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shtirl [24]
3 years ago
8

Solve (3/x-3) =(x/x-3)-(3/2) for x and determine if the solution is extraneous or not

Mathematics
2 answers:
mezya [45]3 years ago
6 0

Answer:

given solution x =3 is extraneous.

Step-by-step explanation:

\frac{3}{x-3} =\frac{x}{x-3} -\frac{3}{2} \\ \frac{3}{x-3} =[tex]\frac{3}{x-3} =\frac{2x-3(x-3)}{2(x-3)}\\=\frac{2x-3x+9)}{2(x-3)}\\= \frac{9-x}{2(x-3)} \\\\\frac{3}{x-3} = \frac{9-x}{2(x-3)} \\\\  \\ 6(x-3) = (9-x)(x-3) \\6(x-3) - (9-x)(x-3) = 0 \\(x-3) (6-9+x) = 0 \\(x-3) (x-3) = 0x=3       [taking LCM  right  hand side] ,we get

gives x =3 as a solution which makes the original equation undefined

therefore the solution extraneous .

Darya [45]3 years ago
3 0
3/(x-3)=x/(x-3)-3/2
6/2(x-3)=2x/2(x-3)-3(x-3)/2(x-3)
6=2x-3(x-3)
6=2x-3x+9
x=3

x=3 makes the denominator x-3 a zero, so the solution is extraneous, 
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