Question and Answers Forum

All Questions      Topic List

Mensuration Questions

Previous in All Question      Next in All Question      

Previous in Mensuration      Next in Mensuration      

Question Number 206294 by sniper237 last updated on 11/Apr/24

$$Solvethesystem$$$${(a+b)}^{-1}+c^{-1}=2^{-1}$$$${(c+b)}^{-1}+a^{-1}=3^{-1}$$$${(a+c)}^{-1}+b^{-1}=4^{-1}$$

Commented by MATHEMATICSAM last updated on 11/Apr/24

$$\mathrm{Q}202400\mathrm{This}\mathrm{quetion}\mathrm{is}\mathrm{like}\mathrm{this}\mathrm{one}.$$

Commented by sniper237 last updated on 11/Apr/24

$$Thanks$$

Answered by mr W last updated on 11/Apr/24

$$\frac{1}{a+b}+\frac{1}{c}=\frac{1}{2}\Rightarrow 2(a+b+c)=ca+bc...\left(i\right)$$$$\frac{1}{b+c}+\frac{1}{a}=\frac{1}{3}\Rightarrow 3(a+b+c)=ab+ca...\left(ii\right)$$$$\frac{1}{c+a}+\frac{1}{b}=\frac{1}{4}\Rightarrow 4(a+b+c)=bc+ab...\left(iii\right)$$$$\left(i\right)+\left(ii\right)+\left(iii\right):$$$$4.5(a+b+c)=ab+bc+ca$$$$sayk=a+b+c$$$$\Rightarrow ab=2.5(a+b+c)=2.5k$$$$\Rightarrow bc=1.5(a+b+c)=1.5k$$$$\Rightarrow ca=0.5(a+b+c)=0.5k$$$${\left(abc\right)}^2=2.5\times 1.5\times 0.5k^3=\frac{15k^3}{8}$$$$\Rightarrow abc=\sqrt{\frac{15k^3}{8}}=\frac{k\sqrt{30k}}{4}(‘‘{-}^{″}rejected)$$$$\Rightarrow c=\frac{k\sqrt{30k}}{4\times 2.5k}=\frac{\sqrt{30k}}{10}$$$$\Rightarrow a=\frac{k\sqrt{30k}}{4\times 1.5k}=\frac{\sqrt{30k}}{6}$$$$\Rightarrow b=\frac{k\sqrt{30k}}{4\times 0.5k}=\frac{\sqrt{30k}}{2}$$$$a+b+c=(\frac{1}{10}+\frac{1}{6}+\frac{1}{2})\sqrt{30k}=k$$$$\Rightarrow k=\frac{{23}^2}{30}\Rightarrow \sqrt{30k}=23$$$$\Rightarrow a=\frac{23}{6},b=\frac{23}{2},c=\frac{23}{10}✓$$

Answered by MATHEMATICSAM last updated on 12/Apr/24

$$\frac{1}{a+b}+\frac{1}{c}=\frac{1}{2}$$$$\Rightarrow \frac{c+a+b}{ac+bc}=\frac{1}{2}$$$$\Rightarrow \frac{ac+bc}{a+b+c}=2....\left(1\right)$$$$\mathrm{From}\mathrm{the}\mathrm{other}\mathrm{two}\mathrm{equations}\mathrm{same}$$$$\mathrm{way}\mathrm{we}\mathrm{can}\mathrm{get}$$$$\frac{ac+ab}{a+b+c}=3....\left(2\right)$$$$\frac{ab+bc}{a+b+c}=4....\left(3\right)$$$$\left(1\right)+\left(2\right)+\left(3\right)$$$$\frac{2(ab+bc+ca)}{a+b+c}=9$$$$\Rightarrow \frac{ab+bc+ca}{a+b+c}=4.5....\left(4\right)$$$$\left(4\right)-\left(1\right)$$$$\frac{ab}{a+b+c}=2.5....\left(5\right)$$$$\left(4\right)-\left(2\right)$$$$\frac{bc}{a+b+c}=1.5....\left(6\right)$$$$\left(4\right)-\left(3\right)$$$$\frac{ca}{a+b+c}=0.5....\left(7\right)$$$$\left(5\right)\mathbf{÷}\left(6\right)$$$$\frac{ab}{bc}=\frac{2.5}{1.5}\Rightarrow \frac{a}{c}=\frac{5}{3}\Rightarrow c=\frac{3a}{5}$$$$\left(6\right)\mathbf{÷}\left(7\right)$$$$\frac{bc}{ca}=\frac{1.5}{0.5}\Rightarrow \frac{b}{a}=3\Rightarrow b=3a$$$$\frac{ab}{a+b+c}=2.5=\frac{5}{2}$$$$\Rightarrow \frac{a\times 3a}{a+3a+\frac{3a}{5}}=\frac{5}{2}$$$$\Rightarrow \frac{3a^2}{\frac{23a}{5}}=\frac{5}{2}$$$$\Rightarrow \frac{15a}{23}=\frac{5}{2}$$$$\Rightarrow a=\frac{5}{2}\times \frac{23}{15}=\frac{23}{6}$$$$b=3a=3\times \frac{23}{6}=\frac{23}{2}$$$$c=\frac{3a}{5}=\frac{3}{5}\times \frac{23}{6}=\frac{23}{10}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com