Which procedure takes the minimum time to solve modulus functions?

[Nousher Ahmed]

1) -|2x-3|+|5-x|+|x-10|=|3-x|
2) |2x-3|-|5-x|-|x-10|-|3-x|=28
3) -|2x-3|+|5-x|+|x-10|≥|3-x|

How can we solve these problems?

The method I know is to plug in the critical values to see which modulus becomes positive and which one becomes negative. Then find out the values of x for which the above conditions become true.

For example, if I am asked to solve the first problem, I will do the following steps.

Step-1: Find out the critical values.
For 2x-3, critical value is 3/2.
For 5-x, critical value is 5.
For x-10, critical value is 10
For 3-x, critical value is 3.

Step-2: When x is less than 3/2, we can write the first equation in the following form:

(2x-3)-(x-5)-(x-10)=-(x-3)
or, 2x-3-x+5-x+10=-x+3
or, x=-9

When 1.5≤x<3,
-(2x-3)-(x-5)-(x-10)=-(x-3)
or, -2x+3-x+5-x+10=-x+3
or, -4x+18=-x+3
or, -3x=-15
or, x=5

x=5 isn't acceptable because we assumed x<3.

We can repeat the step-2 for 3≤x<5, 5≤x<10 and x≥10.

This procedure is very long and a time consumable process.

Is there other way that requires less time and less effort to solve the above problems?

Once I have seen a person to use a number line to solve this sort of problem. However, his explanation works when there is a poisitive sign between two modulus functions, and a value at the right hand side of the equation.

For example, we can consider the following equation.

4) |x-5|+|x-2|=9

We can imagine a number line where two people are standing at two points. One person is 2 units away from zero (center point), and other person is 5 units away from zero. Both persons are standing at the right hand side of center point. They want to meet on such a point for which together they will have to travel total 9 units.

So if the person who is two units away from center point moves right, they can meet on a point which is 8 units away from center point. Hence total distance traveled by them will be now 9 units. So, now we have x=8.

If we repeat this same procedure for the left hand side, we will find x=-1 which is just one unit away from zero (center point). And to reach that point, both of them will have to travel total 9 units.

However, this procedure doesn't work for the equations like 1,2 and 3.

Now I am here to ask the experts of this site to let me know some other ways which can help solve the above equations within shortest possible time.

 

[I like Serena]

Hi Nousher, and welcome to MHB!

First, move all terms to the left side.
That will be the function that we analyze.
The solutions are the values where the function intersects the x-axis. These are the so called 'zeroes'.

Then whatever the sign of each term, the result is a straight line.
Since some signs change at the critical points, we are talking about a set of line segments that are connected to each other.
So if we evaluate the function at those critical points, we can draw the graph by drawing the line segments between the critical points.

More specifically, if 2 neighboring function values are on different sides of the x-axis, there must be a zero in between.
We can then deduce what the sign of each term is and solve the resulting equation.
We don't even have to draw the function for that.
On the other hand, if the neighboring function values are on the same side of the x-axis, there cannot be a zero in between.

That leaves potential zeroes below the lowest critical point, or above the highest critical point.
So we need to check if the corresponding line segments slope towards the x-axis, in which case we need to solve the corresponding equations as well.

 

[Nousher Ahmed]

Hi Nousher, and welcome to MHB!

First, move all terms to the left side.
That will be the function that we analyze.

Then whatever the sign of each term, the result is a straight line.
Since some signs change at the critical points, we are talking about a set of line segments that are connected to each other.
So if we evaluate the function at those critical points, we can draw the graph by drawing the line segments between the critical points.

More specifically, if 2 neighboring function values are on different sides of the x-axis, there must be a zero in between.
We can then deduce what the sign of each term is and solve the resulting equation.
We don't even have to draw the function for that.
On the other hand, if the neighboring function values are on the same side of the x-axis, there cannot be a zero in between.

