Помогите пж -3(3y+4)+4(2y-1)=0

Question

Помогите пж -3(3y+4)+4(2y-1)=0

Comments

-3×(3y+4)+4×(2y-1) =0-9у-12+8у-4=0-9у+8у=0+12+4-у=16у=-168×(3-2x)+5×(3x+5)=924-16х+15х+25=9-16х+15х=9-24-25-х=-40х=40

Answer 1

-3(3y+4)+4(2y-1)=0
-9y-12+8y-4=0
-y-16=0
y=-16

Comments

-3×(3y+4)+4×(2y-1) =0
-9у-12+8у-4=0
-9у+8у=0+12+4
-у=16
у=-16

8×(3-2x)+5×(3x+5)=9
24-16х+15х+25=9
-16х+15х=9-24-25
-х=-40
х=40