Given, (x+y)/xy = 2 => (x+y)=2xy----(i)
(x-y)/xy = 6 => x-y=6xy-----(ii)
Adding equation (i) & (ii),
x+y+x-y = 2xy+6xy = 8xy
=> 2x=8xy => y=1/4
Putting the value of y in eqn (i) we get,
x+1/4=2x*(1/4) => x+1/4=x/2=> x/2=-1/4 => x=-1/2
SO by solving both the equations we get, x=-1/2, y=1/4
Here, *= Multiplication
I hope my answer helps. All the very best for your future endeavors!
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Option 1: $(x+y)^4-x^4$
Option 2: $(x+y)^4-y^4$
Option 3: $(x-y)^4-x^4$
Option 4: $x^4-(x-y)^4$
Question : The third proportional of the following numbers $(x-y)^2, (x^2-y^2)^2$ is:
Option 1: $(x+y)^3(x-y)^2$
Option 2: $(x+y)^4(x-y)^2$
Option 3: $(x+y)^2(x-y)^2$
Option 4: $(x+y)^2(x-y)^3$
Question : The factors of $x^2+4 y^2+4 y-4 x y-2 x-8$ are:
Option 1: $(x-2 y-4)(x-2 y+2)$
Option 2: $\left(x^2-2 y-4\right)\left(x^2-2 y+2\right)$
Option 3: $(x+2 y-4)(x+2 y+2)$
Option 4: $\left(x^2-2 y-4\right)\left(x^2+2 y+2\right)$
Question : Simplify the given expression $\frac{(x^3-y^3)(x+y)}{x^2+x y+y^2}$.
Option 1: $x - y$
Option 2: $x^2-y^2$
Option 3: $x + y$
Option 4: $x^2+y^2$
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