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Question Number 70168 by Askash last updated on 01/Oct/19

Commented by Askash last updated on 01/Oct/19

Please solve ����

Commented by MJS last updated on 01/Oct/19

zillions of possibilities  if we don′t know what′s allowed to use we  cannot solve it

$$\mathrm{zillions}\:\mathrm{of}\:\mathrm{possibilities} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{what}'\mathrm{s}\:\mathrm{allowed}\:\mathrm{to}\:\mathrm{use}\:\mathrm{we} \\ $$$$\mathrm{cannot}\:\mathrm{solve}\:\mathrm{it} \\ $$

Commented by Askash last updated on 01/Oct/19

CAn YOu SoLve it by any easy  method?    If yes then please do.

$$\boldsymbol{\mathcal{C}}{An}\:\boldsymbol{{Y}}{Ou}\:\boldsymbol{{S}}{oLve}\:{it}\:{by}\:{any}\:{easy} \\ $$$${method}? \\ $$$$ \\ $$$${If}\:{yes}\:{then}\:{please}\:{do}. \\ $$

Answered by MJS last updated on 01/Oct/19

ignoring 1, 5  2a+3b=13  9a+4b=17  ⇒ a=−(1/(19))∧b=((83)/(19))  4a+6b=26

$$\mathrm{ignoring}\:\mathrm{1},\:\mathrm{5} \\ $$$$\mathrm{2}{a}+\mathrm{3}{b}=\mathrm{13} \\ $$$$\mathrm{9}{a}+\mathrm{4}{b}=\mathrm{17} \\ $$$$\Rightarrow\:{a}=−\frac{\mathrm{1}}{\mathrm{19}}\wedge{b}=\frac{\mathrm{83}}{\mathrm{19}} \\ $$$$\mathrm{4}{a}+\mathrm{6}{b}=\mathrm{26} \\ $$

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