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

a,b,c ∈R ((b+c+d)/a)=((a+c+d)/b)=((a+b+c)/d)=((a+b+d)/c)=r what is r?

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

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

a+b+c+d = x ((x−a)/a)=((x−b)/b)=((x−d)/d)=((x−c)/c)=r (1)bx−ab=ax−ab ⇒(a−b)x=0 a=b ∨x =0 (2) dx−bd=bx−bd ⇒b=d (3) cx−cd=dx−cd ⇒c=d ∴ a=b=c=d ⇒r = ((b+c+d)/a)=((3a)/a) = 3 for x = 0 ⇒b+c+d = −a r = ((b+c+d)/a)=−1

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

Commented by jagoll last updated on 31/Jan/20

thanks

$${thanks} \\ $$

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

When x=0, r=((x−a)/a)=−1

$${When}\:{x}=\mathrm{0}, \\ $$$${r}=\frac{{x}−{a}}{{a}}=−\mathrm{1} \\ $$

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 Σ: 3(a+b+c+d)=r(a+b+c+d) if a+b+c+d≠0 ⇒r=3 if a+b+c+d=0: ⇒r=−1

$${b}+{c}+{d}={ra} \\ $$$${a}+{c}+{d}={rb} \\ $$$${a}+{b}+{d}={rc} \\ $$$${a}+{b}+{c}={rd} \\ $$$$\Sigma: \\ $$$$\mathrm{3}\left({a}+{b}+{c}+{d}\right)={r}\left({a}+{b}+{c}+{d}\right) \\ $$$${if}\:{a}+{b}+{c}+{d}\neq\mathrm{0}\:\Rightarrow{r}=\mathrm{3} \\ $$$${if}\:{a}+{b}+{c}+{d}=\mathrm{0}:\:\Rightarrow{r}=−\mathrm{1} \\ $$

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

wrong sir. a+b+c+d=0 b+c+d = −a (clear ! ) r = ((b+c+d)/a) =−1

$${wrong}\:{sir}.\:{a}+{b}+{c}+{d}=\mathrm{0} \\ $$$${b}+{c}+{d}\:=\:−{a}\:\left({clear}\:!\:\right) \\ $$$${r}\:=\:\frac{{b}+{c}+{d}}{{a}}\:=−\mathrm{1} \\ $$$$ \\ $$

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

if a+b+c+d=0 ⇒r=−1 this is correct.

$${if}\:{a}+{b}+{c}+{d}=\mathrm{0}\:\Rightarrow{r}=−\mathrm{1} \\ $$$${this}\:{is}\:{correct}. \\ $$

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