Divide the distance of 11/6 into 5 parts to get 11/30 each part. x would be a point 22/30 away from 4/6 i.e. x = 42/30 = 7/5
On the left of 7/5, "three times the distance from 4/6" will be less than "two times the distance from 15/6"

On the left side of 4/6, there will be another such point x such that thrice the distance from 4/6 is equal to twice the distance from 15/6
 3*(4/6 - x) = 2(15/6 - x) 
 x = -3 
On the right of -3, "three time distance from 4/6" will be less than "two times the distance from 15/6"

So the required range is  -3 \leq x \leq 7/5

s3sanghvi · Answer

When I square both sides I get two roots thats x<= 7/5 and x<= -3. And the answer says -3<=x<=7/5. Could anybbody tell me where I am going wrong. Cause I have used the square method till now and never got it wrong. And I am doubting this approach.

Bunuel · Answer

What is the solution set for |3x-2|\leq|2x-5| ? A. x \leq -3 . B. -3 \leq x \leq \frac{7}{5} . C. -3 < x < \frac{7}{5} . D. -3 \leq x . E. x \geq \frac{7}{5} . Square: |3x-2|\leq|2x-5| : 9x^2-12x+4 \leq 4x^2 - 20x+25 5x^2 + 8x -21 \leq 0 Facror: (x + 3)(x - \frac{7}{5})\leq0 The roots are -3 and 7/5. Since the inequality sign is "<", the solution lies between the roots: -3 \leq x \leq \frac{7}{5} . Answer: B. Check these links: https://gmatclub.com/forum/solving-quad ... 70528.html https://www.purplemath.com/modules/solvquad.htm https://www.purplemath.com/modules/factquad.htm Hope it helps.

What is the solution set for |3x - 2| <= |2x - 5|?

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