maths help (1 Viewer)

[youngminii]

Re: inequation

Oh I think I get it
Never learnt that..
Will keep that in mind, thanks

 

[bubblesss]

Re: inequation

can sumone verify my answer
find values of x for these simultaneous equations
x+4/x-6<0 and x-6/x-4>1

my ans was -4<x<4 book says -4<x<6?????????

 

[tommykins]

Re: inequation

x+4/x-6<0

(x+4)(x-6) < 0
Drawing the graph, whatever is negative is your domain.

-4 < x < 6

x-6/x-4>1
(x-6)(x-4) > (x-4)²
0 > (x-4)² - (x-6)(x-4) => 0 > 6(x-4)
Draw line y = 6(x-4 ), less than 0. at x < 4

We have -4 < x < 6
and x<4

thus the answer for BOTH the inequalities is -4 < x < 4. You are correct.

 

[bubblesss]

Re: inequation

any help guys!!!!! ive done the question its just the solutions that aren't coinciding???/

 

[bubblesss]

Re: inequation

x+4/x-6<0

(x+4)(x-6) < 0
Drawing the graph, whatever is negative is your domain.

-4 < x < 6

x-6/x-4>1
(x-6)(x-4) > (x-4)²
0 > (x-4)² - (x-6)(x-4) => 0 > 6(x-4)
Draw line y = 6(x-4 ), less than 0. at x < 4

We have -4 < x < 6
and x<4

thus the answer for BOTH the inequalities is -4 < x < 4. You are correct.

yay !!!!!!!!!!!!!! for once i got it correct!!!!!!

 

[youngminii]

Re: inequation

Couldn't you just have checked by plugging in x=5?
Which doesn't work, would've saved much time?

By the way, is that simultaneously solving them?
I was trying to do it the old fashioned simultaneous way

 

[bubblesss]

Re: inequation

Couldn't you just have checked by plugging in x=5?
Which doesn't work, would've saved much time?

By the way, is that simultaneously solving them?
I was trying to do it the old fashioned simultaneous way

yep thats simultaneously solving them. we could have checked by sub x=5 in both. but yeh i got carried away by fitzpatrick answers lol.

 

[tommykins]

Re: inequation

Couldn't you just have checked by plugging in x=5?
Which doesn't work, would've saved much time?

By the way, is that simultaneously solving them?
I was trying to do it the old fashioned simultaneous way

You can just solve both and graph it on the number line. Where the two domains intersect is your "overall/final" region.

 

[lyounamu]

Re: inequation

Can I just ask,
Why do you times both sides by (2-x)^2 at the start?
Can't you just times both sides by (2-x)?
I know it comes out wrong when you do it by the way I said to, but shouldn't you be able to?

Ok, the reason why we times both sides by (2-x)^2 at the start is to get rid of (2-x) at the bottom without changing the inequation sign. If you times both sides by (2-x), if x is to be in the domain of x>2, (2-x) can be negative. If you times it by negative, the sign must change as well.

 

[Undermyskin]

|x+3| = 2|y+1|/5
x+3 = 2/5(y+1) --> y = 5x/2 + 13/2
x+3 = -2/5 (y+1) --> y = -5/2 -17/2

 

[tacogym27101990]

can someone please help me on this locus question?
the distance of a point from the line x= - 3 is two fifths of its distance from the line y= -1. find the locus?

i swear this is like the 4th time youve asked this question??