Question

how do i solve x/x^2-4 + 1/x+2=3

Answer 1

x/x^2-4 + 1/x+2=3
x/(x-2)(x+2) + 1/x+2=3
x+x-2/(x-2)(x+2)=3
2x-2=3 *(x-2)(x+2)
2x-2=3(x^2 - 4)
3x^2 -2x-2=0
3x^2 -3x + x -2=0
3x(x -1) + 2(x -1)=0
(3x+2)((x - 1) =0
x= -2/3, 1
Hope I did not make any mistake ..let me know if thats correct or not?

Answer 2

oh bad luck...found a mistake...thats the drawback of now writing but typing it

Answer 3

let me correct it again

Answer 4

3x^2 -2x-10=0
x= 2 +- sqrt (4+120) /6
x= 2+- 2 sqrt(31) /6
x= 1+- sqrt (31)/ 3
hope this time its correct...pls reply to let me know

Answer 5

what is sqrt?

Answer 6

square root

Answer 7

well im completely confused because i dont know how to solve the problem..you lost me at the third step

Answer 8

i took the lcm of (x-2)(x+2) and (x+2) which is (x-2)(x+2

Answer 9

and shouldnt you have distributed the lcm throughout both sides?

Answer 10

the lcm which we got , was multiplied with numerator on the right hand side which was 3

Answer 11

fan me pls if you liked my help

Answer 12

ok thanks..

Answer 13

fan me plsssssssssss

Answer 14

tnx ...

Answer 15

welcum..

Answer 16

gud luck with ur maths...