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Mathematics 14 Online
OpenStudy (ammarah):

Solve. Check for extraneous solutions. x^2/x+5=4/x+5

ganeshie8 (ganeshie8):

start by multiplying both sides wid (x+5)

ganeshie8 (ganeshie8):

that gets rid of the denominators

OpenStudy (ammarah):

so far i just cross multiplied and got x^3+5x^=4x+20

OpenStudy (ammarah):

wait so is the answer x= +or -2?

OpenStudy (ammarah):

@ganeshie8

ganeshie8 (ganeshie8):

\(\large \frac{x^2}{x+5}=\frac{4}{x+5}\)

ganeshie8 (ganeshie8):

multiplying (x+5) both sides gives u : \(\large x^2 = 4\)

ganeshie8 (ganeshie8):

taking square-root gives : \(\large x = \pm 2\)

ganeshie8 (ganeshie8):

so, you're right !!

OpenStudy (ammarah):

ok great i have another one im confused on

ganeshie8 (ganeshie8):

And, do check if they satisfy the original equation or not.

OpenStudy (ammarah):

how y plugging in 0?

OpenStudy (ammarah):

by*

ganeshie8 (ganeshie8):

if they DONT satisfy the original equation, then they are called 'extraneous solutions'

ganeshie8 (ganeshie8):

just plugin x = 2 in the given equation, and see if u get same thing on both sides

OpenStudy (ammarah):

ok its satisfies

ganeshie8 (ganeshie8):

wat about x = -2 ?

OpenStudy (ammarah):

no

ganeshie8 (ganeshie8):

why no, i see x=-2 also satisfies

OpenStudy (ammarah):

oops i mean yes

OpenStudy (ammarah):

both satisfy

ganeshie8 (ganeshie8):

cool :) so both are good solutions oly. they're not extraneous/errenous/bad in any way ok

OpenStudy (ammarah):

how about this one: 3/x+2=x-3/2x+4

ganeshie8 (ganeshie8):

look at right side denominator, can u factor the GCF ?

OpenStudy (ammarah):

i simplified 2(x+2)

ganeshie8 (ganeshie8):

\(\large \frac{3}{x+2}=\frac{x-3}{2x+4}\) \(\large \frac{3}{x+2}=\frac{x-3}{2(x+2)}\) \(\large 3 = \frac{x-3}{2}\)

ganeshie8 (ganeshie8):

yes, ive cancelled the x+2 both sides. looks ok ?

OpenStudy (ammarah):

what how?

ganeshie8 (ganeshie8):

at step 2 : look at denominators on both sides x+2 is there on both sides right ?

OpenStudy (ammarah):

ohhh i get it

ganeshie8 (ganeshie8):

good :)

OpenStudy (ammarah):

from here do u just cross multily

ganeshie8 (ganeshie8):

yes !

OpenStudy (ammarah):

x=9

ganeshie8 (ganeshie8):

Perfect !

ganeshie8 (ganeshie8):

u need to check if it is good/bad

OpenStudy (ammarah):

bad

ganeshie8 (ganeshie8):

nope. it is good

OpenStudy (ammarah):

wait nvm

ganeshie8 (ganeshie8):

plug x = 9 in the original equation and see if it satisfies

OpenStudy (ammarah):

yea my bad

ganeshie8 (ganeshie8):

lol okie doke

OpenStudy (ammarah):

i dont get this one: 3/x-1+5/x+1=8x+5/x^-1

OpenStudy (ammarah):

theres like two parts to it

OpenStudy (ammarah):

oh please dont leave i have a couple more prob @ganeshie8

ganeshie8 (ganeshie8):

\(\large \frac{3}{x-1}+\frac{5}{x+1}=\frac{8x+5}{x-1} \)

ganeshie8 (ganeshie8):

like that ?

OpenStudy (ammarah):

i mean x^-1

ganeshie8 (ganeshie8):

then it must be more likely a typo in ur question..

OpenStudy (ammarah):

why

ganeshie8 (ganeshie8):

this doesnt look simple :o

OpenStudy (ammarah):

fine r u good with fractions? i have two questions with complex fractions

OpenStudy (ammarah):

ill open another window k?

ganeshie8 (ganeshie8):

this question il skip... ive no clue how to do it in simple way...

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