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Question Number 53324 by 0955083339 last updated on 20/Jan/19

If 4a + 5b + 9c=36 and 7a + 9b + 17c=66,  then a+b+c=_____.

$$\mathrm{If}\:\mathrm{4}{a}\:+\:\mathrm{5}{b}\:+\:\mathrm{9}{c}=\mathrm{36}\:\mathrm{and}\:\mathrm{7}{a}\:+\:\mathrm{9}{b}\:+\:\mathrm{17}{c}=\mathrm{66}, \\ $$$$\mathrm{then}\:{a}+{b}+{c}=\_\_\_\_\_. \\ $$

Answered by Kunal12588 last updated on 21/Jan/19

let a+b+c=k  4a+4b+4c+b+5c=36  ⇒4k+(b+5c)=36         (1)  7a+7b+7c+2b+10c=66  ⇒7k+2(b+5c)=66       (2)     let b+5c=h  4k+h=36    [multiply by 2]  8k+2h=72  7k+2h=66  ⇒k=a+b+c=72−66=6

$${let}\:{a}+{b}+{c}={k} \\ $$$$\mathrm{4}{a}+\mathrm{4}{b}+\mathrm{4}{c}+{b}+\mathrm{5}{c}=\mathrm{36} \\ $$$$\Rightarrow\mathrm{4}{k}+\left({b}+\mathrm{5}{c}\right)=\mathrm{36}\:\:\:\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{7}{a}+\mathrm{7}{b}+\mathrm{7}{c}+\mathrm{2}{b}+\mathrm{10}{c}=\mathrm{66} \\ $$$$\Rightarrow\mathrm{7}{k}+\mathrm{2}\left({b}+\mathrm{5}{c}\right)=\mathrm{66}\:\:\:\:\:\:\:\left(\mathrm{2}\right)\:\:\: \\ $$$${let}\:{b}+\mathrm{5}{c}={h} \\ $$$$\mathrm{4}{k}+{h}=\mathrm{36}\:\:\:\:\left[{multiply}\:{by}\:\mathrm{2}\right] \\ $$$$\mathrm{8}{k}+\mathrm{2}{h}=\mathrm{72} \\ $$$$\mathrm{7}{k}+\mathrm{2}{h}=\mathrm{66} \\ $$$$\Rightarrow{k}={a}+{b}+{c}=\mathrm{72}−\mathrm{66}=\mathrm{6} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 21/Jan/19

excellent...

$${excellent}... \\ $$

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