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Question-11212




Question Number 11212 by uni last updated on 17/Mar/17
Answered by mrW1 last updated on 17/Mar/17
2x=25−5y=5(5−y)  x=((5(5−y))/2)  5−y must be even:  ⇒5−y=2k, k∈N  y=5−2k>0  x=5k>0  0<k<(5/2)  ⇒k=1 or 2  ⇒y=3 or 1  ⇒x=5 or 10    sum of y′s=3+1=4
$$\mathrm{2}{x}=\mathrm{25}−\mathrm{5}{y}=\mathrm{5}\left(\mathrm{5}−{y}\right) \\ $$$${x}=\frac{\mathrm{5}\left(\mathrm{5}−{y}\right)}{\mathrm{2}} \\ $$$$\mathrm{5}−{y}\:{must}\:{be}\:{even}: \\ $$$$\Rightarrow\mathrm{5}−{y}=\mathrm{2}{k},\:{k}\in\mathbb{N} \\ $$$${y}=\mathrm{5}−\mathrm{2}{k}>\mathrm{0} \\ $$$${x}=\mathrm{5}{k}>\mathrm{0} \\ $$$$\mathrm{0}<{k}<\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\Rightarrow{k}=\mathrm{1}\:{or}\:\mathrm{2} \\ $$$$\Rightarrow{y}=\mathrm{3}\:{or}\:\mathrm{1} \\ $$$$\Rightarrow{x}=\mathrm{5}\:{or}\:\mathrm{10} \\ $$$$ \\ $$$${sum}\:{of}\:{y}'{s}=\mathrm{3}+\mathrm{1}=\mathrm{4} \\ $$

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