Question

Solve the equation
2x/x^2+2x-3+3/x+3=1
Please show steps this one got me the most baffled so far. Thanks in advance

Answer 1

can you factorr
x^2+2x-3 ??

Answer 2

i assume its,
\(\large \dfrac{2x}{x^2+2x-3}+\dfrac{3}{x+3}=1\)

Answer 3

yes that is correct
I was putting it in equation as we speak.

Answer 4

so, factor, x^2+2x-3

Answer 5

uhhh I need to check my notes again... I don't think I can

Answer 6

oh wait just a sec

Answer 7

x^2+x-3

Answer 8

umm....what was that ?
to factor x^2+2x-3, you need to find 2 numbers whose sum is +2 and product = -3
which are those numbers ?

Answer 9

yup then I have no idea... would you be as kind as to show me how I would get those numbers? I have a bunch other I'll try tonight.

Answer 10

" 2 numbers whose sum is +2 and product = -3"
which numbers can have product as -3, very limited set....try to think, its easy

Answer 11

like for example, if i need " 2 numbers whose sum is 5 and product = 6"
it'd be 2 and 3

Answer 12

it would be easier if you list out the factors of 3

Answer 13

how about 3 and -1 ?
required sum was 2

Answer 14

thank you so much for guiding me through this btw...

Answer 15

ok so now after I factor the numbers what next?

Answer 16

x^2+2x-3 = (x+3)(x-1)
right ?

Answer 17

yes I believe so

Answer 18

so we have
\(\large \dfrac{2x}{(x+3)(x-1)}+\dfrac{3}{x+3}=1\)

Answer 19

do you know what we can do, so as to make the denominator of both the terms same ?

Answer 20

did we lost the original exponential x^2 under x2?

Answer 21

oh nvm nvm nvm haha and yes we multiply (x-1) and get rid of it?

Answer 22

or how about this ?
\(\large \dfrac{2x}{(x+3)(x-1)}+\dfrac{3}{x+3}\times \dfrac{x-1}{x-1}=1\)
multiplying and dividing by x-1 to make the denominators same ...

Answer 23

did you notice what i did ?

Answer 24

yes I think so in the first part you divided x-1 and on the second part multiply it, thus creating equal denominator

Answer 25

2x/x-3+3/x-3-1 right?

Answer 26

=1 I mean

Answer 27

now since the denominator are same, we can combine the numerators,
\(\large \dfrac{2x}{(x+3)(x-1)}+\dfrac{3}{x+3}\times \dfrac{x-1}{x-1}=1
\implies \large \dfrac{2x+3(x-1)}{(x+3)(x-1)}=1

\)
ask if any doubts here.

Answer 28

like \(\large \dfrac{☺}{♥}+\dfrac{☻}{♥}=\dfrac{☺+☻}{♥}\)

Answer 29

yeah make it into simple fraction form

Answer 30

now can you proceed further ? know the next step ?

Answer 31

k let me try it

Answer 32

I got 6x-3/x-3

Answer 33

argh I feel I almost got it since x-3/x-3=1

Answer 34

\( \large \dfrac{2x+3(x-1)}{(x+3)(x-1)}=1 \\ 2x+3(x-1)=(x+3)(x-1)\)
just multiplied (x+3)(x-1) on both sides.
now you'll get a quadratic in x

Answer 35

give me a second I need to give my mother her meds be right back

Answer 36

okay.

Answer 37

thank you so much my friend I will have to recheck my notes regardless.

Answer 38

as I am having trouble understanding it at all and I don't want a straight forward answer. All your help was much appreciated :)

Answer 39

hmm.....do that and practice alot, you'll get it soon.