Question 81347
Given:
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{{{(1/(x+2))- (2/x)=3}}}
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I used LCD and got: 
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{{{(1x/(x^2 + 2x))- (2x +4)/(x^2 + 2x) = 3}}} <===
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Then multiplied both sides by {{{x^2 - 2x}}} <=== should be + between terms, not -
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You should now have:
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{{{x - 2x - 4 = 3(x^2 + 2x)}}}
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Multiply out the right side and you get:
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{{{x - 2x - 4 = 3x^2 + 6x}}}
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Combine the two terms on the left side that both contain x to get:
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{{{ -x - 4 = 3x^2 + 6x}}}
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Add x + 4 to both sides to eliminate the terms on the left side.  The result is:
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{{{0 = 3x^2 + 6x + x + 4}}}
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Combine the two terms on the right side that contain x:
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{{{0 = 3x^2 + 7x + 4}}}
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Transpose (switch sides):
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{{{3x^2 + 7x + 4 = 0}}}
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This equation is now in standard quadratic form. It can be solved by graphing, by factoring, 
or by using the quadratic formula.  The more general approach is to use the quadratic
formula, but in this case factoring works.  The left side of this equation factors to:
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{{{(3x + 4)*(x + 1)= 0}}}
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Notice that this equation will be true if either of the factors on the left side equals zero
because zero times anything is zero.  So, one at a time, set the two factors equal to
zero and solve for x.
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{{{3x + 4 = 0}}}
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Subtract 4 from both sides:
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{{{3x = -4}}}
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Divide by 3:
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{{{x = -4/3 }}}
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That's one solution for x.  Now set the other factor equal to zero:
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{{{x + 1 = 0}}}
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Subtract 1 from both sides:
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{{{x = -1}}}
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That's the second solution for x.
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In summary, the two solutions for x are {{{-4/3 }}} and {{{-1}}}
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You can check both of these by substituting them (one at a time) for x in the original
given equation and making sure that the equation still balances on both sides.
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Hope this helps you to understand the problem and also to correct your minor mistake.
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Cheers ...