" (6) "2x+3y=2;x-(y)/(2)=(1)/(2)

Solve the following simultaneous equations using Cramer's rule 2x+3y=2,x-y/2=1/2 .

If ((2x-y)/(x+2y))=(1)/(2), then find the value of ((3x-y)/(3x+y))

Solve the simultaneous equations using Cramer's rule 2x+3y=2, x - y/2 = 1/2

(x + 2) (x + 1) = (x -2) (x -3) (y+3)(y+2)=(y-1)(y-2)

Find the products: 2x+3y)(2x-3y)(x-1)(x+1)(x^(2)+1)(x^(4)+1)

Factorize : (1) (x+y)(2x+3y)-(x+y)(x+1) (2) (x+y)(2a+b)-(3x-2y)(2a+b)

Solve:(1)/(2(2x+3y))+(12)/(7(3x-2y))=(1)/(2)(7)/(2x+3y)+(4)/(3x-2y)=2 where 2x+3y!=0 and 3x-2y!=0

(x+y)^(3)-(x-y)^(3) can be factorized as 2y(3x^(2)+y^(2)) (b) 2x(3x^(2)+y^(2))2y(3y^(2)+x^(2)) (d) 2x(x^(2)+3y^(2))