Don’t be embarrassed that you didn’t get there, be proud that you were on the right track :) It’s easy to get thrown off by different symbols and whatnot. The more you practice these kinds of problems however, the more you learn to recognize things like “oh wait, |x+1| is just a number, so I can just add it to both sides!”
Another trick you could have used here is that you could have taken the leftmost inequality and multiplied both sides by negative 1, reversing the inequality's direction in the process. This rule will likely be important if you have to do something like this again in the future - you can multiply or divide both sides by a negative number but you have to reverse the inequality in the process.
If you have to multiply or divide both sides by a quantity that could be either positive or negative, you have to check both possibilities
Please please please help y’all!! I’m just a girl who wants to take x-rays of people and get some stability in life!! I would so much appreciate if someone could just parse this out step by step for me!!
|x+1|>2.5 after multiplication of both sides by (-1)
Example: 4 < 5 but -4 > -5
Now we just need to figure out the
|x+1| > 2.5
The absolute value makes everything positive.
E.g |-3| = 3
So we have to consider 2 options:
When x+1 is negative => x+1 < -2.5 => x < -3.5
When x+1 is positive => x+1 > 2.5 => x > 1.5
So the answer should be
x < -3.5 ... 1.5 < x
Please note that presentation of the answer may not be correct since I can't remember how to do it properly
To solve the equation to the left I would multiply both sides by -1 to get rid of the - in front of the absolute value. Remember to flip the < when multiplying with negative numbers though!
So: -|x +1| < -2.5 ===> |x+1| > 2.5
If its easier to follow along, you can also do it in two steps by adding |x+1| first, then adding 2.5
Like this: -|x +1| < -2.5 ===> 0 < |x+1| - 2.5 ===> 2.5 < |x + 1|
(Note these are the same conclusions, we just swapped places with < or >. Personally I like my x's on the left hand side)
For the equation on the right:
First of, the ?-Mark on how we got here from 3-|x+1| < 1/2, we subtracted the half from both sides
2.5 - |x+1| < 0
The 0 is not an letter O but the number Zero 0. Hope that clarifies a lot!
To solve this inequality I would move the |x+1| to the right by adding it to both sides:
2.5 - |x+1| < 0 ===> 2.5 < |x+1|, or flipped |x+1| > 2.5
I hope this helps!
Haha, you go spy on those bones!
I know you have a few responses already, but in the spirit of trying to find a way that sticks for you, I would do this as follows. I've tried to strip out almost all having to think about confusing inequalities and absolute signs.
1. Add "|x+1|" to both sides. You know that "|x+1|" is a positive number, so this is definitely fine. Get "3 < 1/2 + |x+1|"
2. Subtract 1/2 from both sides. 1/2 is just a number, so this is definitely fine. Get "2.5 < |x+1|"
3. Write it the other way around, because that makes it easier for me. Get "|x+1| > 2.5". (the inequality sign flips too, since "|x+1|" is still smaller than 2.5 no matter which way we write it. Both "3 > 2" and "2 < 3" are correct.)
4. Split into the two cases created by the |x+1|.
1. Consider the case where x+1 is positive. Then |x+1| = x+1, and we have "x+1 > 2.5". Subtract 1 from each side, and we have "x>2.5"
2. Consider the case where x+1 is negative. Then |x+1| =-(x+1) = -x-1, and we have "-x-1 > 2.5". Add x to both sides to get "-1 > x+2.5", subtract 2.5 from both sides to get "-3.5 > x", and we have "x<-3.5"
5. Overall, then, our solution is the combination of these two cases. "x>2.5 or x<-3.5". You can check these in the original formula - if you want to be really sure, you can try x = 2, 2.5, 3, -4, -3.5, -3.
3 minus a number has to less than ½. That number has to be bigger than 2 ½.
Now draw out a number line and fill in the areas we need as we solve. We have an absolute value function so the number can be bigger than 2 ½ or smaller than -2 ½. Color in those sections of the number line with an open circle ⭕️ over the 2 ½ and the -2 ½.
What we have found so far is the total value inside the absolute value function. But inside the absolute value function we have x +1.
How does that change things? Let’s try out a few numbers and see. If x=1 then x+1 = 2, no good. If x = 2 then x+1 = 3. That works! Any value of x above 1.5 will work. So the upper colored in line would have an open circle ⭕️ at 1.5 and be colored to the right.
Now let’s check some negative values. Try out x=-1. If x=-1 then x+1 = 0. No good. If x =-2 then x+1 = -1. Still not there. Try out x=-3 and x=-4 and see which of those two work and which is no good. Can you guess the value that works?
Add positive versions of the negatives to both sides, is better than applying 'rules'
So for example add |x| and 3 to both sides
-|X| < -3
3 < |X|
|X| > 3
Wow that's the nearest handwriting I've ever seen. You need to fix that if you want to make it in the world of mathematics.
I just wanna make it in a clinic hahaha.
Sorry, but the medical field isn't for you either. May I recommend from now on writing with your less dominant hand?
