Question

If \[a(y + z) = b(z + x) = c(x + y)\] and out of a, b, c no two of them are equal then show that, \[\dfrac{{y - z}}{{a(b - c)}} = \dfrac{{z - x}}{{b(c - a)}} = \dfrac{{x - y}}{{c(a - b)}}.\]

Answer 1

Hint: We can solve this by assuming \[a(y + z) = b(z + x) = c(x + y) = K\] . Meaning that each term is equal to K. By subsuming each term to K and subtracting one with the other we get \[y - z\] , \[z - x\] and \[x - y\] values. After making some changes and simplification we can show that \[\dfrac{{y - z}}{{a(b - c)}} = \dfrac{{z - x}}{{b(c - a)}} = \dfrac{{x - y}}{{c(a - b)}}\] .

Complete step-by-step answer:
Given, \[a(y + z) = b(z + x) = c(x + y)\]
Assume that it is equal to ‘K’.
That is, \[a(y + z) = b(z + x) = c(x + y) = K\]
Then we have, \[a(y + z) = K\] , \[b(z + x) = K\] and \[c(x + y) = K\] . And dividing this with ‘a’, ‘b’ and ‘c’ respectively we have,
 \[y + z = \dfrac{K}{a}{\text{ - - - - - (1)}}\] , \[z + x = \dfrac{K}{b}{\text{ - - - - - - - (2)}}\] and \[x + y = \dfrac{K}{c}{\text{ - - - - - - - (3)}}\] .
If we see the problem we need \[y - z\] , \[z - x\] and \[x - y\] . To obtain this we subtract the above equation one with the other.
Now subtract equation (2) with equation (1). We have,
 \[ \Rightarrow z + x - (y + z) = \dfrac{K}{b} - \dfrac{K}{a}{\text{ }}\]
Taking L.C.M. in the right hand side of the equation,
 \[ \Rightarrow z + x - y - z = \dfrac{{Ka - Kb}}{{ab}}\]
Cancelling ‘z’ on the left hand side of the equation, and taking ‘K’ common on the right hand side of the equation, we have:
 \[ \Rightarrow x - y = \dfrac{{K(a - b)}}{{ab}}\]
In the problem we have \[c(a - b)\] in the denominator of \[x - y\] , so divide the above equation with \[c(a - b)\] on both sides.
 \[ \Rightarrow \dfrac{{x - y}}{{c(a - b)}} = \dfrac{K}{{abc}}{\text{ - - - - - - - (4)}}\] .
Now subtracting equation (3) with equation (2). We have,
 \[ \Rightarrow x + y - (z + x) = \dfrac{K}{c} - \dfrac{K}{b}{\text{ }}\]
Taking L.C.M. in the right hand side of the equation,
 \[ \Rightarrow x + y - z - x = \dfrac{{Kb - Kc}}{{bc}}\]
Cancelling ‘x’ on the left hand side of the equation, and taking ‘K’ common on the right hand side of the equation, we have:
 \[ \Rightarrow y - z = \dfrac{{K(b - c)}}{{bc}}\]
In the problem we have \[a(b - c)\] in the denominator of \[y - z\] , so divide the above equation with \[a(b - c)\] on both sides.
 \[ \Rightarrow \dfrac{{y - z}}{{a(b - c)}} = \dfrac{K}{{abc}}{\text{ - - - - - - - - (5)}}\] .
Similarly, now subtracting equation (1) with equation (3). We have,
 \[ \Rightarrow y + z - (x + y) = \dfrac{K}{a}{\text{ }} - \dfrac{K}{c}\]
Taking L.C.M. in the right hand side of the equation,
 \[ \Rightarrow y + z - x - y = \dfrac{{Kc - Ka}}{{ac}}\]
Cancelling ‘y’ on the left hand side of the equation, and taking ‘K’ common on the right hand side of the equation, we have:
 \[ \Rightarrow z - x = \dfrac{{K(c - a)}}{{ac}}\]
In the problem we have \[b(c - a)\] in the denominator of \[z - x\] , so divide the above equation with \[b(c - a)\] on both sides
 \[ \Rightarrow \dfrac{{z - x}}{{b(c - a)}} = \dfrac{K}{{abc}}{\text{ - - - - - - - (6)}}\]
Now from equation (4), equation (5) and equation (6) we have,
 \[\dfrac{{x - y}}{{c(a - b)}} = \dfrac{K}{{abc}}\] , \[\dfrac{{y - z}}{{a(b - c)}} = \dfrac{K}{{abc}}\] and \[\dfrac{{z - x}}{{b(c - a)}} = \dfrac{K}{{abc}}\] .
Then we have,
 \[\dfrac{{y - z}}{{a(b - c)}} = \dfrac{{z - x}}{{b(c - a)}} = \dfrac{{x - y}}{{c(a - b)}}\] .
Hence proved.
So, the correct answer is “ \[\dfrac{{y - z}}{{a(b - c)}} = \dfrac{{z - x}}{{b(c - a)}} = \dfrac{{x - y}}{{c(a - b)}}\] .”.

Note: We simplify the above problem based on what we need to prove. As we can see we divided \[x - y\] term with \[c(a - b)\] because in the equation that we need to prove have \[\dfrac{{x - y}}{{c(a - b)}}\] . That is what we did for the rest of the terms. Careful in the calculation part especially in the sign multiplication while expanding the brackets.