Question

Решите пожалуйста номер 895 (2,4)

Answer 1

$\frac{(10-y)}{6} + \frac{3y+8}{3} = \frac{y+6}{2} \frac{10-y}{6}+ \frac{6y+16}{6}= \frac{3y+18}{6}$  
60-6y+36y+96=18y+108
-6y+36y-18y=108-60-96
12y=-48
y=$\frac{-48}{12}$ 
y=-4
----
$\frac{5-9x}{8}- \frac{6+10x}{8}= \frac{20-12x}{8}$  
40-72х-48-80х=160-96х
-72х-80х+96х=160-40+48
-56х=168
х=$\frac{-168}{56}$ 
х=-3