[" 42."x+y=a+b,3(x+3y)=11xy],[" ar "-by=a^(2)-b^(2)," 41"(x)/(a)+(y)/(b)=2]

[" 2.If two positive integers "a" and "b" are written as "a=x^(3)y^(2)" and "b=xy^(3);x,y" are prime "],[[" numbers,then "HCF(a,b)" is "," (c) "x^(3)y^(3)],[" (a) "xy," (b) "xy^(2)," (d) "x^(2)y^(2)]]

2x + y = (7xy) / (3), x + 3y = (11xy) / (3)

2a(x+y)-3b(x+y)

Factorise 25x^(2) - 30xy + 9y^(2) . The following steps are involved in solving the above problem . Arrange them in sequential order . (A) (5x - 3y)^(2) " " [ because a^(2) - 2b + b^(2) = (a-b)^(2)] (B) (5x)^(2) - 30xy + (3y)^(2) = (5x)^(2) - 2(5x)(3y) + (3y)^(2) (C) (5x - 3y) (5x - 3y)

The following steps are involved in expanding (x+ 3y)^(2) . Arrange them in sequential order from the first to the last . (A) (x + 3y)^(2) = x^(2) + 6xy + 9y^(2) (B) (x + 3y)^(2) = (x)^(2) + 2(x)(3y) + (3y)^(2) (C) Using the identify (a + b)^(2) = a^(2) + 2ab + b^(2) , where a = x and b = 3y.

If a : b = x : y , then show that (a^(2) + b^(2)) : a^(3)/(a + b) : : (x^(2) + y^(2)) : x^(3)/(x + y)

If a=(x)/(x+y) and b=(y)/(x-y), then (ab)/(a+b) is equal to (a) (xy)/(x^(2)+y^(2)) (b) (x^(2)+y^(2))/(xy)( c) (x)/(x+y) (d) ((y)/(x+y))^(2)

Factorise the following : (i) 9a^(2)-b^(2) " " (ii) 81x^(3)-x " " (iii) x^(3)-49xy^(2) " " (iv) a^(2)-(b-c)^(2) (v) (x-y)^(3)-x+y " " (vi)x^(2)y^(2)+1-x^(2)-y^(2) " " (vii) 25(a+b)^(2)-49(a-b)^(2) " " (viii) xy^(5)-x^(5)y (ix) x^(8)-256 " " (x) x^(8)-81y^(8)