# Thread: Problem Check - Absolute Value: solving |x+1| + |x−2| = 7

1. ## Problem Check - Absolute Value: solving |x+1| + |x−2| = 7

Hey all. Im back. Quick question to test my understanding. The problem is, Find all numbers x satisfying the given equation.

The equation is: $|x+1|+|x-2| = 7$.

3 Cases:
a) $x < -1$.
b) $x > 2$.
c) $-1 \le x \ge 2$.

a) $x < -1$. In this case $x+1 < 0$ which means that $|x+1| = -(x+1)$ and $x-2 < 0$ which means that $|x-2| = -(x-2)$.

solve: $-(x+1)-(x-2) = 7 \rightarrowtail$(because i wanted to)$\rightarrowtail -x+1-x+2 = 7 \rightarrowtail -2x+3 = 7 \rightarrowtail -2x = 6 \rightarrowtail x = -3$.

check: $|x+1|+|x-2| = 7 \rightarrowtail |-3+1|+|-3-2| = 7 \rightarrowtail |-2|+|-5| = 7 \rightarrowtail 2+5 = 7 \rightarrowtail 7 = 7$.

Thus, x<-1.

b) $x > 2$. In this case $x+1 > 0$ which means that $|x+1| = (x+1)$ and $x-2 > 0$ which means that $|x-2| = (x-2)$.

solve: $(x+1)+(x-2) = 7 \rightarrowtail x+1+x-2 = 7 \rightarrowtail 2x-1 = 7 \rightarrowtail 2x = 8 \rightarrowtail x = 4$.

check: $|x+1|+|x-2| = 7 \rightarrowtail |4+1|+|4-2| = 7 \rightarrowtail |5|+|2| = 7 \rightarrowtail 7 = 7$.

Thus, x >2.

c) $-1 \le x \ge 2$. In this case $x+1 > 0$ which means that $|x+1| = (x+1)$ and $x-2 < 0$ which means that $|x-2| = -(x-2)$.

solve: $(x+1)-(x-2) = 7 \rightarrowtail x+1-x+2 = 7 \rightarrowtail 3 \neq 7$. Not a Solution.

Thus the set of numbers x such that $|x+1|+|x-2| = 7$ is $\{-3,4\}$.

2. Originally Posted by doughishere
c) $1 \le x \ge 2$.
c) $1 \le x \ge 2$.
If you hadn't written it twice, I would have thought it a typo. You seem to think this means something. It doesn't.

Generally, your presentation is quite thorough. Try not to invent new notation. It's okay to write vertically and let the reader sort out a detail or two.

Note, also, that you can throw out many extraneous solutions simply by looking at the Domain. When you define x in (-1,2) and you get x = 7, there is no need to "check" that. Just throw it out.

3. Originally Posted by tkhunny
If you hadn't written it twice, I would have thought it a typo. You seem to think this means something. It doesn't.

Generally, your presentation is quite thorough. Try not to invent new notation. It's okay to write vertically and let the reader sort out a detail or two.

Note, also, that you can throw out many extraneous solutions simply by looking at the Domain. When you define x in (-1,2) and you get x = 7, there is no need to "check" that. Just throw it out.

Im trying to formulate it in my mind and thus write it down so that theres some order to it. Logically this is how i see the problem/answer and logically i wanted to present the 3 cases at the top because i though this was the hardest part....next i wanted to take each of the cases at the top and present them so that the reader and i have the same understanding of the problem and its solution.

.The goal is to be as complete and through as possible but also a template for the dozen or so problems that are similar. The arrows i admit were for flair.

The book specifically states that we need to check our answers...I assume for completeness thats also the case....in other words, "its nice to check the answer."

4. Originally Posted by doughishere
The book specifically states that we need to check our answers...I assume for completeness thats also the case....in other words, "its nice to check the answer."
You missed my point. If an algebraic manipulation leads you to a result that is not in the specified Domain, then it ISN'T "an answer". Thus, there is no requirement, even using the book's language, to check it. You should feel a mandate to "check" things that might be valid. There is no responsibility to check things that cannot possibly be valid. On the other hand, simply checking that something violates the Domain, may constitute "checking". Just don't do unnecessary things.

5. Originally Posted by doughishere
$1 \le x \ge 2$
You need to fix this notation.

Are you trying to refer to values of x between 1 and 2, inclusive?

If so, your cases do not address values of x from -1 up to but not including 1.

$1 \le x \ge 2$. In this case $x+1 > 0$ which means that $|x+1| = (x+1)$ and $x-2 < 0$ which means that $|x-2| = -(x-2)$.
Not when x = 2.

6. Originally Posted by mmm4444bot
You need to fix this notation.

Are you trying to refer to values of x between 1 and 2, inclusive?

If so, your cases do not address values of x from -1 up to but not including 1.

Not when x = 2.

Ok i get whats going on now. Its a simple typo....it should have been -1.

The 3 cases should be

a) $x < -1$.
b) $x > 2$.
c) $-1 \le x \ge 2$.

7. Originally Posted by doughishere
Ok i get whats going on now. Its a simple typo....it should have been -1.

The 3 cases should be

a) $x < -1$.
b) $x > 2$.
c) $-1 \le x \ge 2$.
Well, that's one correction made, yes. But what the other posters were trying to point out is that one of your signs is pointing the wrong way in case c. What you wrote doesn't at all say what you think it says. Think about what the two halves of the inequality are saying. $-1 \le x$ tells you that -1 is less than or equal to x. Or, in other words, x is greater than or equal to -1. All good so far, but now let's look at the second half. $x \ge 2$. This says that x is greater than or equal to 2.

Putting those together, the whole inequality says that x is greater than to equal to -1, and x is greater than or equal to 2. Oops! What you wrote is technically accurate, but it's vacuously true. Given $x \ge 2$, it automatically follows that $x \ge -1$ by the property of transitivity (because $2 \ge -1$). Also, aside from not representing the interval you desire, it's just poor, very confusing notation. If you were to punch that into a graphing calculator, like, say, Desmos, it would throw up an error: "Double inequalities must both go the same way, e.g. $1 \le y \le 2$."

8. Originally Posted by ksdhart2
Well, that's one correction made, yes. But what the other posters were trying to point out is that one of your signs is pointing the wrong way in case c. What you wrote doesn't at all say what you think it says. Think about what the two halves of the inequality are saying. $-1 \le x$ tells you that -1 is less than or equal to x. Or, in other words, x is greater than or equal to -1. All good so far, but now let's look at the second half. $x \ge 2$. This says that x is greater than or equal to 2.

Putting those together, the whole inequality says that x is greater than to equal to -1, and x is greater than or equal to 2. Oops! What you wrote is technically accurate, but it's vacuously true. Given $x \ge 2$, it automatically follows that $x \ge -1$ by the property of transitivity (because $2 \ge -1$). Also, aside from not representing the interval you desire, it's just poor, very confusing notation. If you were to punch that into a graphing calculator, like, say, Desmos, it would throw up an error: "Double inequalities must both go the same way, e.g. $1 \le y \le 2$."

I get it now. Thanks. And thanks to everyone.

9. Heres a question...if I have the equation $|x+1|+|x-2| +|x-5| = 7$ assuming i just wanted to look at the "cases", this time they would be

a) $x < -1$.
b) $-1 \le x \le 2$.
c) $2 < x \le 5$.
d) $5 < x$

is that right?

10. Other than the fact that it doesn't matter where you include "=", you seem to have the idea.

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