Grade 12th passAlgebra
If x/b+c=y/c+a=z/a+b then show (b-c) x+(c-a) y+(a-b) z=0
3 Answers
Arun
Dear Aiswarya
I think you have done a midtake in the given condition.
x/(b+c-a) = y/( c+a-b) = z/(a+b-c) = k
There are three fractions.
Use the following properties of fractions.
1. We know that if both numerator and denominator of a fraction p/q are multiplied by the same term, it remains the same
p/q = px / qx = py / qy = pz/qz
-
2. a/b = c/d = e/f
Then adding all the numerators and denominators, the fraction doesn't change.
a/b = c/d = e/f = (a + c + e) / (b + d + f)
-
Using above two properties together
Multiply first fraction by (b - c)/(b - c), second by (c -a)/(c - a), third by (a - b)/(a - b) then adding them
Numerator = (b - c)x + (c -a)y + (a - b)z
Denominator = (b - c)(b + c - a) + (c - a)(c + a - b) + (a - b)(a + b - c)
Denominator = b2 - c2 - a(b - c) + c2 - a2- b(c - a) + a2 - b2 - c(a - b)
Denominator = - a(b - c)- b(c - a) - c(a - b)
Denominator = - ab + ac - bc + ab - ca +bc
Denominator = 0
Therefore
x/(b+c-a) = y/( c+a-b) = z/(a+b-c) = [(b - c)x + (c -a)y + (a - b)z ] / 0 = k
(b - c)x + (c -a)y + (a - b)z = 0
Answer is zero
Regards
Arun (askIITians forum expert)
Hemant Patel
Let x/b+c=y/c+a=z/a+b = k
then
x=(b+c)k ...(1)
y=(c+a)k ...(2)
z=(a+b)k ...(3)
to prove
(b-c) x+(c-a) y+(a-b) z=0
L.H.S. = (b-c) x+(c-a) y+(a-b) z
= (b-c) (b+c) k + (c-a) (c+a)k + (a-b) (a+b)k by eq (1), (2) & (3)
= k [(b-c) (b+c) + (c-a) (c+a) + (a-b) (a+b) ]
=k[b^2 – c^2 + c^2 – a^2 + a^2 – b^2]
= k [0]
= 0
=R.H.S.
Rishi Sharma
Dear Student,
Please find below the solution to your problem.
Here are three fractions.
Use the following properties of fractions.
1. We know that if both numerator and denominator of a fraction p/q are multiplied by the same term, it remains the same
p/q = px / qx = py / qy = pz/qz
2. a/b = c/d = e/f
Then adding all the numerators and denominators, the fraction doesn't change.
a/b = c/d = e/f = (a + c + e) / (b + d + f)
Using above two properties together
Multiply first fraction by(b - c)/(b - c), second by(c -a)/(c - a), third by(a - b)/(a - b)then adding them
Numerator = (b - c)x + (c -a)y + (a - b)z
Denominator = (b - c)(b + c - a) + (c - a)(c + a - b) + (a - b)(a + b - c)
Denominator = b2- c2- a(b - c) + c2- a2- b(c - a) + a2- b2- c(a - b)
Denominator = - a(b - c)- b(c - a) - c(a - b)
Denominator = - ab + ac - bc + ab - ca +bc
Denominator = 0
Therefore
x/(b+c-a) = y/( c+a-b) = z/(a+b-c) = [(b - c)x + (c -a)y + (a - b)z ] / 0 = k
(b - c)x + (c -a)y + (a - b)z = 0
Answer is zero
Thanks and Regards
Please find below the solution to your problem.
Here are three fractions.
Use the following properties of fractions.
1. We know that if both numerator and denominator of a fraction p/q are multiplied by the same term, it remains the same
p/q = px / qx = py / qy = pz/qz
2. a/b = c/d = e/f
Then adding all the numerators and denominators, the fraction doesn't change.
a/b = c/d = e/f = (a + c + e) / (b + d + f)
Using above two properties together
Multiply first fraction by(b - c)/(b - c), second by(c -a)/(c - a), third by(a - b)/(a - b)then adding them
Numerator = (b - c)x + (c -a)y + (a - b)z
Denominator = (b - c)(b + c - a) + (c - a)(c + a - b) + (a - b)(a + b - c)
Denominator = b2- c2- a(b - c) + c2- a2- b(c - a) + a2- b2- c(a - b)
Denominator = - a(b - c)- b(c - a) - c(a - b)
Denominator = - ab + ac - bc + ab - ca +bc
Denominator = 0
Therefore
x/(b+c-a) = y/( c+a-b) = z/(a+b-c) = [(b - c)x + (c -a)y + (a - b)z ] / 0 = k
(b - c)x + (c -a)y + (a - b)z = 0
Answer is zero
Thanks and Regards
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