How do you solve $(3x)/(x+1)+6/(2x)=7/x$ ?

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Answer 1 · 2016-10-01T16:29:59

$x=-2/3$ or $x=2$

$(3x)/(x+1)+6/(2x)=7/x$ can be rewritten as

$(3x)/(x+1)=7/x-6/(2x)$

or $(3x)/(x+1)=7/x-3/x=4/x$

Hence, $(3x)/((x+1))xx x(x+1)=4/x xx x(x+1)$

or $(3x)/(cancel(x+1))xx xcancel((x+1))=4/cancelx xx cancelx(x+1)$

or $3x^2=4x+4$

or $3x^2-4x-4=0$

or $3x^2-6x+2x-4=0$

or $3x(x-2)+2(x-2)=0$

or $(3x+2)(x-2)=0$

Hence either $3x+2=0$ i.e. $x=-2/3$

or $x-2=0$ i.e. $x=2$

How do you solve (3x)/(x+1)+6/(2x)=7/x(3x)/(x+1)+6/(2x)=7/x?