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Question Number 92880 by fath035990 last updated on 09/May/20

solve 8ϰ+4=3(ϰ−1)+7

$$\mathrm{solve}\:\mathrm{8}\varkappa+\mathrm{4}=\mathrm{3}\left(\varkappa−\mathrm{1}\right)+\mathrm{7} \\ $$

Answered by niroj last updated on 09/May/20

   8ϰ+4=3(ϰ−1)+7    or, 8ϰ+4=3ϰ−3+7    or,  8ϰ−3ϰ= 4−4       or,   5ϰ=0       ∴  ϰ=0

$$\:\:\:\mathrm{8}\varkappa+\mathrm{4}=\mathrm{3}\left(\varkappa−\mathrm{1}\right)+\mathrm{7} \\ $$$$\:\:\mathrm{or},\:\mathrm{8}\varkappa+\mathrm{4}=\mathrm{3}\varkappa−\mathrm{3}+\mathrm{7} \\ $$$$\:\:\mathrm{or},\:\:\mathrm{8}\varkappa−\mathrm{3}\varkappa=\:\mathrm{4}−\mathrm{4}\:\:\: \\ $$$$\:\:\mathrm{or},\:\:\:\mathrm{5}\varkappa=\mathrm{0} \\ $$$$\:\:\:\:\:\therefore\:\:\varkappa=\mathrm{0} \\ $$$$ \\ $$

Commented by fath035990 last updated on 10/May/20

thanks

$$\mathrm{thanks} \\ $$

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