Fraction proportionality-linearity

Question

If $$\frac {x_1} {y_1} =\frac {x_2} {y_2}=\dotsb=\frac {x_n} {y_n}=k$$ Then $$\frac {\alpha_1 x_1+\alpha_2x_2+\dotsb +\alpha_nx_n } {\alpha_1y_1 + \alpha_2y_2 + \dotsb + \alpha_ny_n}=k $$

Can someone explain why this is true. I ve heard that this can be proved with mathematical induction but that doesnt give a sense of understanding the result

Edit:Also if there are other ways of understanding this equality please post

Answer 1

Write $x_j=ky_j$ and simpilify.

Answer 2

This is an immediate consequence of linearity, i.e. if a fraction $f$ is a root of $n$ linear equations then it is a root of any linear combination of these equations.

$$\begin{align} y_1 f = x_1\,&\Rightarrow\ \ \alpha_1 y_1\ f\ =\ \alpha_1 x_1\\ &\ \ \,\vdots\\ y_n f = x_n\,&\Rightarrow\ \ \alpha_n y_n\ f\ =\ \alpha_n x_n\!\!\!\!\!\!\!\!\!\!\phantom{I_{I_{I_I}}}\\ \hline {\rm\color{#c00}{adding}}\,\ \!&\!\Rightarrow\! (\color{#c00}\Sigma \alpha_i y_i)f = \color{#c00}\Sigma \alpha_i x_i\!\!\!\!\!\!\!\!\!\!\phantom{I^{I^I}} \end{align}\qquad\qquad$$

Answer 3

Using induction.

For $n=1$: $\frac {x_1} {y_1} =\frac {\alpha_1 x_1}{\alpha_1y_1}$ is true.

Assume it is true for $n=k$: $$\color{blue}{\frac {x_1} {y_1} =\frac {x_2} {y_2}=\dotsb =\frac{x_k}{y_k}=\frac {\alpha_1 x_1+\alpha_2x_2+\dotsb +\alpha_kx_k } {\alpha_1y_1 + \alpha_2y_2 + \dotsb + \alpha_ky_k}}$$

Prove it for $n=k+1$: $$\color{blue}{\frac {x_1} {y_1} =\frac {x_2} {y_2}=\dotsb =\frac{x_k}{y_k}}=\frac{x_{k+1}}{y_{k+1}}= \\ \color{blue}{\frac {\overbrace{\alpha_1 x_1+\alpha_2x_2+\dotsb +\alpha_kx_k}^{a}}{\underbrace{\alpha_1y_1 + \alpha_2y_2 + \dotsb + \alpha_ky_k}_{b}}} =\frac{\overbrace{\alpha_{k+1}x_{k+1}}^{c}}{\underbrace{{\alpha_{k+1}y_{k+1}}}_{d}}=\\ =\frac {\alpha_1 x_1+\alpha_2x_2+\dotsb +\alpha_kx_k +\alpha_{k+1}x_{k+1}} {\alpha_1y_1 + \alpha_2y_2 + \dotsb + \alpha_ky_k+a_{k+1}y_{k+1}},$$ because: $$\frac{a}{b}=\frac{c}{d} \iff \frac ab=\frac{a+c}{b+d}.$$