Algebraic Equation

3((x+5)/(x-1)+4) .
4((x+5)/(x-1))+1

I think you have an simplification error

Start by simplifying the denominator-----
3((x+5)/(x-1)+4)=3[(x+5)+4(x-1)]/(x-1) =
3[x+5+4x-4]/(x-1) =3[5x+1]/(x-1)=
[15x+3]/(x-1) ……

Now simplify the numerator-----
4((x+5)/(x-1))+1=[4(x+5)+1(x-1)]/(x-1) =
[4x+20+x-1]/(x-1)=[5x+19]/(x-1)

Put back in form

{[15x+3]/(x-1)}/{[5x+19]/(x-1)} =
[15x+3]/ [5x+19]

At least that's what I come up when I performing the order of operations as written

Rap c∫:;?/

Hmmm, looks a bit confusing, how exactly did you go about simplifying the numerator??

I apologize I got the terms denominator and numerator reversed. Embarrassing

The way I read it there are two pairs of parenthesis in the expression

3((x+5)/(x-1)+4), first you take the expression (x+5)/(x-1)+4 simplyify it and multiply by three

(x+5)/(x-1)+4 = (x+5)/(x-1)+4(x-1)/(x-1) =

((x+5)+4(x-1))/(x-1) =

(x+5+4x-4)/(x-1) = (5x-1)/(x-1)

then multiply this expression by 3

3(5x-1)/(x-1) = (15x-3)/(x-1)

Rap c∫/

Yeh, you solved the first equation. Now how can you show that it's equal to:

3x+11
5x+19

Why not just simplify it by saying let

A = x+5 and B = x-1 (for x<>1) so B = A-6 (for x <>1)

Giving show

3(A/B + 4) = (3A - 4)/(5A -6)
4A/B + 1

So multiple all top and bottom terms on LHS by (A-6) [i.e. = B], to change the proof to be show:

3(A + 4(A-6)) = (3A - 4)/(5A -6)
4A + (A-6)

Note demominator is correct 5A - 6, or 5x + 25 - 6 = 5x + 19

But the numerator looks wonky - mistake in parenthesis somewhere?

As 3(A + 4(A-6)) = 15A - 72 = 15x + 75 - 72 = 15x + 3 = 3(5x + 1) = 3 (5A - 24)

Giving a RHS term

15x + 3
5x + 19

Are the parenthesis correct on you numerator?

Basically, it's saying

3* x+5 + 4
...... x-1

By the way instead of 3x+11, it's 3x+111

Sorry??? What is 3x+111 the numerator of the RHS???

I get 15x + 3 <> 3x + 11 or 3x + 111?

The numerators are the same on both sides. Mult through by 1/1 and eliminate. Then

15X+3=3X+11
12X=8
X=2/3

or

15X+3=3X+111
12X=108
X=9

Rap c∫/

Lol - that's cute!

But we are looking at inequalities trying to show they are equivalent for all discrete (x<>1) values of x, not trying to find a single x were the equations intersect!

Hmm, seems correct...Anyways, Here is my working so far:

Numerator for first value:
=(x+5+4x-4)/(x-1)
3(5x+1)/(x-1)
=(15x+3)/(x-1)

Denominator for first value:
=4(x+5)/(x-1)
=(4x+20)/(x-1)
(4x+20+x-1)/(x-1)
=(5x+19)/(x-1)

First Value:

(15x+3)/(x-1)
.(5x+19)/(x-1)

Overall:

(15x+3) = ....... 3x+111
(5x+19) ......... 5x+19

Soo eliminate the denominators and then solve for x, which is 9 and then substitute to prove they're equal!

THANKS TO ALL THAT HELPED, esp. RapRap

Err, as I said before - that doesn't show they are equal - which was the question you asked - it shows they intersect!

They question therefore that you should have asked for future sake is simply the LHS and find a value for x that solves the inequalities

and

if you are going to change the numerator of the RHS - do it clearly in the first post (and all following posts were you are referencing your working) - not cryptically in the seventh post!

It wasn't until you solved this that I understood the question you were actually trying to ask and the adjusted inequalities you were evaluating.

Good luck with your assignment!