I'm having some trouble working out some questions for absolute values:

|4y - 7| = |3y - 2|

1 / |y-2| is greater than or equal to 3

|5y - 2| = |3x + 4|

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- Jan 26th 2008, 04:13 PMOgsta101Absolute Value Issues
I'm having some trouble working out some questions for absolute values:

|4y - 7| = |3y - 2|

1 / |y-2| is greater than or equal to 3

|5y - 2| = |3x + 4| - Jan 26th 2008, 04:40 PMDivideBy0
Question 2

$\displaystyle \frac{1}{|y-2|}\geq 3$

Since the LHS must be positive, we can reciprocate both sides, reversing the inequality sign:

$\displaystyle |y-2| \leq \frac{1}{3}$

$\displaystyle \Rightarrow -\frac{1}{3}\leq y-2\leq \frac{1}{3}$

$\displaystyle \Rightarrow \frac{5}{3} \leq y \leq \frac{7}{3}$ - Jan 26th 2008, 05:16 PMJhevon
bearing in mind the definition of absolute values (look this up of you don't know), we realize the solution will be given by:

$\displaystyle 4y - 7 = \pm (3y - 2)$

so we have $\displaystyle 4y - 7 = 3y - 2$

or

$\displaystyle 4y - 7 = 2 - 3y$

solve those for your two solutions.

Quote:

|5y - 2| = |3x + 4|

- Jan 26th 2008, 06:55 PMThePerfectHacker
Just square both sides to get right rid of absolute values. Then see which solutions to the quadradic satisfy the original equation.

- Feb 1st 2008, 08:25 AMstarrynight
Absolute value is when anything in between the || is going to end up in the positive state of a number. Your first question states :

|4y - 7| = |3y - 2|

so when you look at it, it should be like this:

4y + 7 = 3y + 2

4y-3y=-7+2

If you are supposed to solve the answer would be:

y= -5 :) - Feb 2nd 2008, 02:48 PMJhevon
you seem to be confused about absolute values. it does not mean you just change all the signs to positive. first of all, there are two answers, second of all, none of them is the answer you gave (plug it into the equation and you'll see it is wrong). do as i did or what TPH did (i prefer his method, and it is more standard than mine) to solve the problem. and look up the definition of what it means to take absolute values