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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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united07 | Student, Undergraduate | (Level 1) eNoter

Posted July 23, 2011 at 6:28 PM via web

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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)

3 Answers | Add Yours

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted July 24, 2011 at 1:12 AM (Answer #1)

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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 ]).

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united07 | Student, Undergraduate | (Level 1) eNoter

Posted July 24, 2011 at 2:13 AM (Answer #2)

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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..

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beckden | High School Teacher | (Level 1) Educator

Posted July 24, 2011 at 4:30 AM (Answer #3)

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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 + b^2)/(a^2+b^2) = b/v
v = (a^2+b^2)/(ac + b)

u+v = 2x and u-v=2y so

2x = (a^2 + b^2)/(bc + a) + (a^2+b^2)/(ac+b)
x = 1/2(a^2+b^2)(1/(bc+a)+1/(ac+b))
x = 1/2(a^2+b^2)(((ac+b)+(bc+a))/((ac+b)(bc+a)))
x = 1/2(a^2+b^2)(4ab(a+b)(c+1))/(4ab(ac+b)(bc+a))
x = (a^2+b^2)(a+b)(2abc + 2ab)/((2abc+2b^2)(2abc+2a^2))

Now 2abc = a^2+b^2 so
x = (a^2+b^2)(a+b)(a^2 + 2ab + b^2)/((a^2 + 3b^2)(3a^2 + b^2))
x = (a^2+b^2)(a+b)^3/((a^2+3b^2)(3a^2+b^2))

2y = (a^2 + b^2)/(bc + a) - (a^2 + b^2)/(ac+b)
y = 1/2 (a^2+b^2)(1/(bc+a) - 1/(ac+b))
y = 1/2(a^2+b^2)(((ac+b)-(bc+a))/((ac+b)(bc+a)))
y = 1/2(a^2+b^2)(4ab(a-b)(c-1))/(4ab(ac+b)(bc+a))
y = (a^2+b^2)(a-b)(2abc - 2ab)/((2abc+2b^2)(2abc+2a^2))

Now 2abc = a^2+b^2 so
y = (a^2+b^2)(a-b)(a^2 - 2ab + b^2)/((a^2 + 3b^2)(3a^2 + b^2))
y = (a^2+b^2)(a-b)^3/((a^2+3b^2)(3a^2+b^2))

So the answer is

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

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