MedVision ad

inequality problem (2 Viewers)

Tryingtodowell

Active Member
Joined
Jan 18, 2024
Messages
282
Gender
Female
HSC
2025
1710840769103.png

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
 
Joined
Oct 22, 2023
Messages
79
Gender
Male
HSC
2024
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
Joined
Jan 18, 2024
Messages
282
Gender
Female
HSC
2025
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
Joined
Feb 22, 2009
Messages
4,255
Gender
Male
HSC
N/A
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: . 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

Edit: typo, last line, corrected.

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

Average Boreduser

Rising Renewal
Joined
Jun 28, 2022
Messages
3,158
Location
Somewhere
Gender
Female
HSC
2026
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
Joined
Feb 22, 2009
Messages
4,255
Gender
Male
HSC
N/A
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.
 
Last edited:

Tryingtodowell

Active Member
Joined
Jan 18, 2024
Messages
282
Gender
Female
HSC
2025
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: . 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

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

Tryingtodowell

Active Member
Joined
Jan 18, 2024
Messages
282
Gender
Female
HSC
2025
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: . 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

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: . 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
 

Tryingtodowell

Active Member
Joined
Jan 18, 2024
Messages
282
Gender
Female
HSC
2025
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 😭
 

Drongoski

Well-Known Member
Joined
Feb 22, 2009
Messages
4,255
Gender
Male
HSC
N/A
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:
 
Last edited:

Tryingtodowell

Active Member
Joined
Jan 18, 2024
Messages
282
Gender
Female
HSC
2025
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:
I feel bad but I still dont get yours -> like using '0.5' and stuff
 

Drongoski

Well-Known Member
Joined
Feb 22, 2009
Messages
4,255
Gender
Male
HSC
N/A
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.
 
Last edited:

Users Who Are Viewing This Thread (Users: 0, Guests: 2)

Top