solving modulus equations

Hi guys,
I'm really struggling to understand how to sketch the modulus graphs and also use that graph to solve the equations . in the example blow part a, should i just sketch the graph of 2x+1 first and then reflect the negative y-values in x-axis?

You can ignore the 4 and the minus first, and just think about the graph |2x + 1|. This graph is quite simple: for all x < -1/2 you will have that 2x + 1 is negative, right? And for all x > -1/2, 2x + 1 is non-negative. Hence their will be a "cusp" where x = -1/2, i.e. you reflect in the line x = -1/2. Then to obtain the new graph all you have to do is reflect that in the y axis, since its minus, and translate it, to account for the +4.

do you think my way of dealing with this is also correct as well?first sketch the graph of 2x+1 and then reflect the negative y-valus in the x-axis to get |2x+1|, the multiply y-coordinates of any turning point of the graph by -1 to get -|2x+1| and at the end just translate by 4 vertically

oops I meant reflect in the x-axis in my previous post by the way. Yes, that is right I think
Well I mean you multiply everything by -1 I'm not sure what you mean by multiplying the coordinates of the turning point

Like every single point on the graph should be multiplied by -1?

Well yeah once you've got |2x + 1|, to get -|2x +1|, you flip everything in the x-axis. And then add the 4 to get the final graph

how about solving the equation in part b?im not sure what method would be the best to use as i keep getting wrong answers for part b

Do it by case analysis. Always think of |x| as a function made up of pieces

for x >/= 0, |x| = x
for x < 0, |x| = -x

And you can do the same thing for |2x + 1|, where the critical point is x = -1/2 rather than x = 0. Then, depending on the value of x, you have an explicit, direct equation to solve, without moduli in the way. You just need to check that whatever answer you get is actually in the range you're considering.

So for an example, if you had 3 - |x| = 2, you'd say first look at x >/= 0, this gives the equation 3 - x = 2 giving the solution x = 1, which is in the range you're considering, so is a solution. Then you'd look at x < 0, giving the equation 3 + x = 2, yielding x = -1, which again is in the range, so is a solution. This problem is no different, just the expression is a tiny bit more complicated.

(by the way to clarify I know that is a dumb example because you can rearrange, but I use it to illustrate a simple case..)

so should i start by considering x>-1.5 and x<-1.5?

well the important value here is x = -1/2 as this is where 2x + 1 = 0.

so for x>-1.5 we have 2x+1=4 to solve and for x<-1.5 we have -2x-1=4 to solve ?

well it looks like you're trying to solve for RHS = 0 there (also as I said we should be looking at -1/2 not -1.5). You have either

x = 4 - (2x + 1)
or
x = 4 + (2x + 1)

depending on which side of -1/2 you are on

Oh sorry i meant to say -0.5 . so we've got x=1.5 and x=-1.5 as our two solutions?

solutions I get are 1 and -5

Dont you have to rearrange the two above equations to find the value of x?

yes, I don't see where you get x = +/-1.5 from though

Move 4 and 1 to the other side and then devide by 2 to make x the subject?

x = 4 - (2x + 1) gives x = 4 - 2x - 1 so rearranging 3x = 3 hence x = 1
x = 4 + (2x + 1) gives x = 5 + 2x so rearranging x = -5

My mistake thanks for you time