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Given-a-b-5ab-b-c-7bc-c-a-6ac-where-a-b-c-0-Find-abc-




Question Number 134912 by bramlexs22 last updated on 08/Mar/21
Given  { ((a+b=5ab)),((b+c=7bc)),((c+a=6ac)) :}  where a,b,c ≠ 0.  Find abc
$$\mathrm{Given}\:\begin{cases}{\mathrm{a}+\mathrm{b}=\mathrm{5ab}}\\{\mathrm{b}+\mathrm{c}=\mathrm{7bc}}\\{\mathrm{c}+\mathrm{a}=\mathrm{6ac}}\end{cases} \\ $$$$\mathrm{where}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\neq\:\mathrm{0}. \\ $$$$\mathrm{Find}\:\mathrm{abc} \\ $$
Answered by Ñï= last updated on 08/Mar/21
(1/a)+(1/b)=5⇔A+B=5  (1/b)+(1/c)=7⇔B+C=7  (1/a)+(1/c)=6⇔A+C=6  ⇒A=2    B=3   C=4  ⇒  abc=(1/(ABC))=(1/(24))
$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=\mathrm{5}\Leftrightarrow{A}+{B}=\mathrm{5} \\ $$$$\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\mathrm{7}\Leftrightarrow{B}+{C}=\mathrm{7} \\ $$$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{c}}=\mathrm{6}\Leftrightarrow{A}+{C}=\mathrm{6} \\ $$$$\Rightarrow{A}=\mathrm{2}\:\:\:\:{B}=\mathrm{3}\:\:\:{C}=\mathrm{4} \\ $$$$\Rightarrow\:\:{abc}=\frac{\mathrm{1}}{{ABC}}=\frac{\mathrm{1}}{\mathrm{24}} \\ $$
Commented by Rasheed.Sindhi last updated on 10/Mar/21
Nice sir!
$${Nice}\:{sir}! \\ $$

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