inequality problem (1 Viewer)

Tryingtodowell

Active Member

Please someone help me solve this. They did it graphically but I don't get a thing they did and if theres an easier way than graphing method then please tell me howw

kendricklamarlover101

Member
u can consider cases as |x+1| = -(x+1) for x< -1 and |x-5| = -(x-5) for x < 5.
then from there u just solve the inequality for each case (x<-1, -1<x<5 , x > 5)
for example -1<x<5 we have:
x+1-(x-5) > 7 => -4 > 7 which isnt true so theres no value of x between -1 and 5 that satisfies the inequality
for x<-1:
-(x+1)-(x-5) > 7 => -2x-6>7 => x<1/2 but x is less than -1 so all values less than -1 are greater than 7
a similar process can be done for x>5

Tryingtodowell

Active Member
u can consider cases as |x+1| = -(x+1) for x< -1 and |x-5| = -(x-5) for x < 5.
then from there u just solve the inequality for each case (x<-1, -1<x<5 , x > 5)
for example -1<x<5 we have:
x+1-(x-5) > 7 => -4 > 7 which isnt true so theres no value of x between -1 and 5 that satisfies the inequality
for x<-1:
-(x+1)-(x-5) > 7 => -2x-6>7 => x<1/2 but x is less than -1 so all values less than -1 are greater than 7
a similar process can be done for x>5
for each cases how do you know which one will be negative or positive

Drongoski

Well-Known Member
To me graphing is easier like this:

Draw the number line and mark off the 2 numbers "-1" and "5". The distance between these 2 numbers = 5-(-1) = 6. Now read the inequality this way: the distance of the number x from "-1" plus the distance of x from the number "5" is greater than 7. Now x cannot be a number within the interval: $\bg_white -1 \leq x \leq 5$. So x must be outside this interval, to the right of 5 or to the left of -1. In fact we only need x to be greater than 0.5 (2 x 0.5 = 1 = 7 - 6)) to the right of "5" or to the left of "-1", that is $\bg_white x \leq -1.5 or x \geq 5.5$

Edit: typo, last line, corrected.

This wordy explanation may make this sound complicated. A simple diagram would show you how easy it is.

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Average Boreduser

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but I dont understand how to do it
Quickest way is by adding y values of both graphs by plotting points and then using those points, you can make a general graph. (this case makes it rlly easy as its an absolute value function that is translated on the y-axis).

Drongoski

Well-Known Member
but I dont understand how to do it
Follow what i said. Draw a line and mark off -1 and 5. any number x between and including -1 and 5 will have |x+1| + |x-5| = 6 (a fixed sum - so we need at least 1 more for this sum) So x must be outside this interval; x now only needs to be more than 0.5 beyond the 2 numbers -1 and 5; i.e. at least 0.5 to the left of -1 or to the right of 5 along the (number) line.

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Tryingtodowell

Active Member
To me graphing is easier like this:

Draw the number line and mark off the 2 numbers "-1" and "5". The distance between these 2 numbers = 5-(-1) = 6. Now read the inequality this way: the distance of the number x from "-1" plus the distance of x from the number "5" is greater than 7. Now x cannot be a number within the interval: $\bg_white -1 \leq x \leq 5$. So x must be outside this interval, to the right of 5 or to the left of -1. In fact we only need x to be greater than 0.5 (2 x 0.5 = 1 = 7 - 6)) to the right of "5" or to the left of "-1", that is $\bg_white -1.5 \leq x or x \geq 5.5$

This wordy explanation may make this sound complicated. A simple diagram would show you how easy it is.
uhhh

Tryingtodowell

Active Member
To me graphing is easier like this:

Now x cannot be a number within the interval:
Whattt???

Tryingtodowell

Active Member
To me graphing is easier like this:

Draw the number line and mark off the 2 numbers "-1" and "5". The distance between these 2 numbers = 5-(-1) = 6. Now read the inequality this way: the distance of the number x from "-1" plus the distance of x from the number "5" is greater than 7. Now x cannot be a number within the interval: $\bg_white -1 \leq x \leq 5$. So x must be outside this interval, to the right of 5 or to the left of -1. In fact we only need x to be greater than 0.5 (2 x 0.5 = 1 = 7 - 6)) to the right of "5" or to the left of "-1", that is $\bg_white -1.5 \leq x or x \geq 5.5$

This wordy explanation may make this sound complicated. A simple diagram would show you how easy it is.
Im trying so hard to understand this but I dont get the rest of it from...

