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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    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?

  2. anonymous
    • 5 years ago
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    oh bad luck...found a mistake...thats the drawback of now writing but typing it

  3. anonymous
    • 5 years ago
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    let me correct it again

  4. anonymous
    • 5 years ago
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    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

  5. anonymous
    • 5 years ago
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    what is sqrt?

  6. anonymous
    • 5 years ago
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    square root

  7. anonymous
    • 5 years ago
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    well im completely confused because i dont know how to solve the problem..you lost me at the third step

  8. anonymous
    • 5 years ago
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    i took the lcm of (x-2)(x+2) and (x+2) which is (x-2)(x+2

  9. anonymous
    • 5 years ago
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    and shouldnt you have distributed the lcm throughout both sides?

  10. anonymous
    • 5 years ago
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    the lcm which we got , was multiplied with numerator on the right hand side which was 3

  11. anonymous
    • 5 years ago
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    fan me pls if you liked my help

  12. anonymous
    • 5 years ago
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    ok thanks..

  13. anonymous
    • 5 years ago
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    fan me plsssssssssss

  14. anonymous
    • 5 years ago
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    tnx ...

  15. anonymous
    • 5 years ago
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    welcum..

  16. anonymous
    • 5 years ago
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    gud luck with ur maths...

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