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\[\frac{ 2 }{ x}+\frac{ 5 }{ x-2 }=\frac{ 2 }{ x-1 }\]
\[\frac{ \left( 2 \right)\left( x-2 \right) }{ x } + \frac{ 5\left( x \right) }{ x-2 } =\frac{ 2 }{ x-1 }\] \[\frac{ 2x-4+5x }{ x(x-2) }=\frac{ 2 }{ x-1 }\] Then cross over and solve the equation. \[\left( 7x-4)(x-1 \right)=(2(x^{2}-2x)\] Continue solving:)
5x^2-15x+4=0
do i leave it like this?
no, you have to solve for x
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