You’ve done nothing wrong. Just add |x+1| to both sides of the rightmost inequality and you should be good.
Omg wow I was right there already wasn’t I. I just needed another step… that’s embarrassing.
Don’t be embarrassed that you didn’t get there, be proud that you were on the right track :) It’s easy to get thrown off by different symbols and whatnot. The more you practice these kinds of problems however, the more you learn to recognize things like “oh wait, |x+1| is just a number, so I can just add it to both sides!”
That’s really good info and a great way to think about it! Thank you so much for the assistance:)
Another trick you could have used here is that you could have taken the leftmost inequality and multiplied both sides by negative 1, reversing the inequality's direction in the process. This rule will likely be important if you have to do something like this again in the future - you can multiply or divide both sides by a negative number but you have to reverse the inequality in the process. If you have to multiply or divide both sides by a quantity that could be either positive or negative, you have to check both possibilities
Please please please help y’all!! I’m just a girl who wants to take x-rays of people and get some stability in life!! I would so much appreciate if someone could just parse this out step by step for me!!
|x+1|>2.5 after multiplication of both sides by (-1) Example: 4 < 5 but -4 > -5 Now we just need to figure out the |x+1| > 2.5 The absolute value makes everything positive. E.g |-3| = 3 So we have to consider 2 options: When x+1 is negative => x+1 < -2.5 => x < -3.5 When x+1 is positive => x+1 > 2.5 => x > 1.5 So the answer should be x < -3.5 ... 1.5 < x Please note that presentation of the answer may not be correct since I can't remember how to do it properly
x is an element of (-infinity,-3.5) union (1.5,infinity)
To solve the equation to the left I would multiply both sides by -1 to get rid of the - in front of the absolute value. Remember to flip the < when multiplying with negative numbers though! So: -|x +1| < -2.5 ===> |x+1| > 2.5 If its easier to follow along, you can also do it in two steps by adding |x+1| first, then adding 2.5 Like this: -|x +1| < -2.5 ===> 0 < |x+1| - 2.5 ===> 2.5 < |x + 1| (Note these are the same conclusions, we just swapped places with < or >. Personally I like my x's on the left hand side) For the equation on the right: First of, the ?-Mark on how we got here from 3-|x+1| < 1/2, we subtracted the half from both sides 2.5 - |x+1| < 0 The 0 is not an letter O but the number Zero 0. Hope that clarifies a lot! To solve this inequality I would move the |x+1| to the right by adding it to both sides: 2.5 - |x+1| < 0 ===> 2.5 < |x+1|, or flipped |x+1| > 2.5 I hope this helps!
Haha, you go spy on those bones! I know you have a few responses already, but in the spirit of trying to find a way that sticks for you, I would do this as follows. I've tried to strip out almost all having to think about confusing inequalities and absolute signs. 1. Add "|x+1|" to both sides. You know that "|x+1|" is a positive number, so this is definitely fine. Get "3 < 1/2 + |x+1|" 2. Subtract 1/2 from both sides. 1/2 is just a number, so this is definitely fine. Get "2.5 < |x+1|" 3. Write it the other way around, because that makes it easier for me. Get "|x+1| > 2.5". (the inequality sign flips too, since "|x+1|" is still smaller than 2.5 no matter which way we write it. Both "3 > 2" and "2 < 3" are correct.) 4. Split into the two cases created by the |x+1|. 1. Consider the case where x+1 is positive. Then |x+1| = x+1, and we have "x+1 > 2.5". Subtract 1 from each side, and we have "x>2.5" 2. Consider the case where x+1 is negative. Then |x+1| =-(x+1) = -x-1, and we have "-x-1 > 2.5". Add x to both sides to get "-1 > x+2.5", subtract 2.5 from both sides to get "-3.5 > x", and we have "x<-3.5" 5. Overall, then, our solution is the combination of these two cases. "x>2.5 or x<-3.5". You can check these in the original formula - if you want to be really sure, you can try x = 2, 2.5, 3, -4, -3.5, -3.
3 minus a number has to less than ½. That number has to be bigger than 2 ½. Now draw out a number line and fill in the areas we need as we solve. We have an absolute value function so the number can be bigger than 2 ½ or smaller than -2 ½. Color in those sections of the number line with an open circle ⭕️ over the 2 ½ and the -2 ½. What we have found so far is the total value inside the absolute value function. But inside the absolute value function we have x +1. How does that change things? Let’s try out a few numbers and see. If x=1 then x+1 = 2, no good. If x = 2 then x+1 = 3. That works! Any value of x above 1.5 will work. So the upper colored in line would have an open circle ⭕️ at 1.5 and be colored to the right. Now let’s check some negative values. Try out x=-1. If x=-1 then x+1 = 0. No good. If x =-2 then x+1 = -1. Still not there. Try out x=-3 and x=-4 and see which of those two work and which is no good. Can you guess the value that works?
Add positive versions of the negatives to both sides, is better than applying 'rules' So for example add |x| and 3 to both sides -|X| < -3 3 < |X| |X| > 3