Verified answer

1.

(2/x) - (2/x-1) + (2/x-2)

Take LCM of x, x-1 and x-2,the three denominators(Dr) which is, multiplying Nr(numerator) and Dr of each term by the other two..

it means,

2(x-1)(x-2)/x(x-1)(x-2) - 2x(x-2)/x(x-1)(x-2) + 2x(x-1)/x(x-1)(x-2)

as the Dr of each term is same we can club all the three terms

[2(x-1)(x-2) - 2x(x-2) + 2x(x-1)] / [x(x-1)(x-2)]

Simplifying Nr. we get

2(x-1)(x-2) - 2x(x-2) + 2x(x-1)

= 2x²-6x +4 - (2x²-4x) + (2x²-2x)

= 2x²-6x +4 - 2x²+4x + 2x²-2x

=2x²-4x +4

Dr. we get x(x-1)(x-2) = x(x²-3x+2) = x³-3x²+2x

So simplified expression is Nr/Dr = (2x²-4x +4)/(x³-3x²+2x)

2. Part 1:

[(x² - x - 20)/(4)]/[(x - 5)/(10)]

=[(x² -5x +4x - 20)/(4)]*[(10)/(x-5)]

=[{x(x-5) +4(x-5)} /(4)]*[(10)/(x-5)]

=[(x+4)(x-5) /(4)]*[(10)/(x-5)]

=[(x+4)(x-5)*10 /(4*(x-5))]

= 5(x+4)/2

= (5x/2) + 10

and

[(x^2 - x - 20)/(x - 5)]/[(4)/(10)]

=[(x² -5x +4x - 20)/(x-5)]*[(10)/(4)]

=[{x(x-5) +4(x-5)} /(x-5)]*[(5)/(2)]

=[(x+4)(x-5) /(x-5)]*[(5)/(2)]

=(x+4)*(5/2)

=(5x/2) +10

Part 2: Yes the complex fractions are equivalent as both the expressions have same terms in the numberator and the denominator ultimately.

3.

1/x + 2/9 = 1/3

Taking LCM on left hand side,

9/9x + 2x/9x = 1/3

(9+2x)/9x = 1/3

By cross multiplication

3(9+2x)=1*9x

27 + 6x = 9x

27 = 9x-6x

27 = 3x

Dividing both sides by 3

9 = x

so x = 9

Hope it helped ... ... ... :)>

sparkling up for the two x or y in the two equation then plug what you get in the different equation for that comparable variable. as an occasion: a million.12x=3.sixty 4+a million.4y->divide the two aspects by a million.12 after which you get x=(3.sixty 4/a million.12)+(a million.4y/a million.12) (Sorry, i've got not got a calculator). Plug that final equation in for x and then slove for y. once you sparkling up for y in the surprising equation plug it in in the different equation and discover x.