Solving Mods using diagrams

Hi Karishma,

Is there any other method to solve these kind of problems ?

Thx much,

You can certainly use algebra to solve these questions. Divide the number line, check for positive negative regions and solve for various cases etc. I have given this approach because a member wanted the diagrammatic approach to this question (the diagrammatic approach is much faster and intuitive but only after you have a thorough understanding of what you are supposed to do and, more importantly, why). Search for mods in the forum and you will find many examples discussing algebraic approaches to such questions.

Though you cant find better response than the Karishma's one, here is the algebraic approach -

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

Since it is inequality, you must remember that you cannot square both sides until and unless you are sure about the signs of both sides.
We know that modulus is always positive, so both l.h.s and r.h.s can be squared and sign of inequality will remain same.

This is what I remember when I try to solve such questions -

It is okay to square both sides of the equation if you are sure that each side is greater than or equal to zero
i.e

|2x-5|^2 >= |3x-2|^2

Open the square and subtract one equation from another, you will get

5x^2+8x-21<=0

i.e

5(x-7/5)(x+3)<=0

ie. x belongs to (-3,7/5)

let me know if somthing is not clear

Hi Karishma,
I tend to solve such questions using algebraic approach. Also, I know the basics of graphical approach and can apply well if the equation is follows pattern such as |x-a|+|x-b| = variable.
However, for solving below questions, I usually fall flat. The solution that you have provided has given me some insights, but I failed to understand the below colored portion.
Could you please explain that part a bit.
i.e How did you infer this-

The third red line will be equal to twice the distance of 11/6 i.e. this point will be 2*11/6 = 22/6 away from 4/6.

Thanks
H

Look at the second diagram very carefully. We want that the sum of the lengths of the red lines to be equal to the sum of the lengths of the green lines. Note that the red lines are shorter than the green lines. Then how will the sum of the length of the red lines be equal to the sum of the length of the green lines? Thankfully, we have an extra red line (3 red lines vs 2 green lines). This extra red line will need to make up for the extra length of the green lines as compared to the length of the red lines. This one extra red line needs to cover the extra length of both the green lines. What is the extra length of both the green lines together? It is 2* (11/6) (look at the diagram). Hence, the length of the red line = 22/6
So the leftmost point must be at a distance 22/6 to the left of 4/6.

Awesum Post. Kudos. Thanks a lot for the explanation.

Karishma - I couldnt understand how we got 4/6 and 11/15. Please help me understand little more detail.

And one more question in the notes there is one more approach based on algebra - 5X^2+8X-21<=0 how do we solve this?

The given inequality is \(3|x-2/3| \leq 2|x-5/2|\)

So transition points are 2/3 and 5/2. It is hard to work with fractions until and unless they have the same denominator. So
2/3 = 2*2/3*2 = 4/6
5/2 = 5*3/2*3 = 15/6

Now they both have the same denominator so it is easier to handle. The distance between these two points is 15/6 - 4/6 = 11/6
The logic behind this method is given in these posts:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/01 ... edore-did/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/01 ... s-part-ii/


Also, \(5X^2+8X-21<=0\)
\(5X^2 + 15X - 7X - 21 <= 0\)
\((5X -7)(X + 3) <= 0\)
\(5(X - 7/5)(X + 3) <= 0\)
\(-3 <= X <= 7/5\)

This inequality was solved very easily using the method discussed in these posts:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/06 ... e-factors/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/07 ... ns-part-i/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/07 ... s-part-ii/

The given inequality is \(3|x-2/3| \leq 2|x-5/2|\)

So transition points are 2/3 and 5/2. It is hard to work with fractions until and unless they have the same denominator. So
2/3 = 2*2/3*2 = 4/6
5/2 = 5*3/2*3 = 15/6

Now they both have the same denominator so it is easier to handle. The distance between these two points is 15/6 - 4/6 = 11/6
The logic behind this method is given in these posts:

Also, \(5X^2+8X-21<=0\)
\(5X^2 + 15X - 7X - 21 <= 0\)
\((5X -7)(X + 3) <= 0\)
\(5(X - 7/5)(X + 3) <= 0\)
\(-3 <= X <= 7/5\)

This inequality was solved very easily using the method discussed in these posts:


This particular question can also be solved without using diagrams. Here is how we approach it:

We know that if two number a and b are positive and a>=b, then we can also say

a2>=b2

Since we have modulus on both sides, we can make use of this algebric expression

|3x – 2| <= |2x – 5|

This implies:

|3x - 2|2 <= |2x - 5|2

And we also know |a|2 = a2

thus, we will get a simple inequality as:

(3x-2)2 <= (2x-5)2

Or

(3x-2)2 - (2x-5)2 <=0

using a2 - b2 = (a-b)(a+b)

(3x - 2 - 2x +5)(3x - 2 + 2x - 5) <=0
(x+3)(5x-7) <=0
(x+3)(x-7/5)<=0
-3<=x<=7/5

I hope it helps!!

