solving modulus equations

Hi Guys ,
I'm trying to solve a modulus equation but my answer keep coming different that what the text book suggests . any help?

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Reply 1

Farhan.Hanif93

Original post by Alen.m

Hi Guys ,
I'm trying to solve a modulus equation but my answer keep coming different that what the text book suggests . any help?

That's because you aren't being careful enough about where the functions inside the moduli are positive and negative. One thing I'd advise is to sketch the LHS and the RHS on the same graph (being careful to not sketch

y=|2x+1|

instead), to get a feel for which particular sections of each of the curves intersect.

Recall that the definition of the modulus is

Unparseable latex formula:

|f(x)|=\begin{cases} f(x) & \text{if} \ f(x) \geq 0 \\ -f(x) & \text{if} \ f(x)<0

.

So for the LHS, this means that

|3x+2| = 3x+2

whenever

x\geq -2/3

. Otherwise it's

-(3x+2)

for

x<-2/3

.

Can you write the RHS in a similar way, by identifying where it has to flip from +x to -x for

|x|

? Then you need to examine the 3 possible regions that arise for any intersections.

Reply 2

Alen.m

Original post by Farhan.Hanif93

y=|2x+1|

instead), to get a feel for which particular sections of each of the curves intersect.

Recall that the definition of the modulus is

Unparseable latex formula:

|f(x)|=\begin{cases} f(x) & \text{if} \ f(x) \geq 0 \\ -f(x) & \text{if} \ f(x)<0

.

So for the LHS, this means that

|3x+2| = 3x+2

whenever

x\geq -2/3

. Otherwise it's

-(3x+2)

for

x<-2/3

.

Can you write the RHS in a similar way, by identifying where it has to flip from +x to -x for

|x|

? Then you need to examine the 3 possible regions that arise for any intersections.

im ok with sketching graphs of two functions of f(x) and g(x) also finding the point of intersection but after that i dont know how to form the equations and solve to find x . Dont know how to realise weather f(x) is positive or negative or about g(x) weather it's positive or negative on the point where it intersects with f (x) :frown:

Reply 3

Farhan.Hanif93

Original post by Alen.m

I see. Just so I know where we're at, do you understand what I've done with

|3x+2|

in my post above? In that it's

+(3x+2)

wherever

3x+2

is positive/non-negative, and

-(3x+2)

whenever

3x+2

is negative?

If not, lets look at the RHS instead. Do you see that

|x|=x

for

x\geq 0

, and

-x

for

x<0

? Then

2|x|+1 = 2x+1

for

x\geq 0

and

2(-x)+1

for

x<0

?

The reason this helps is because then you can search for solutions in regions that define where the functions are positive and negative. Let's start from the far left, what do the LHS and the RHS look like if

x < -2/3

?

P.S. I'm about to head out so I may not be able to respond again until later. If anyone would like to continue where I've left off, please feel free, otherwise I'll respond when I return.

(edited 7 years ago)

Reply 4

Alen.m

Original post by Farhan.Hanif93

I see. Just so I know where we're at, do you understand what I've done with

|3x+2|

in my post above? In that it's

+(3x+2)

wherever

3x+2

is positive/non-negative, and

-(3x+2)

whenever

3x+2

is negative?

If not, lets look at the RHS instead. Do you see that

|x|=x

for

x\geq 0

, and

-x

for

x<0

? Then

2|x|+1 = 2x+1

for

x\geq 0

and

2(-x)+1

for

x<0

x < -2/3

?

P.S. I'm about to head out so I may not be able to respond again until later. If anyone would like to continue where I've left off, please feel free, otherwise I'll respond when I return.

Can i just ask by LHS you mean g (x) and by right hand side you mean f(x) right? Sorry im so bad at this but appreciate your time alot mate if you can expand it a bit for me please

Reply 5

Farhan.Hanif93

Original post by Alen.m

Can i just ask by LHS you mean g (x) and by right hand side you mean f(x) right? Sorry im so bad at this but appreciate your time alot mate if you can expand it a bit for me please

But LHS, I literally mean anything to the left of the equals sign as written. So it would be

|3x+2|

(which you called f(x) in your working) and the RHS is

2|x|+1

(which you called g(x) in your working).

Reply 6

Alen.m

Original post by Farhan.Hanif93

But LHS, I literally mean anything to the left of the equals sign as written. So it would be

|3x+2|

(which you called f(x) in your working) and the RHS is

2|x|+1

(which you called g(x) in your working).

How about this: pick any point in a given range where the point of intersection occurs and put it in the equation to find the sign of it? Would this work?