Given that x, `y in R`. Solve: `x/(1+2i)+y/(3+2i)=(5+6i)/(8i-1)`

Solve: i x^2-3x-2i=0,

Express the following in the form of a+bi, (i) (-5i) (1/8 i) , (ii) (-i)(2i)(-1/8i)^(3)

If x + iy = (3 + 5i)/(7-6i), they y =

solve for x and y . (5)/(x-1)+(1)/(y-2) =2 and ( 6) / ( x-1) -(3)/ (y-2) =1

Find the values of the real numbers x and y, if the complex numbers (3 - i) x - (2-i) y + 2i + 5 and 2x + (-1 + 2i)y + 3 + 2i are equal

Find x and y for which of the following is satisfied. ((1+i)x-2i)/(3+i)+((2-3i)y+i)/(3-i)=i

Solve by cross-multiplication method. (i) 8x-3y=12,5x=2y +7 (ii) 6x+ 7y -11 =0, 5x+ 2y =13 (iii) (2)/(x)+(3)/(y) =5, (3)/(x) -(1)/(y) +9=0

A path of length n is a sequence of points (x_(1),y_(1)) , (x_(2),y_(2)) ,…., (x_(n),y_(n)) with integer coordinates such that for all i between 1 and n-1 both inclusive, either x_(i+1)=x_(i)+1 and y_(i+1)=y_(i) (in which case we say the i^(th) step is rightward) or x_(i+1)=x_(i) and y_(i+1)=y_(i)+1 ( in which case we say that the i^(th) step is upward ). This path is said to start at (x_(1),y_(1)) and end at (x_(n),y_(n)) . Let P(a,b) , for a and b non-negative integers, denotes the number of paths that start at (0,0) and end at (a,b) . Number of ordered pairs (i,j) where i ne j for which P(i,100-i)=P(i,100-j) is

A path of length n is a sequence of points (x_(1),y_(1)) , (x_(2),y_(2)) ,…., (x_(n),y_(n)) with integer coordinates such that for all i between 1 and n-1 both inclusive, either x_(i+1)=x_(i)+1 and y_(i+1)=y_(i) (in which case we say the i^(th) step is rightward) or x_(i+1)=x_(i) and y_(i+1)=y_(i)+1 ( in which case we say that the i^(th) step is upward ). This path is said to start at (x_(1),y_(1)) and end at (x_(n),y_(n)) . Let P(a,b) , for a and b non-negative integers, denotes the number of paths that start at (0,0) and end at (a,b) . The sum P(43,4)+sum_(j=1)^(5)P(49-j,3) is equal to