[12sqrt(3x-y+7)+2=10],[(2y+(x+11)/(7))/(10)=10]

3x-(y+7)/(11)+2=10 and 2y+(x+11)/(7)=10

Solve: 3x - (y+7)/11+2 =10 and 2y+ (x+11)/7 =10

x+7y=10,3x-2y=7

{:((10)/(x + y) - (4)/(x - y)= -2),((15)/(x + y) + (7)/(x - y) = 10):}

{:((10)/(x + y) - (4)/(x - y)= -2),((15)/(x + y) + (7)/(x - y) = 10):}

sqrt3x^2+10x+7sqrt3

Rationalise the denominator: (a) (1)/(root(3)(3) + root(3)(2)) , (b) (2)/(sqrt5 + sqrt3 + sqrt2) , (c) (x^(2))/(sqrt(x^(2) + y^(2)) - y) , (d) (1)/(sqrt6 + sqrt5 - sqrt11) (e) (sqrt(x + 2y) - sqrt(x -2y))/(sqrt(x + 2y) + sqrt(x - 2y)) , (f) (sqrt10 + sqrt5 - sqrt3)/(sqrt10 - sqrt5 + sqrt3)

If A = (3,-4) , B= (7,0) and C = (14 , -7) are the three consecutive vertices of a parallelogram ABCD , then find the slope of the diagonal BD . The following are the steps involved in solving the above problem . Arrange them is sequential order (A) ((x + 7)/(2) , (y + 0)/(2)) = ((3 + 14)/(2) , (-4-7)/(2)) . (B) The slope of BD = (-11-0)/(10-7) = (-11)/(3) . (C) (x+7)/(2) = (17)/(2) and (y+0)/(2) - (-11)/(2) implies x = 10 , y = -11 therefore D = (10 , -11) . (D) Let the fourth vertex be D(x , y) . we know that the diagonals of a parallelogram bisect each other .

{:((x)/(2) + y = 0.8),((7)/(x + (y)/(2)) = 10):}