Now x cannot be a number within the interval: $\bg_white -1 \leq x \leq 5$. So x must be outside this interval, to the right of 5 or to the left of -1. In fact we only need x to be greater than 0.5 (2 x 0.5 = 1 = 7 - 6)) to the right of "5" or to the left of "-1", that is $\bg_white -1.5 \leq x or x \geq 5.5$

Tryingtodowell

Active Member
Follow what i said. Draw a line and mark off -1 and 5. any number x between and including -1 and 5 will have |x+1| + |x-5| = 6 (a fixed sum - so we need at least 1 more for this sum) So x must be outside this interval; x now only needs to be more than 0.5 beyond the 2 numbers -1 and 5; i.e. at least 0.5 to the left of -1 or to the right of 5 along the (number) line.
I still dont get itt

Tryingtodowell

Active Member
Try viewing it on desmos.
Thats easy for you to say but it doesnt help when I cant do literally anything at all

kkk579

hello
Just split it up into different domains and take the specific positive and negative values for that, then solve each section like a normal inequality

kkk579

hello
You should probably search it up and watch a yt vid on how to do it it would be easier to understand

Drongoski

Well-Known Member
Whattt???
Say you choose any x between -1 and 5, x = 2, say. then |x+1| + |x-5| = |2+1| + |2 - 5| = 6. Choose another such number, say x = -0.7; then |x+1| + |x-5| = |-0.7 + 1| + |-0.7 -5| =0.3 + 5.7 = 6 (again). If you choose x = -3 (which is more than 0.5 to the left of -1), then |-3+1| + |-3-5| = 2 + 8 = 10 (this is greater than 7). If x = 5.2 (which is NOT more than 0.5 to the right of 5), |5.2 + 1| + |5.2-5| = 6.2 + 0.2 = 6.4 (not more than 7); so 5.2 is not a solution. You only need to look at the question geometrically; no algebra of inequalities needed.

Note my correction to typo: $\bg_white x \leq -0.15 or x \geq 5.5$

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Tryingtodowell

Active Member
omgg I think I understand it now tyy

Tryingtodowell

Active Member
Say you choose any x between -1 and 5, x = 2, say. then |x+1| + |x-5| = |2+1| + |2 - 5| = 6. Choose another such number, say x = -0.7; then |x+1| + |x-5| = |-0.7 + 1| + |-0.7 -5| =0.3 + 5.7 = 6 (again). If you choose x = -3 (which is more than 0.5 to the left of -1), then |-3+1| + |-3-5| = 2 + 8 = 10 (this is greater than 7). If x = 5.2 (which is NOT more more than 0.5 to the right of 5), |5.2 + 1| + |5.2-5| = 6.2 + 0.2 = 6.4 (not more than 7); so 5.2 is not a solution. You only need to look at the question geometrically; no algebra of inequalities needed.

Note my correction to typo: $\bg_white x \leq -0.15 or x \geq 5.5$
I feel bad but I still dont get yours -> like using '0.5' and stuff

Drongoski

Well-Known Member
I feel bad but I still dont get yours -> like using '0.5' and stuff
OK. Say you take x = 5.7 say. Then this number is |5.7 + 1| = 1 + 5.7 = 6.7 away from the number -1, and |5.7 -5| = 0.7 away from the number 5; so it is = the constant 6 plus 2 x 0.7 from the 2 numbers. Remember, any x outside the closed interval [-1,5] is 6 + twice its distance from the nearer of the 2 numbers -1 and 5. Remember |x+1| is |x -(-1)| and is its distance (how far away from) from the number -1, just as |x-5| is the distance of x from the number 5.
Any x within the interval [-1,5] has a total distance of 6 (always) from -1 and from 5; so we are short 1 to make it at least 7. So any x outside this interval only needs an additional 0.5 or more from each of the 2 numbers -1 and 5 and this additional distance occurs twice, one from -1 and the other from 5.

I think I've confused you even more now. So from a very easy geometrical concept, I've succeeded in making it look really hard.

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