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solve 2x-3/x + 4 = x - 3/x
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\[\frac{2x-3}{x+4}=\frac{x-3}{x}\] start by cross multiplying to get \[x(2x-3)=(x+4)(x-3)\] and solve that one
sorry i think u misinterpreted my equation. it is (2x-3) / (x) + (4) = (x) - 3/(x)
i'm in pre calculus
my directions are solve the equation
perhaps \[\frac{2x-3}{x}+4=x-\frac{3}{x}\] like that?
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exactly
since \(x\) cannot be equal to zero, multiply both sides by \(x\) and start with \[2x-3+4x=x^2-3\] and solve that one
6x = x^2
what do i do from here?
r u there?
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\[6x = x^2 \] \[x^2-6x=0\] \[x(x-6)=0\] so \(x=0\) which is impossible or \(x=6\)
got it! THANKS
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