please check my work if i done correctly. I'm little confused.

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- February 2nd 2011, 12:58 AMjam2011Absolute value inequalities.
please check my work if i done correctly. I'm little confused.

- February 2nd 2011, 01:31 AMProve It
Remember that , so I would rewrite this equation as

.

Solve for . - February 2nd 2011, 03:28 AMBAdhi
is not , it is

- February 2nd 2011, 04:15 AMArchie Meade
and are both linear functions.

One has a positive slope and the other has a negative slope, so the lines intersect at a single point.

The modulus of these are V-shaped as we invert the part of the graph below the x-axis,

hence there are now two points of intersection if the original lines do not intersect on the x-axis.

Hence, we can look for the points of intersection of 4 lines, while eliminating points of intersection below the y-axis

(refer to the dashed lines on the attachment).

Therefore the lines do not intersect above the x-axis

and are therefore not a solution of

although the value of x is a solution to the equation.

so these lines also do not intersect above the x-axis.

therefore these lines**do intersect above the x-axis.**

therefore**these lines also intersect above the x-axis.**

Hence the solution is the union of

and

- February 2nd 2011, 05:51 AMjam2011
since Mr. prove it give (2x- 6)^2 = (4 - 5x)^2 then

(2x-6-4+5x)(2x-6+4-5x) by x^2-y^2 = (x+y)(x-y)

7x-10 = 0 -3x-2 = 0

x = 10/7 x = -2/3

Can we consider this a short cut method.?thank you a lot for your effort.God bless - February 2nd 2011, 06:32 AMArchie Meade
That's a "cute" calculation you made!

Yes, it's a good shortcut and locates the values of x that are the solutions of the modulus function.

This is because the x co-ordinates of the points of intersection of the lines

are the same as the points of intersection of the parabolas.

The method I showed earlier will give the x and y co-ordinates if required,

however only the x co-ordinates are requested. - February 2nd 2011, 11:37 AMArchie Meade
Without reference to graphs, you can also try this way......

When x>3, both 2x-6 and 5x-4 are both positive

and when 5x<4, both 2x-6 and 5x-4 are both negative.

**Case 1:**both positive within modulus

....No, since the minimum value of 3x is 9.

**Case 2:**

**Case 3:**Both negative within modulus

- February 2nd 2011, 05:34 PMjam2011
wow thats great...thanks a lot for all of your information. i need to study more further. God bless you sir.