|5x+4| - |3x+7| = 5 (1 Viewer)

[coroneos]

How do you graph this?

 

[Rahul]

you could try graphing say y = |5x+4| and y = |3x+7| + 5 and finding the points of interception.

 

[freaking_out]

Originally posted by coroneos
How do you graph this?

i think by addition of ordinates...:s

 

[Rahul]

Originally posted by freaking_out
i think by addition of ordinates...:s

wft??? nvm, i would stick to my methods way too late now to change

 

[freaking_out]

now that i look at it, the question seems beyond 3u!!!

maybe try plotting.

 

[...]

Originally posted by freaking_out
now that i look at it, the question seems beyond 3u!!!

oo..thanks for that

 

[Rahul]

not really beyond 3u...

graph y = |5x+4|, then label the two lines with the equations.
then graph y = |3x+7| + 5 , and label the two lines. and then equate the ones that meet and find the points. that will give you the values for x

 

[haz_it_all]

wats wrong wit jus subbin in values??? it'll get the marks

 

[freaking_out]

Originally posted by coroneos
How do you graph this?

what i meant is that u can't be asked to "graph" it coz, where is the y then in the equation??? i reckon u'll prolly only b asked to solve it, in where (like rahul said) you'll individually draw LHS and RHS and find the points of intersection(note this is not graphing!!).

but ofcourse correct me if i'm wrong.

edit: btw, what does the book have as a solution?

 

[redslert]

Originally posted by haz_it_all
wats wrong wit jus subbin in values??? it'll get the marks

you can sub in all the values u want and draw the pretties graph.....but if the question asks you to determine the values and equations of each line......YOU ARE SCREWED!

the drawing of inequality graphs is a long process.....
let me get out pen and paper to do it first

 

[freaking_out]

Originally posted by haz_it_all
wats wrong wit jus subbin in values??? it'll get the marks

actually how u gonna sub in a value for the above...coz there's no y in the equation!!!

na like i said i can't c how u can b asked to graph that.

 

[redslert]

bah

wrong solution

 

[Leasha23]

Originally posted by redslert
ok here is my solution to:
|5x + 4| - |3x + 7| = 5

for |5x + 4|:
= 5x-4 if 5x-4>=0 ---> .'.x>= 4/5
= -(5x-4) if 5x-4<0 ---> .'. x < 4/5

for |3x + 7|:
= 3x+7 if 3x+7>=0 ---> .'.x>= -7/3
= -(3x+7) if 3x+7<0 ---> .'. x < -7/3

if x<0 : |5x + 4| - |3x + 7| = 5
(-)(5x + 4) - (-)(3x + 7) = 5
x = 3

if 0<=x<4/5 : |5x + 4| - |3x + 7| = 5
(-)(5x + 4) - (3x + 7) = 5
x = -1

if x>=4/5 : |5x + 4| - |3x + 7| = 5
(5x + 4) - (3x + 7) = 5
x = 8

.'. the inequality is the graph of:
x= -1
x= 3
x= 8

and then u draw this simple graph

i believe it is right....not 100% sure......
tell me if it is wrong

this question is NO WAY beyond 3u!

Hmm theoretically correct though you left out one little bit:

x = 3 is not within the region of x<0 (as it is in your answer)
x = -1 is not with in the region of 0<=x<4/5
However, x=8 IS within the region of x>4/5
Therefore, the only correct solution is x= 8

 

[redslert]

Originally posted by Leasha23
Hmm theoretically correct though you left out one little bit:

x = 3 is not within the region of x<0 (as it is in your answer)
x = -1 is not with in the region of 0<=x<4/5
However, x=8 IS within the region of x>4/5
Therefore, the only correct solution is x= 8

ARRRRRRR
thanks for pointing this out!

I MADE TWO BIG MISTAKES!

firstly when i copy the question onto paper i mixed up the + and - signs

ok ill do it again

 

[redslert]

OK SECOND TIME, should be rite!

solution to:
|5x + 4| - |3x + 7| = 5

for |5x + 4|:
= 5x+4 if 5x+4>=0 ---> .'.x>= -4/5
= -(5x+4) if 5x+4<0 ---> .'. x < -4/5

for |3x + 7|:
= 3x+7 if 3x+7>=0 ---> .'.x>= -7/3
= -(3x+7) if 3x+7<0 ---> .'. x < -7/3

if x<-7/3 : |5x + 4| - |3x + 7| = 5
(-)(5x + 4) - (-)(3x + 7) = 5
x = -1
BUT DOES NOT MATCH CONDITION (x<-7/3)
.'. x cannot equal -1

if -7/3<=x<-4/5 : |5x + 4| - |3x + 7| = 5
(-)(5x + 4) - (3x + 7) = 5
x = -2 <-----matches condition (-7/3<=x<-4/5)

if x>=-4/5 : |5x + 4| - |3x + 7| = 5
(5x + 4) - (3x + 7) = 5
x = 4 <-----matches condition (x>=-4/5)

.'. the inequality is the graph of:
x= -2
x= 4

and then u draw this simple graph
OK THIS IS 110% correct!
i checked with grafeq

 

[xiao1985]

can't we jsut assume cases like when x < - 7/3 and -7/3 < x < - 4/5, and x > 4/5???

cuz that how i would do it, simple and fast

 

[redslert]

yes u can if the question doesn't worth much

but when it is worth 2-3marks you are expected to show working....

 

[eeyore]

graph 5x+4, and -5x-4
graph 3x+7 and -3x-7
then just subtract the 2nd graph from the first
then graph y=5 and see where it crosses your new graph
you'd probably have to equate them to find exact solutions

 

[freaking_out]

Originally posted by eeyore
graph 5x+4, and -5x-4
graph 3x+7 and -3x-7
then just subtract the 2nd graph from the first
then graph y=5 and see where it crosses your new graph
you'd probably have to equate them to find exact solutions

yeah, graphing is the best way for this question i reckon, just graph y=|5x+4| and y=|3x+7| + 5, and find the points of intersection...this method i reckon is less prone to mistakes.