SOLUTION: If X/(b+c-a) = Y/(c+a-b) = Z/(a+b-c) then (b-c)x + (c-a)y + (a-b)Z is?

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Question 635103: If X/(b+c-a) = Y/(c+a-b) = Z/(a+b-c) then (b-c)x + (c-a)y + (a-b)Z is?
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!
 =  = 

Taking the first two fractions

 = 

Cross multiplying:

{x(c+a-b) = y(b+c-a)

x(-b+c+a) = y[b+(c-a)]

x[(-b+c)+a] = yb+y(c-a)

x[-(b-c)+a] = yb+y(c-a)

-x(b-c)+xa = yb+y(c-a)

xa-yb = x(b-c)+y(c-a)     <--- equation (1)

Taking the first and third fractions:

 = 

x(a+b-c) = z(b+c-a)

x[a+(b-c)] = z(-a+b+c)

xa+x(b-c) = z[(-a+b)+c]

xa+x(b-c) = z[-(a-b)+c]

xa+x(b-c) = -z(a-b)+zc

x(b-c)+z(a-b) = zc-xa     <--- equation (2)

Taking the second and third fractions:

 = 

Cross-multiplying:

y(a+b-c) = z(c+a-b)

y(-c+a+b) = z[c+(a-b)]

y[(-c+a)+b] = zc+z(a-b)

y[-(c-a)+b] = zc+z(a-b)

-y(c-a)+yb = zc+z(a-b)

yb-zc = y(c-a)+z(a-b)     <--- equation (3)

Putting equations (1), (2), and (3) together

xa-yb = x(b-c)+y(c-a)     <--- equation (1)
x(b-c)+z(a+b) = zc-xa     <--- equation (2)
yb-zc = y(c-a)+z(a-b)     <--- equation (3)

Reversing equation (2)

xa-yb = x(b-c)+y(c-a)     <--- equation (1)
xc-xa = x(b-c)+z(a-b)     <--- equation (2)
yb-zc = y(c-a)+z(a-b)     <--- equation (3)

Adding all three equation (equals added to equals give equals)

xc-zc =2x(b-c)+2y(c-a)+2z(a-b)

Dividing through by 2

 = x(b-c)+y(c-a)+z(a-b)

or if you like:

(b-c)x+(c-a)y+(a-b)z = 

Edwin

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