Solve : `{:((a)/(x)-(b)/(y)=0),((ab^(2))/(x)+(a^(2)b)/(y)=a^(2)+b^(2)):}`

{:((a^(2))/(x) - (b^(2))/(y) = 0),((a^(2)b)/(x)+(b^(2)a)/(y) = "a + b, where x, y"ne 0.):}

Solve (x)/(a) + (y)/(b) = 1 and a(x - a)- b(y + b) = 2a^(2) + b^(2) .

Show that the tangent of an angle between the lines (x)/(a)+(y)/(b)=1 and (x)/(a)-(y)/(b)=1 and (2ab)/(a^(2)-b^(2)) .

If a,b,x,y are real number and x,y gt 0 , then (a^(2))/(x)+(b^(2))/(y) ge ((a+b)^(2))/(x+y) so on solving it we have (ay-bx)^(2) ge 0 . Similarly, we can extend the inequality to three pairs of numbers, i.e, (a^(2))/(x)+(b^(2))/(y)+(c^(2))/(z) ge ((a+b+c)^(2))/(x+y+z) Now use this result to solve the following questions. If abc=1 , then the minimum value of (1)/(a^(3)(b+c))+(1)/(b^(3)(a+c))+(1)/(c^(3)(a+b)) is

Solve : a(x+y)+b(x-y)=a^2-a b+b^2 , a(x+y)-b(x-y)=a^2+a b-b^2

Solve the following system of equations in x and y : a/x-b/y=0,\ \ \ \ (a b^2)/x+(a^2b)/y=a^2+b^2 , where x ,\ y!=0 .

Find the angle between the lines , (a^(2) - ab) y = (ab+b^(2))x+b^(3) , and (ab+b^(2)) y = (ab-a^2)x+a^(3) where a lt b lt 0

Using Cramers rule, solve the following system of linear equations: (a+b)x-(a-b)y=4a b , (a-b)x+(a+b)y=2(a^2-b^2)

If the origin is shifted to the point ((a b)/(a-b),0) without rotation, then the equation (a-b)(x^2+y^2)-2a b x=0 becomes (A) (a-b)(x^2+y^2)-(a+b)x y+a b x=a^2 (B) (a+b)(x^2+y^2)=2a b (C) (x^2+y^2)=(a^2+b^2) (D) (a-b)^2(x^2+y^2)=a^2b^2

Using Cramers rule, solve the following system of linear equations: (a+b)x-(a-b)y=4a b (a-b)x+(a+b)y=2(a^2-b^2)