simple a lv inequality eqn

kavvv SG

QNS: by considering the graphs of y = l 2x-1 l and y = 2+ l x+1 l. hence or otherwise, solve the inequality l 2x-1 l - l x+1 l > 2, x is a real no.

SOLn:

using gc,( gc plots both curves separately, y = l 2x-1 l and y = 2+ l x+1 l, in the same screen)

given that:

l 2x-1 l - l x + 1 l > 2

l 2x-1 l > 2 + l x+1 l

the intersections occurred at x= - 0.667 and x= 4.

therefore from the graphs, l 2x-1 l - l x + 1 l > 2 --> x< -0.667 or x > 4.

What i do not understand: why is the intersection of the 2 equations related to solution of l 2x-1 l - l x + 1 l > 2 . the "conventional" way to find the solution for any inequality is to find the roots right? maybe the intersection is related to the roots of l 2x-1 l - l x + 1 l ? if so, how is it related to it?

Uncertain SG

I'm confused of your qn.

Anyway, the idea is simple. Ignoring the inequality sign, you can consider the graph

y = |2x-1| - |x+1| <---- you have to draw this graph out

Now, consider the case where there is a line called y = 2 (basically a horizontal line intersecting y-axis at 2)

y = |2x-1| - |x+1| ------ (1)

y = 2 ----- (2)

Solving this simultaneous equation is similar to solving inequality:

since (1) = (2):

|2x-1| - |x+1| = 2 ---> use your gc, u will get x = -0.667 or 4

But when inequality kicks in, u will get the solution above.

|2x-1| - |x+1| > 2 ---> x< -0.667 or x>4

kavvv SG

erm. thx.but what i meant was that why the solution could be found by finding the intersection between the 2 eqn( y = l 2x-1 l and y = 2+ l x+1 l) instead of y = |2x-1| - |x+1| as one and then finding the roots of the equation: y = |2x-1| - |x+1|. but nvm, i kinda realised the answer to my question from urs lolx. thx again.

Uncertain SG

you are welcome.

The roots of y = |2x-1| - |x+1| can only be found if y = 0 (aka the x-axis). It is the same as intersection between the graph y = |2x-1| - |x+1| and line y = 0.

When intersection occurs, |2x-1| - |x+1| = y = 0

making |2x-1| - |x+1| = 0 and then find x from here.