Question

Suma a doua nr este 24 intregi si 3 pe 8. Aflati nr stiind ca unul dintre ele este mai mare decat celalalt cu 3 intregi si 5 pe 6.
Urgent va rog.

Answer 1

$a+b=24 \frac{3}{8}$

$a=3 \frac{5}{6} +b= \frac{6*3+1}{6} +b= \frac{18+5}{6} +b= \frac{23}6} + b$

$a+b=24 \frac{3}{8}$

$a+b=\frac{8*24+3}{8}$

$\frac{23}{6}+b +b=\frac{195}{8}$ 

$2b=\frac{195}{8}- \frac{23}{6}$
 
$2b= \frac{3*195-4*23}{24}$
 
$2b= \frac{585-92}{24}$
 
$2b= \frac{493}{24}$

$b= \frac{493}{24} * \frac{1}{2} = \frac{493}{48}$

$a= \frac{23}{6} + b= \frac{23}{6} + \frac{493}{48} = \frac{23*8+493}{48} =\frac{184+493}{48}= \frac{677}{48}$

Verificare
$a+b= \frac{677}{48} + \frac{493}{48} = \frac{677+493}{48} = \frac{1170}{48} = \frac{195}{8} =24 \frac{3}{8}$