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2014-08-21T19:30:14+08:00

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1) x+ \frac{8}{x-2} =1+ \frac{4x}{x-2}
Multiply both sides by x-2
 x^{2} -2x+8=x-2+4x
 x^{2} -6x+10=x
 x^{2} -7x+10=0
We can factor this out
 x^{2} -7x+10=(x-2)(x-5)

If (x-2)=0 then x=2
but if x=2 then in the original equation when we substitute x with 2 then 2+ \frac{8}{2-2} =1+ \frac{8}{2-2} . 0 cannot be the denominator.

If (x-5)=0 then x=5 then the equation would be true

2)  \frac{2 x^{2} }{5} + \frac{5x}{4} =10
 \frac{8 x^{2} +25x}{20} =10
8 x^{2} +25x-200=0
0
2014-08-21T20:31:27+08:00