|2y - 3|| - 7 < 0

Question

in standard form

Answer 1

Nicole, you have a typo in your problem somewhere. It needs an *even* number of absolute value bars, to be a proper equation -- otherwise, it's undefined! Assuming that third bar is extra, you'd break the problem down the same way you must for any absolute value problem: for each absolute value term, there are *two* possibilities that you MUST consider separately, in turn. Either the argument (what's inside the bars) is zero or positive, and then you can drop the bars and solve the inequality, OR it is negative and then the bars reverse the value (that is, turn it from negative to positive) and you replace the bars with a set of parentheses instead, but with a leading minus sign outside, and you solve that inequality, too. So your *two* equations you have to solve are:

(2y - 3) - 7 .lt. 0  (I'm on a Nook, it doesn't have "less than") where 2y - 3 .gt. or .eq. 0

And -(2y - 3) - 7 .lt. 0 where 2y - 3 .lt. 0

So, these two possibilities reduce to 3/2 .lt. or .eq. y .lt. 5

And -2 .lt. y .lt. 3/2

So, together, these two inequality solutions make the region:

-2 .lt. y .lt. 5

Note a couple of things here! First, *each* absolute value in an equation splits the problem into two SETS of problems without the absolute value. In each set of problems, one equation comes directly from your initial problem equation, and the other equation "checks" your hypothesis as to the + or - sign of the absolute value argument. So, if you had an equation with two absolute values in it, it could split into 2 x 2 = 4 SETS of equations. It's not that the equations are necessarily hard to solve, but you must stay organized about it, so that you don 't forget to apply ALL the limits that apply when you break down the absolute values.

And second, don't forget to always carry along the equation "proving" your assumption about + or - for each absolute value. It's like checking your work -- sometimes solutions to equations give absurd solutions that don't make sense for the original problem, and that's the case with solutions to absolute value problems.