- -|2x-3|+|5-x|+|x-10|=|3-x|
- |2x-3|-|5-x|-|x-10|-|3-x|=28
- -|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.
- |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 travelled 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 rewrite your first equation as $$|x-5|+|x-10|=|x-3| + |2x-3| $$ For $x\ge 10,$ get $2x-15=3x-6,$ impossible,
For $5\le x \le 10,$ lhs $=5,$ but $|2x-3|$ is much bigger.
For $3\le x\le 5,\ \ 15-2x=3x-6,$ which gives the solution $$x=4.2$$