Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x + 1| = 2x – 1

Can someone explain this to me
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thanks that helped your explanation is more simpler and fundamental!

the below explanation i did not get especially the last part where we need to substitute and check

(1) SUFFICIENT: Here, we are told that |x + 3| = 4x – 3. When dealing with equations containing variables and absolute values, we generally need to consider the possibility that there may be two values for the unknown that could make the equation work.
We can find the two possible solutions by setting the expression inside the absolute value equal to ± the expression on the right side of the equation. In this case, our two solutions are:

x + 3 = +(4x – 3) and x + 3 = –(4x – 3)
x + 3 = 4x – 3 x + 3 = –4x + 3
6 = 3x 5x = 0
2 = x x = 0

However, you need to be careful. When there is a variable outside the absolute value, both solutions are not always valid. We need to plug both x = 2 and x = 0 back into the original equation and test them.

|(2) + 3| = 4(2) – 3 and |(0) + 3| = 4(0) – 3
|5| = 8 – 3 |3| = –3
5 = 5 3 = –3

Therefore, x = 0 is not a valid solution and we know that 2 is the only possible solution. x is definitely positive.
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Whenever i see this type of question i start to solve for X. Now in this case, from st-1 i got X = 2 or 0 From st-2 also the same answer and i ticked E. Can someone explain how i should start to think for this type of questions?? however i forgot to put back tha values and check.

This appears every so often on absolute value questions. It relies on you not testing the possible solution. You must always verify the solutions you get in your first step. I've made this mistake several times

I did same and got tricked. Can someone explain why we are getting x=0 and it doesn't fit the equation. Are we doing anything wrong on absolute problems?
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Solving for x is not the best way to deal with this problem.

But if you do this way you should know that checking validity of the solutions is very important for absolute value questions.

When you solve these kind of questions with check point method (discussed at: some-inequalities-questions-93760.html?hilit=substitute%20method), you test validity of solution on the stage of obtaining these values (by checking whether the value is in the range you are testing at the moment) and if the value obtained IS in the range you are testing at the moment, you don't need to substitute it afterwards to check, you've already done the checking and if the value obtained IS NOT in the range you are testing at the moment you also don't need to substitute it afterwards to check, you've already done the checking.

When you just expand the absolute value once with negative sign and once with positive sign then you should check whether obtained solution(s) satisfy equation be substituting solution(s) back to the equation.

Check 3-steps approach in Wlaker's post on Absolute Value for more on this: math-absolute-value-modulus-86462.html

Hope it helps.
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