Question

0,6(4y-18)-0,4(5-7y) при y=2 целых 4/13

Answer 1

$0,6(4y-18)-0,4(5-7y)=2,4y-10,8-2+2,8y=5,2y-12,8\\\\\\y=2\frac4{13}=\frac{30}{13}:\\\\5,2y-12,8=5,2\cdot\frac{30}{13}-12,8=0,4\cdot30-12,8=12-12,8=-0,8$