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NatalieLove

Solve for x:

  • 2 years ago
  • 2 years ago

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  1. NatalieLove
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    • 2 years ago
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  2. Mertsj
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    Multiply all terms by the common denominator which is 21x(x+4)

    • 2 years ago
  3. Mertsj
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    Remember that x cannot be 0 or -4

    • 2 years ago
  4. NatalieLove
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    then no solution

    • 2 years ago
  5. Mertsj
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    I got x = 3

    • 2 years ago
  6. JopHP
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    lest's satrt with the left hand side L.H.S=(3x+12+3x)/(3x^2+12x) then we multiply l.h.s with r.h.s to get 21x^2+84x+21x^2=30x^2+120x therefore 42x^2+84x=30x^2+120x so 12x^2=36x ................ therefore........... 12x^2-36x=0 so.... 12x(x-3)=0 therfore (X=0) Refused.............or........... x=3

    • 2 years ago
  7. Mertsj
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    \[21x(x+4)\times\frac{3}{3x}+21x(x+4)\times\frac{1}{x+4}=\frac{10}{7x}\times \times21x(x+4)\]

    • 2 years ago
  8. Mertsj
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    \[7(x+4)(3)+21x(1)=10(3)(x+4)\]

    • 2 years ago
  9. Mertsj
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    \[21x+84+21x=30x+120\]

    • 2 years ago
  10. Mertsj
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    \[42x+84=30x=120\]

    • 2 years ago
  11. Mertsj
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    \[12x=36\]

    • 2 years ago
  12. Mertsj
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    \[x=3\]

    • 2 years ago
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