Mathematics.?

1a) Solve the equation 1 + 2/x = 3/2.





1b) Given that a = square root b + 2 / c,


express b in terms of a and c.





1c) Express the following as a single fraction, in its simplest form,





(i) K^2 - k / K^2 - 1.


(ii) 1 / x - 3 - 2 / 3 - x.





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Mathematics.?
1 + 2/x = 3/2





Multiply both sides of the equation by 2x





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





2x + 4 = 3x





Transpose 2x





2x + 4 - 2x = 3x - 2x





Collect like terms





4 = x





- - - - - - - -s-
Reply:1a) 1=+2/x=2/3


multiply the equation by 2x:


2x+4=3x


substract 2x from both sides:


4= 3x -2x


therefore x = 4


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1b)a = square root b+2/c


by squaring both sides:


a^2 = b+2/c


substract 2/c from both sides:


a^2 - 2/c =b


therefore b = a^2 - 2/c


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1c)


(i)k^2 - k/k^2 - 1


by analysing:


k(k-1) / (k-1)(k+1)


= k/k+1


------------------------


(ii)1/x - 3 - 2/3 - x


by uniting the denomiator making it 3x:


3 - 9x - 2x - 3x^2 / 3x


= 3 - 11x - 3x^2/ 3x





hope i helped you......
Reply:%26gt;1a) Solve the equation 1 + 2/x = 3/2.


2/x = 3/2 - 1 = 1/2


Multiply across by x


2 = x/2


4 = x





%26gt;1b) Given that a = √b + 2 / c,


express b in terms of a and c.





Make b the subjest of the equation =%26gt; isolate the b term on the RHS by subtracting 2/c from both sides


a = √b + 2 / c


a -2/c = √b


Square both sides


b = (a -2/c)²


= (ac-2)²/c² if you prefer





%26gt;1c) Express the following as a single fraction, in its simplest form





(i) K^2 - k / K^2 - 1


Are 'k' and 'K' supposed to be the same variable or different?


And there are no brackets on anything?


If they are the same:


k² - k/k² - 1 = k² - k/k² - 1


= k² - 1/k - 1


= (k³ - k -1)/k





%26gt;(ii) 1 / x - 3 - 2 / 3 - x


Again you wrote no brackets e.g around the "3 - x", so the answer to what you wrote is...


1/x - 3 - 2/3 - x = 1/x - 11/3 - x


= (1 - 11x/3 - x²)/x


= (3 - 11x - 3x²)/3x





(Take more care with your brackets...)
Reply:1a.(x+2)/x=3/2


2x+4=3x


x=4








1b. a=sqroot(b+2/c)


a^2=b+2/c


a^2c=bc+2


(a^2c-2)/c=b








c.i





k(k-1)/(k-1)(k+1)


=k/(k+1)





c(ii)








on taking LCM


we get


(3-x-2x+6)/(-x^2+6x-9)


(-3x+9)/-(x-3)^2


3/(x-3)
Reply:1)


1 + 2/x = 3/2.


2/x = 3/2 - 1


2/x = 1/2


x = 4





2)


a = square root b + 2 / c,


a - 2/c = sqrt b


(a - 2/c)^2 = b


b = a^2 - 4a/c + 4/c^2





3)


(i) K^2 - k / K^2 - 1


= k(k-1)/(k-1)(k+1)


= k/(k+1)


.


(ii) 1 / x - 3 - 2 / 3 - x.


= (3-x)/[(x-3)(3-x)] - (2x-6)/[(3-x)(x-3)]


= (3 - x - 2x + 6)/[(3-x)(x-3)]


= (9-3x)/[(3-x)(x-3)]


= 3(3-x)/[(3-x)(x-3)]


= 3/(x-3)