Question

a, b and c are different numbers. Find x + y + z, when x/(b-c) = y(c-a) = z/(a - b). Answer is 0.

Answer 1

the 2nd term supposed to be y/(c-a) right?

Answer 2

It's this:
\(\frac{x}{b-c}=\frac{y}{c-a}=\frac{z}{a-b}\)

Answer 3

xc-xa=yb-yc
xa-xb=zb-zc
ya-yb=zc-za

adding them up-
x(c-a+a-b) +y(a-b)=y(b-c)+z(b-c+c-a)
x(c-b) +y(a-b)-y(b-c)=z(b-a)
(c-b)(x+y)=z(b-a) -y(a-b)
(c-b)(x+y)=(b-a)(z+y)

Answer 4

ehh..use hit and trial from here?

Answer 5

or maybe use componedo/dividnedo

Answer 6

Well, I don't think that I should use any trial and error. It seems like it's simplification problem. And the answer is 0.

Answer 7

Rewriting the equation as a system of three:
\[\begin{cases}
(c-a)x+(c-b)y=0\\(a-b)x+(c-b)z=0\\(a-b)y+(a-c)z=0
\end{cases}\iff\underbrace{\begin{bmatrix}c-a&c-b&0\\a-b&0&c-b\\0&a-b&a-c\end{bmatrix}}_{\mathbf A}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}\]Carry on with row reduction. The coefficient matrix has only two independent rows. You can see this by adding the first row to the second, and subtracting the first row from the third:
\[\left[\begin{array}{ccc|c}c-a&c-b&0&0\\(a-b)+(c-a)&0+(c-b)&(c-b)+0&0\\0-(c-a)&(a-b)-(c-b)&(a-c)-0&0\end{array}\right]\\
\implies \left[\begin{array}{ccc|c}c-a&c-b&0&0\\c-b&c-b&c-b&0\\a-c&a-c&a-c&0\end{array}\right]\implies\left[\begin{array}{ccc|c}c-a&c-b&0&0\\1&1&1&0\\0&0&0&0\end{array}\right]\]This means the matrix has rank \(2\), so the dimension of the nullspace is \(1\) (because \(\mathrm{rank} A+\dim\mathrm{null}\mathbf A=n\), where \(n\) is the number of columns of \(\mathbf A\)). This means there is one non-trivial vector in the nullspace, which means there is some vector \(\mathbf x=\begin{bmatrix}x\\y\\z\end{bmatrix}\neq0\) that solves \(\mathbf {Ax}=\mathbf 0\).

Reducing \(\mathbf A\) further into RREF, you get
\[\left[\begin{array}{ccc|c}
1&0&\dfrac{c-b}{a-b}&0\\[1ex]
0&1&\dfrac{a-c}{a-b}&0\\[1ex]
0&0&0&0
\end{array}\right]\]whose nullspace has a basis consisting of the vector \(\mathbf x=\begin{bmatrix}b-c\\c-a\\a-b\end{bmatrix}\). Add up these components and you find that \(x+y+z=0\), as desired.

Answer 8

bravo!!