Math, asked by LYCONTRIX, 1 year ago

★ STANDARD QUESTION 9 ★

★ CONTENT QUALITY SUPPORT REQUIRED ★

Solve for " x "
$\frac{(m + n)x - (a - b)}{(m - n)x - (a + b)} = \frac{(m + n)x + (a + c)}{(m - n)x+ (a - c)}$
★ LEVEL - 100 ★

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ShuchiRecites: Bhaiya, I am in 8th and I heartly wanted to answer it but as its of not my course so I am passing this question to others bcz I am also a part of LEVEL-100

LYCONTRIX: Yeah sure , but hey , uh can answer anything available here , and thanks a lot for passing the questions ;)

ShuchiRecites: Ok bhaiya, actually this is of 9th and I will come to 9th in April

LYCONTRIX: no dear , according to me , it's of 7th STANDARD

LYCONTRIX: uh were taught indices in 7th class probably , so use your brain at maximum

ShuchiRecites: Hmn, bhaiya but we are not taught such questions even in not 8th, may be my school is not much forward like the schools in which u have studied

LYCONTRIX: I got my knowledge from my passion , All the schools are same buddy

ShuchiRecites: Nice bhaiya, mine favouriti is also maths

ShuchiRecites: can u give me hint, as I have never done this question ever before, so its new to me

Answers

Answered by sivaprasath

Solution:

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Given :

$\frac{(m+n)x-(a-b)}{(m-n)x-(a+b)} = \frac{(m+n)x+(a+c)}{(m-n)x + (a-c)}$

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To find :

The value of x,.

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$\frac{(m+n)x-(a-b)}{(m-n)x-(a+b)} = \frac{(m+n)x+(a+c)}{(m-n)x + (a-c)}$

⇒ $[(m+n)x-(a-b)] [(m-n)x + (a-c)] = [ (m+n)x+(a+c)] [(m-n)x-(a + b)]$

⇒ $[(m+n)x][(m-n)x] + [(a-c)(m+n)x] - [(m-n)x(a-b)] - (a-b)(a-c) = [(m+n)x][(m-n)x] + [(m-n)x][(a+c)]-[(a+b)(m+n)x]-[(a+c)(a+b)]$

⇒ $ax(m+n)-cx(m+n)-ax(m-n)+bx(m-n)-a(a-c)+b(a-c) = ax(m-n)+cx(m-n)-ax(m+n)-bx(m+n)-a(a+c)-b(a-c)$

⇒ $ax(m+n-(m-n)) + bx(m-n) - cx(m+n) - a(a-c)+b(a-c) = ax(m-n -(m+n)) - bx(m+n) + cx(m-n)-a(a+c)-b(a+c)$

⇒ $bx(m-n+(m+n)) - cx(m+n+(m-n)) -a(a-c-(a+c)) + b(a-c+(a+c)) = 0$

⇒ $bx(2m)-cx(2m) -a(-2c)+b(2a) = 0$

⇒ x(2mb - 2mc) + 2ac + 2ab = 0

⇒ 2x(mb - mc) + 2(ac + ab) = 0

⇒ 2(x(mb - mc) + (ac + ab)) = 0

⇒ x(bm - cm) + (ac + ab) = 0

⇒ x (bm - cm) = - (ac + ab)

⇒ $x = \frac{- (ac + ab)}{(bm - cm)}$

⇒ $x = - \frac{a(b + c)}{m(b-c)}$ (I'm not sure this answer is correct,.)

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Hope it Helps!!

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Answered by Anonymous

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