You are Here: Home

# C3 Modulus, can you do this? Tweet

Maths and statistics discussion, revision, exam and homework help.

Announcements Posted on
Important: please read these guidelines before posting about exams on The Student Room 28-04-2013
1. C3 Modulus, can you do this?
Ok how would one solve this?

Please put worked solutions in spoiler please.

10|x|= |x-6|
3|x-4|< 2|x+3|

I'm unsure how do I work out these ones with a number outside the mod. Please clarify to aid my understanding. Much appreciated.
2. Re: C3 Modulus, can you do this?
Treat them as if they were brackets. First work it out for the positive one and then negative.

Example

Solve

x - 5= 3x - 18

and then x -5 = -3x + 18

-x + 5 = 3x - 18

and finally -x + 5= -3x +18

So you have four solutions, possibly.
3. Re: C3 Modulus, can you do this?
you could sketch them, and then work out the solutions.
I dont think you need all the solutions because some will be for the normal graphs intercepting.
4. Re: C3 Modulus, can you do this?
(Original post by hcam)
Treat them as if they were brackets. First work it out for the positive one and then negative.

Example

Solve

x - 5= 3x - 18

and then x -5 = -3x + 18

-x + 5 = 3x - 18

and finally -x + 5= -3x +18

So you have four solutions, possibly.
but you only need to do one with one + and one with both + or both - [spoiler] one gives 13/2 and the other 23/2 [spoiler]
5. Re: C3 Modulus, can you do this?
One method which will work with any modulus question is squaring both sides - a square of any real number is always positive, so it will remove the modulus sign and enable you to treat it as a normal quadratic:

e.g.
abs(x-5) = 3abs(x-6)
square both sides:
(x-5)2 = (3(x-6))2
(x-5)2 = (3x-18)2

from there you can multiply out the brackets, collect your terms on one side and solve using factorising or the formula:

Spoiler:
Show
x2-10x+25 = 9x2-108x+324
8x2-98x+299 = 0
(4x+23)(2x+13) = 0
x = 6.5 or 5.75

The others can be solved similarly
Last edited by DavidH20; 09-06-2012 at 11:54.