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Find the value of X and Y. (a/(x+y))-(b/(x-y))=1 (b/(x+y))+(a/(x-y))=((a²-b²)/2ab)

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a/(x+y) - b/(x-y) = 1
b/(x+y) + a/(x-y) = (a^2-b^2)/(2ab)

Let u = (x+y) and v = (x-y) and c = (a^2+b^2)/(2ab)

(1) a/u - b/v = 1
(2) b/u +a/v = c so  multiply (1) by a and (2) by b

(3) a^2/u - ab/v = a
(4) b^2/u +ab/v = bc   Add together to get

(a^2+b^2)/u =  bc+a or

u = (a^2 + b^2)/(bc + a)

Substituting into (1) we get

a/((a^2+b^2)/(bc+a)) - b/v = 1
a(bc+a)/(a^2+b^2) - 1 = b/v
(abc +a^2 - a^2 + b^2)/(a^2+b^2) = b/v
(abc +...

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united07 | Student

but the Answer is (a-b;a+b)

 

and also you have a mistake in 1st equation. there must be - not + between a and b.

so please could you solve it once more. i will be thankfull for that..

giorgiana1976 | Student

To add or subtract two fractions, they must have the same denominator.

For the first and for the 2nd equations, the common denominator is (x-y)(x+y). This product returns the difference of squares x^2 - y^2.

The 1st equation will become:

a(x-y) + b(x+y) = x^2 - y^2

The 2nd equation will become:

b(x-y) + a(x+y) = (a^2 - b^2)(x^2 - y^2)/2ab => x^2 - y^2 = 2ab*[b(x-y) + a(x+y)]/(a^2 - b^2)

(a^2 - b^2)*[a(x-y) + b(x+y)] = 2ab*[b(x-y) + a(x+y)]

a^3*(x-y) + a^2*b(x+y) - ab^2*(x-y) - b^3*(x+y) = 2ab^2(x-y) + 2a^2*b(x+y)

-3ab^2*(x-y) + a^3*(x-y) = a^2*b(x+y) + b^3*(x+y)

(x-y)(a^3 - 3ab^2) = (x+y)(b^3+ a^2b)

a^3*x - 3ab^2*x - a^3*y + 3ab^2*y = b^3*x+ a^2b*x + b^3*y+ a^2b*y

a^3*x - 3ab^2*x - b^3*x- a^2b*x = b^3*y+ a^2b*y + a^3*y - 3ab^2*y

x(a^3 - a^2b - 3ab^2 - b^3) = y(b^3 + a^2b - 3ab^2 + a^3)

x = y(b^3 + a^2b - 3ab^2 + a^3)/(a^3 - a^2b - 3ab^2 - b^3)

Therefore, the value of x will be substituted in the 1st equation:

ay[(b^3 + a^2b - 3ab^2 + a^3)/(a^3 - a^2b - 3ab^2 - b^3)-1] + by[(b^3 + a^2b - 3ab^2 + a^3)/(a^3 - a^2b - 3ab^2 - b^3)+1] = y^2[(b^3 + a^2b - 3ab^2 + a^3)/(a^3 - a^2b - 3ab^2 - b^3) - 1][(b^3 + a^2b - 3ab^2 + a^3)/(a^3 - a^2b - 3ab^2 - b^3) + 1]

y[a(b^3 + a^2b - 3ab^2 + a^3)/(a^3 - a^2b - 3ab^2 - b^3)-a + b(b^3 + a^2b - 3ab^2 + a^3)/(a^3 - a^2b - 3ab^2 - b^3)+b ] = y^2[(b^3 + a^2b - 3ab^2 + a^3)/(a^3 - a^2b - 3ab^2 - b^3) - 1][(b^3 + a^2b - 3ab^2 + a^3)/(a^3 - a^2b - 3ab^2 - b^3) + 1]

y^2[(b^3 + a^2b - 3ab^2 + a^3)/(a^3 - a^2b - 3ab^2 - b^3) - 1][(b^3 + a^2b - 3ab^2 + a^3)/(a^3 - a^2b - 3ab^2 - b^3) + 1] - y[a(b^3 + a^2b - 3ab^2 + a^3)/(a^3 - a^2b - 3ab^2 - b^3)-a + b(b^3 + a^2b - 3ab^2 + a^3)/(a^3 - a^2b - 3ab^2 - b^3)+b ] = 0

y1 = 0

y[(b^3 + a^2b - 3ab^2 + a^3)/(a^3 - a^2b - 3ab^2 - b^3) - 1][(b^3 + a^2b - 3ab^2 + a^3)/(a^3 - a^2b - 3ab^2 - b^3) + 1] = [a(b^3 + a^2b - 3ab^2 + a^3)/(a^3 - a^2b - 3ab^2 - b^3)-a + b(b^3 + a^2b - 3ab^2 + a^3)/(a^3 - a^2b - 3ab^2 - b^3)+b ]

y2 = [a(b^3 + a^2b - 3ab^2 + a^3)/(a^3 - a^2b - 3ab^2 - b^3)-a + b(b^3 + a^2b - 3ab^2 + a^3)/(a^3 - a^2b - 3ab^2 - b^3)+b ]/[a(b^3 + a^2b - 3ab^2 + a^3)/(a^3 - a^2b - 3ab^2 - b^3)-a + b(b^3 + a^2b - 3ab^2 + a^3)/(a^3 - a^2b - 3ab^2 - b^3)+b ]

x = 0

x = y2*(b^3 + a^2b - 3ab^2 + a^3)/(a^3 - a^2b - 3ab^2 - b^3)

The values of x and y are: (0;0) or (y2*(b^3 + a^2b - 3ab^2 + a^3)/(a^3 - a^2b - 3ab^2 - b^3) ; [a(b^3 + a^2b - 3ab^2 + a^3)/(a^3 - a^2b - 3ab^2 - b^3)-a + b(b^3 + a^2b - 3ab^2 + a^3)/(a^3 - a^2b - 3ab^2 - b^3)+b ]/[a(b^3 + a^2b - 3ab^2 + a^3)/(a^3 - a^2b - 3ab^2 - b^3)-a + b(b^3 + a^2b - 3ab^2 + a^3)/(a^3 - a^2b - 3ab^2 - b^3)+b ]).