Question

suma a doua nr este 24intregi si 3 supra 8. aflati nr, stiind ca unul este mai mare decat celalalt cu 3 intregi si 5 supra 6.

Answer 1

$a+b=24 \frac{3}{8} =\ \textgreater \ a+b=9 \\ \\ a=b+ 3\frac{5}{6} =\ \textgreater \ a=b+ \frac{5}{2} \\ \\ b+ \frac{5}{2} +b=9 \\ \\ 2b=9- \frac{5}{2} amplificam \\ \\ 4b=18-5=\ \textgreater \ b= \frac{13}{4} \\ \\ a= \frac{13}{4} + \frac{5}{2} amplificam=\ \textgreater \ a= \frac{13}{4} + \frac{10}{4} =\ \textgreater \ a= \frac{23}{4} \\ \\ \\ verificare:a+b= \frac{23}{4} + \frac{13}{4} = \frac{36}{4} =9$