Question

OK i still need help with this problem i get the concept of a/b=c/d-> ad=bc but I'm having trouble applying that to this problem can someone help ill post the link

Answer 1

start by multiplying both sides of the equation by (x+3) (to get rid of the fraction on the left hand side)

then multiply by (x-1) (to get rid of the fraction on the right hand side)


what do you get?

Answer 2

after these steps , and expansion, you should have a quadratic expression on the left hand side, and a different quadratic on the right hand side,


solve for x

Answer 3

that what i did I got x^2-2x+1 = x^2+5x+6 this is where i started having trouble solving for x i first got rid of the x^2 by subtracting them to cancel them out so I would be left with -2x+1=5x+6 which this is where it got missed up i don't know what i did wrong

Answer 4

that is good so far

x^2-2x+1 = x^2+5x+6

-2x+1 = 5x+6

now add 2x to both sides, and take away 6

Answer 5

yes so i would have 7x and then i would subtract the 6 to get -5 and the answer would be x= -5/7 right?

Answer 6

I also have another problem do you think you can help with it?

Answer 7

that problem ill post here in case you can help
I know it has a solution I'm just having trouble getting one that works i know it has one because it marks it correct on the software that says it has one solution

Answer 8

x= -5/7 yes!


now lets check it to be sure

(x-1)/(x+3) = (x+2)/(x-1)

plugging in x=-5/7

LHS = (-5/7-1)/(-5/7+3)
= (-12/7)/(16/7)
= -12/16
= -3/4

RHS = (-5/7+2)/(-5/7-1)
= (9/7)/(-12/7)
= 9/-12
=-3/4

LHS=RHS \(\checkmark\)

Answer 9

ok yes thanks for your help on that problem

Answer 10

first factor the denominator of the left hand side