Can Someone help me solve these two problems
Absolute value of 2x+1 = x-1 (two solutions i think?)
and
square root of 3x+1 = 3 + square root of x-4
please help me solve
In order to solve an absolute value problem, you have to solve two separate equations. The first will be $\displaystyle 2x + 1 = x - 1$ and the second will be $\displaystyle -2x - 1 = x - 1$ or equivalently, $\displaystyle 2x + 1 = -x + 1$.
For the square root problem you begin with $\displaystyle \sqrt{3x + 1} = 3 + \sqrt{x-4}$. What you need to do is square both sides, then put everything that's not under a square root on one side, and square both sides again.
$\displaystyle |2x+1| = x-1$
You can pull the one out since it is always positive:
$\displaystyle |2x|+1 = x-1$
$\displaystyle |2x|-x = -2$
From here it should be clear that, as you have already discovered, there are no solutions, because no matter the value of x, |2x| will always be greater than it, and so |2x|-x will always be greater than or equal to zero, and thus can never equal -2.
Be sure to see that the method that icemanfan showed you still works; you just have to be careful about your signs:
$\displaystyle \left\lvert2x + 1\right\rvert = x - 1$
$\displaystyle \Rightarrow\left(x\ge-\frac12\text{ and }2x + 1 = x - 1\right)\text{ or }\left(x<-\frac12\text{ and }-2x - 1 = x - 1\right)$
Considering the two cases separately:
$\displaystyle 2x + 1 = x - 1\Rightarrow x = -2$, but this contradicts our requirement that $\displaystyle x\ge-\frac12,$ so it is not a solution.
$\displaystyle -2x - 1 = x - 1\Rightarrow x = 0$, but this contradicts our requirement that $\displaystyle x<-\frac12,$ so it is also not a solution.
These are the only two possibilities (from the trichotomy law), so the original equation has no solutions.
It is, however, always a good idea to check your values in the original equation, because unless you are very careful in performing operations you can easily end up with extraneous (and sometimes missing) solutions.
Glad you caught that, Moo.
Tread carefully, angel.white . In general,
$\displaystyle \left\lvert a+b\right\rvert\ne\left\lvert a\right\rvert + \left\lvert b\right\rvert,$
but you can say that
$\displaystyle \left\lvert a+b\right\rvert\le\left\lvert a\right\rvert + \left\lvert b\right\rvert$
as Moo pointed out. This is known as the triangle inequality.