$$\cfrac{2x+3y}{a-2b} = \cfrac{4y+7z}{3b-c} = \cfrac{6z+5x}{2c-3a}$$Find: $11x+17y+20z$
I tried to solve this through the normal and boring substitution method and the calculation is becoming extensively lengthy and ugly with $6$ variables.
Since this is an contest question (previous year's), I believe the examiners wouldn't make the students work so hard and consume so much time. So, I'd like to know if this question has a cleverer approach to it. I tried through componendo-dividendo too, but it doesn't seem to be working. I am not aware of matrices/determinants as of yet, so if the solution could be without the application of these, I'd be grateful. Thanks!
Best Answer
$t=\dfrac{6x+9y}{3a-6b} = \dfrac{8y+14z}{6b-2c} = \dfrac{5x+6z}{2c-a}= \dfrac{11x+17y+20z}{2a}\implies 11x+17y+20z = 2at$ .
Note: This answer is before OP's edit. Even after OP edited his question, the method still works. You just note that: $11x+17y+20z = 3(2x+3y)+ 2(4y+7z) + 6z+5x$.