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Question Number 80108 by jagoll last updated on 31/Jan/20

$$a,b,c\in \mathbb{R}$$$$\frac{b+c+d}{a}=\frac{a+c+d}{b}=\frac{a+b+c}{d}=\frac{a+b+d}{c}=r$$$$whatisr?$$

Commented by john santu last updated on 31/Jan/20

$$\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=\mathrm{x}$$$$\frac{\mathrm{x}-\mathrm{a}}{\mathrm{a}}=\frac{\mathrm{x}-\mathrm{b}}{\mathrm{b}}=\frac{\mathrm{x}-\mathrm{d}}{\mathrm{d}}=\frac{\mathrm{x}-\mathrm{c}}{\mathrm{c}}=r$$$$\left(1\right)\mathrm{bx}-\mathrm{ab}=\mathrm{ax}-\mathrm{ab}\Rightarrow (\mathrm{a}-\mathrm{b})\mathrm{x}=0$$$$\mathrm{a}=\mathrm{b}\vee \mathrm{x}=0$$$$\left(2\right)\mathrm{dx}-\mathrm{bd}=\mathrm{bx}-\mathrm{bd}\Rightarrow \mathrm{b}=\mathrm{d}$$$$\left(3\right)\mathrm{cx}-\mathrm{cd}=\mathrm{dx}-\mathrm{cd}\Rightarrow \mathrm{c}=\mathrm{d}$$$$\therefore \mathrm{a}=\mathrm{b}=\mathrm{c}=\mathrm{d}\Rightarrow \mathrm{r}=\frac{b+c+d}{a}=\frac{3a}{a}=3$$$$forx=0\Rightarrow b+c+d=-a$$$$r=\frac{b+c+d}{a}=-1$$$$$$

Commented by jagoll last updated on 31/Jan/20

$$thanks$$

Commented by $@ty@m123 last updated on 31/Jan/20

$$Whenx=0,$$$$r=\frac{x-a}{a}=-1$$

Answered by mr W last updated on 31/Jan/20

$$b+c+d=ra$$$$a+c+d=rb$$$$a+b+d=rc$$$$a+b+c=rd$$$$\mathrm{\Sigma }:$$$$3(a+b+c+d)=r(a+b+c+d)$$$$ifa+b+c+d\ne 0\Rightarrow r=3$$$$ifa+b+c+d=0:\Rightarrow r=-1$$

Commented by john santu last updated on 31/Jan/20

$$wrongsir.a+b+c+d=0$$$$b+c+d=-a(clear!)$$$$r=\frac{b+c+d}{a}=-1$$$$$$

Commented by mr W last updated on 31/Jan/20

$$ifa+b+c+d=0\Rightarrow r=-1$$$$thisiscorrect.$$

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