# Absolute value inequalities.

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• Feb 2nd 2011, 12:58 AM
jam2011
Absolute value inequalities.
please check my work if i done correctly. I'm little confused.
• Feb 2nd 2011, 01:31 AM
Prove It
Remember that $\displaystyle |X| = \sqrt{X^2}$, so I would rewrite this equation as

$\displaystyle \sqrt{(2x-6)^2} = \sqrt{(4-5x)^2}$

$\displaystyle (2x- 6)^2 = (4 - 5x)^2$.

Solve for $\displaystyle x$.
• Feb 2nd 2011, 03:28 AM
BAdhi
$|2x -6|=|4 -5x|$ is not $-(2x-6)=-(4-5x)$, it is $(2x-6)=\pm(4-5x)$
• Feb 2nd 2011, 04:15 AM
Archie Meade
Quote:

Originally Posted by jam2011
please check my work if i done correctly. I'm little confused.

$2x-6$ and $4-5x$ 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).

$2x-6=4-5x\Rightarrow\ 7x=10\Rightarrow\ x=\frac{10}{7}$

$2\left(\frac{10}{7}\right)-6<0$

Therefore the lines do not intersect above the x-axis
and are therefore not a solution of

$|2x-6|=|4-5x|$

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

$2x-6=-(4-5x)\Rightarrow\ 2x-6=5x-4\Rightarrow\ -2=3x\Rightarrow\ x=-\frac{2}{3}$

$2\left(-\frac{2}{3}\right)-6<0$

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

$-(2x-6)=4-5x\Rightarrow\ 6-2x=4-5x\Rightarrow\ 3x=-2\Rightarrow\ x=-\frac{2}{3}$

$-\left[2\left(-\frac{2}{3}\right)-6\right]>0$

therefore these lines do intersect above the x-axis.

$-(2x-6)=-(4-5x)\Rightarrow\ 6-2x=5x-4\Rightarrow\ 7x=10\Rightarrow\ x=\frac{10}{7}$

$-\left[2\left(\frac{10}{7}\right)-6\right]>0$

therefore these lines also intersect above the x-axis.

Hence the solution is the union of

$-(2x-6)=(4-5x)$

and

$-(2x-6)=-(4-5x)$
• Feb 2nd 2011, 05:51 AM
jam2011
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
• Feb 2nd 2011, 06:32 AM
Archie Meade
Quote:

Originally Posted by jam2011
Since Mr. Prove It gives (2x- 6)^2 = (4 - 5x)^2

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

7x-10 = 0 so x = 10/7

-3x-2 = 0 so x = -2/3

Can we consider this a short cut method? thank you a lot for your effort. God bless

That's a "cute" calculation you made!

$(2x-6)^2-(4-5x)^2=0\Rightarrow\ [(2x-6)+(4-5x)][(2x-6)-(4-5x)]=0$

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.
• Feb 2nd 2011, 11:37 AM
Archie Meade
Without reference to graphs, you can also try this way......

$|2x-6|=|5x-4|$

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

$x\ge3,\;\;5x\ge4\Rightarrow\ x\ge3$

$(2x-6)=(5x-4),\;\;\;x\ge3$

$-2=3x\;\;?\;\;\;x\ge3$ ....No, since the minimum value of 3x is 9.

Case 2:

$2x-6\le0,\;\;5x-4\ge0\;\;\;\Rightarrow\ 5x\ge4,\;\;x\le3$

$-(2x-6)=5x-4,\;\;\;\frac{4}{5}\le x\le3$

$6-2x=5x-4\Rightarrow\ 10=7x\;\;?\;\;\;\frac{4}{5}\le x\le3$

$x=\frac{10}{7}$

Case 3: Both negative within modulus

$x\le\frac{4}{5}$

$-(2x-6)=-(5x-4)\Rightarrow\ 6-2x=4-5x\Rightarrow\ 3x=-2\;\;?\;\;\;x\le\frac{4}{5}$

$x=-\frac{2}{3}$
• Feb 2nd 2011, 05:34 PM
jam2011
wow thats great...thanks a lot for all of your information. i need to study more further. God bless you sir.