That leaves potential zeroes below the lowest critical point, or above the highest critical point.
So we need to check if the corresponding line segments slope towards the x-axis, in which case we need to solve the corresponding equations as well.

I am very much thankful to you for elaborate explanation.

I am afraid to say that I hardly understand your explanation. From a discription, it is impossible for me to visualize the whole situation. Will you please draw graphs to solve the above stated problems? The above problems are new to me. I beg you to provide more information as much as possible with the graphs, so that I can understand the whole procedure.

I beg pardon as I have asked you to do something that requires a lot of time.

 

[I like Serena]

I am afraid to say that I hardly understand your explanation. From a discription, it is impossible for me to visualize the whole situation. Will you please draw graphs to solve the above stated problems? The above problems are new to me. I beg you to provide more information as much as possible with the graphs, so that I can understand the whole procedure.

Let's pick your $|x-5|+|x-2|=9$.
We define the function $f(x) = |x-5|+|x-2|-9$.
So we want to know for which values of $x$ we have that $f(x)=0$.

Critical points are $x=2$ and $x=5$ for which we have $f(2)=-6$ and $f(5)=-6$.
Let's pick $x=0$ and $x=6$ as they are below respectively above the critical points. Those have $f(0)=-2$ and $f(6)=-4$.
From those we can deduce that the graph looks like this:

That is, the critical points are on the same side of the x-axis, so there cannot be a zero in between.
Below $x=2$ the graph slopes towards the x-axis, so there must be a zero there.
Above $x=5$ the graph slopes towards the x-axis as well, so there must be a zero there as well.

For values below $x=2$ all signs are negative, so we need to solve $-(x-5)+-(x-2)-9=0 \implies -2x-2=0 \implies x=-1$.
For values above $x=5$ all signs are positive, so we need to solve $+(x-5)++(x-2)-9=0 \implies 2x-16=0 \implies x=8$.

Therefore the solutions are $x=-1$ and $x=8$.

 

[Nousher Ahmed]

Let's pick your $|x-5|+|x-2|=9$.
We define the function $f(x) = |x-5|+|x-2|-9$.
So we want to know for which values of $x$ we have that $f(x)=0$.

Critical points are $x=2$ and $x=5$ for which we have $f(2)=-6$ and $f(5)=-6$.
Let's pick $x=0$ and $x=6$ as they are below respectively above the critical points. Those have $f(0)=-2$ and $f(6)=-4$.
From those we can deduce that the graph looks like this:
View attachment 11139

That is, the critical points are on the same side of the x-axis, so there cannot be a zero in between.
Below $x=2$ the graph slopes towards the x-axis, so there must be a zero there.
Above $x=5$ the graph slopes towards the x-axis as well, so there must be a zero there as well.

For values below $x=2$ all signs are negative, so we need to solve $-(x-5)+-(x-2)-9=0 \implies -2x-2=0 \implies x=-1$.
For values above $x=5$ all signs are positive, so we need to solve $+(x-5)++(x-2)-9=0 \implies 2x-16=0 \implies x=8$.

Therefore the solutions are $x=-1$ and $x=8$.

I didn't face any problem to solve the equation you solved. But I faced to solve the first three equations. Will you please let me know whether the method you have shown is applicable for the equation 1,2 and 3?
Equation 1,2 and 3 are below:
1) -|2x-3|+|5-x|+|x-10|=|3-x|
2) |2x-3|-|5-x|-|x-10|-|3-x|=28
3) -|2x-3|+|5-x|+|x-10|≥|3-x|

How can I use the method for these three equations?

 

[skeeter]

(1) move all terms to one side …

$f(x) = -|2x-3|+|5-x|+|x-10|-|3-x| = 0$

f(-10) = -1

f(3/2) = 21/2

f(3) = 6

f(5) = -4

f(10) = -19

f(15) = -24

f(x) has two zeros, one in the interval -10 < x < 3/2 and the other in the interval 3 < x < 5

continue …