Question

If (a+b+c)p = (b+c-a)q = (c+a-b)r = (a+b-c)s then show that qrs=p(qr+rs+qs)​

Answer 1

★ Concept

We would make use of the given equation and from that we will derive the equation to be proved. Firstly, the given equation will be equated to a constant term 'k'. Then the equation will be divided into its constituent parts such that each of its part would be equal to k(assumed constant). Thus, we will form different equations in terms of that constant and finally we'll add all these equations then on the application of basic algebra we'll get the equation to be proved.

Let's begin the operation on the given equation

$\rule{180pt}{1pt}$

Elucidation

(a+b+c)p=(b+c-a)q=(c+a-b)r=(a+b-c)s = k

(where K is a constant)

⇢ (a+b+c)p = k (Shift 'p' on RHS)

eq(1) a + b + c = k/p

⇢ (b+c-a)q = k (Shift 'q' on RHS)

eq(2) b + c - a = k/q

⇢ (c+a-b)r = k (Shift 'r' on RHS)

eq(3) c + a - b= k/r

⇢ (a+b-c)s = k (Shift 's' on RHS)

eq(4) a + b - c= k/s

Adding equation 1 , 2 , 3 and 4 we'll get

$\sf \: 2(a+b+c) = k {\huge[}\dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{r} + \dfrac{1}{s} {\huge]}$

From equation (1)

$\sf \: 2 {\huge(}\dfrac{k}{p} {\huge)} = k {\huge[}\dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{r} + \dfrac{1}{s} {\huge]}$

On Cancellation of k on both sides

$\sf \dfrac{2}{p} - \dfrac{1}{p} = {\huge[} \dfrac{1}{q} + \dfrac{1}{r} + \dfrac{1}{s} {\huge]}$

$\sf \dfrac{1}{p} = {\huge[} \dfrac{rs + qs + qr}{qrs} {\huge]}$

On Cross Multiplication

⇢ qrs = p (qr+rs+qs)

Hence, proved ✔

$\underline{\rule{180pt}{2pt}}$

☞ Additional Information

This type of question can also be solved without taking constant but in that case the proving may become complicated. Also, on using that method you've to avoid addition of all equation rather use the concept of simultaneous linear equation.

Answer 2

Answer: let's solve the equation .

$\rm(a+b+c)p=(b+c-a)q=(c+a-b)r=(a+b-c)s = k$

(K is constant)

• (a+b+c)p = k (Shift 'p' on RHS)

eq(1) a + b + c = k/p

•(b+c-a)q = k (Shift 'q' on RHS)

eq(2) b + c - a = k/q

• (c+a-b)r = k (Shift 'r' on RHS)

eq(3) c + a - b= k/r

• (a+b-c)s = k (Shift 's' on RHS)

eq(4) a + b - c= k/s

Add the equations 1 , 2 , 3 and 4 we'll get

$\sf \: 2(a+b+c) = k {\huge[}\dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{r} + \dfrac{1}{s} {\huge]}</p><p>$

From equation (1)

$\sf \: 2 {\huge(}\dfrac{k}{p} {\huge)} = k {\huge[}\dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{r} + \dfrac{1}{s} {\huge]}$

By Cancelling of k on both sides

$\sf \dfrac{2}{p} - \dfrac{1}{p} = {\huge[} \dfrac{1}{q} + \dfrac{1}{r} + \dfrac{1}{s} {\huge]}$

$\sf \dfrac{1}{p} = {\huge[} \dfrac{rs + qs + qr}{qrs} {\huge]}$

On Cross Multiplication

⇢ qrs = p (qr+rs+qs)

Hence, proved

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