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|
Question 2
$\displaystyle \frac{1}{|y-2|}\geq 3$
Since the LHS must be positive, we can reciprocate both sides, reversing the inequality sign:
$\displaystyle |y-2| \leq \frac{1}{3}$
$\displaystyle \Rightarrow -\frac{1}{3}\leq y-2\leq \frac{1}{3}$
$\displaystyle \Rightarrow \frac{5}{3} \leq y \leq \frac{7}{3}$
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.
what do you want to do here? you have two unknowns. which do you want to solve for...? is this a typo? in any case, try a similar method to what i did above|5y - 2| = |3x + 4|
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
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