Solve for x, such that `A*X=C and A xx X=B` with `C ne0`.

The number of three digit numbers of the form xyz such that x lt y , z le y and x ne0 , is a ) 240 b) 244 c) 276 d) 285

Solve the cryptarithm: A B \ X\ A B =\ A C B

Statement-1: If a, b, c are distinct real numbers, then a((x-b)(x-c))/((a-b)(a-c))+b((x-c)(x-a))/((b-c)(b-a))+c((x-a)(x-b))/((c-a)(c-b))=x for each real x. Statement-2: If a, b, c in R such that ax^(2) + bx + c = 0 for three distinct real values of x, then a = b = c = 0 i.e. ax^(2) + bx + c = 0 for all x in R .

Let a,b,c , d ar positive real number such that a/b ne c/d, then the roots of the equation: (a^(2) +b^(2)) x ^(2) +2x (ac+ bd) + (c^(2) +d^(2))=0 are :

Solve: a x+b y=c ,\ \ \ \ b x+a y=1+c

Solve: a x+b y=c ,\ \ \ \ b x+a y=1+c

Solve: (x)/(a) - (a+b)/(x)= (b(a-b))/(ax) , when x ne 0 and a ne 0 .

Solve by factorization: a b x^2+(b^2-a c)x-b c=0

Solve for x and y : (2)/(x) + (3)/(y ) = 13, (5)/(x) - ( 4)/( y) = - 2 ( x ne 0 and y ne 0 )

Solve for x and y : (2)/(x) + (3)/(y ) = 13, (5)/(x) - ( 4)/( y) = - 2 ( x ne 0 and y ne 0 )