Mod questions are generally trickier as there could always be a case which one forgets. Depending on the question, one should be able to make a call (which will come with practice) whether to go with algebric way or with diagrams or BOTH. I personally prefer using a combination of both. Diagrams to understand the data and algebra to solve equations quickly.

Question: Solution set for |3x-2| <= |2x-5|

First lets create lines for 3x-2 = 0 (line A) and 2x-5 = 0 (line B)



It is easy to see that |3x-2|= the distance of a point from line A; and |2x-5| = distance of a point from line B; We have to find region (range of x) for which the above condition holds true. Meaning the value of x for which distance from line A is smaller than distance from line B.

1) Lets take a point right next to line B (X>5/2) but on the right side of it (region K). It is easy to see that its distance from line A is greater than distance from line B. Now as we move further right, the distance from line A will increase 3 times (because its 3x-2) but only 2 times from line B. Hence, the difference in distance will keep on increasing. Conclusion: No points possible in this region

2) Lets take a point right next to line A (X<2/3) but on the left of it (region I). It is easy to see that its distance from line A is less than distance from line B. Now as we move further left, the distance from line A will increase 3 times (because its 3x-2) but only 2 times from line B. Hence, there will be a point beyond which distance from line A will be greater than distance from line B. To get that point simply solve for 3x-2 = 2x-5. You can remove the mod sign because both sides will have the same sign (negative) for X<2/3. This will give x = -3. Hence for X< -3 distance from line A will be greater than line B. Alternatively, for X>= -3 the distance from line A will be smaller than B.

3) Now, lets take the middle region - region J (2/3<X<5/2). A point right next to line A (but on the right side) is closer to line A than B. As one moves towards line B the distance from line A will increase and decrease from line B. Resulting in a point beyond which the distance from A will become greater than B. To find that point solve for 3x-2 = -(2x-5)
This is because on the right side of A, 3x-2> 0; and on the Left side of B, 2x-5 < 0;
Solving it will give x = 7/5. Hence for X > 7/5 distance from A will be greater than distance from line B. Alternatively, for X <= 7/5 the distance from line A will be smaller than B.

Answer for -3<=X<= 7/5 , distance from A will be smaller than B and above equations holds true.

PS: With practice you will see that it is not required to consider case 1 at all. One can solve only using case 2 and 3.

I’ll try to help you with an easy-to-understand and more visual approach to solve this question.

Quickly plot the 2 lines on x-y axes:
Y= 3x-2
Y= 2x-5

We find that they intersect in the 3rd Quadrant.



Now comes the key point that you must understand: what does |3x-2| mean? For any given value of x, the modulus sign |3x-2| denotes the distance of the point (3x-2) from the x-axis.

Once we understand this, the problem becomes pretty easy to solve.

We can observe from the graph that for x<-3, the line 3x-2 is at a greater distance from the x-axis than the line 2x-5. Or, to put it mathematically,
|3x-2|>|2x-5|

At x=-3, both lines are at the same distance from the x-axis. i.e. |3x-2|= |2x-5|

As we move rightwards from x=-3, we observe that the line 3x-2 is closer to the x-axis than the line 2x-5. How do we say it mathematically? Yeah, you guessed it:
|3x-2|<|2x-5|

Tada! This is the zone we wanted to be in!

But, is this true for all values of x>-3? Again, take a look at the graph.

We observe that, after some point, let’s call it Point G, in the 1st Quadrant, the line 3x-2 again starts getting more distant from the x-axis than the line 2x-5.

How do we find this point G?

At the point G, both the lines will be equidistant from the x-axis; one will be above the x-axis, and the other will be below it. To put it in other words, the magnitude of y (y= 3x-2 and y= 2x-5) will be same for both lines at the point G; but the signs of y will be opposite.

So, the following equation will be true for point G:
3x-2 = -(2x-5)
Solving it, we get x= 7/5.

So, we can conclude that between x>-3 and x<7/5, the line 3x-2 is closer to the x-axis than the line 2x-5. We say it mathematically as under:
|3x-2|<= |2x-5| for -3<=x<=7/5

A quick refresher on basic formulas in. mod function. May not be helpful in the pertinent problem but still important in speedy solutions.

Following 6 basic formulas are commonly used (Last 2 are derived from first 4):


(A) |x| ≤ a
---> -a ≤ x ≤ a
(B) |x| ≥ a
---> x ≤ -a or x ≥ a
(C) |x| No solution (Remember :- |x|≤ 0 can have a solution for x=0)
(D) |x| ≥ 0
---> -∞ (a – r) r --->  x a + r

Rgds,
Rajat