How do you solve the equation #25- [2+ 5y - 3(y+2)]= - 3( 2y - 5) - 15( y - 1) - 3y + 3#?

1 Answer
Jul 3, 2018

#y=-13/11#

Explanation:

Expand some expressions as needed.

#25-[2+5ycolor(red)(-3(y+2))]=color(blue)(-3(2y-5))color(orange)(-15(y-1))-3y+3#
#->25-[2+5y+color(red)((-3*y)+(-3*2))=color(blue)((-3*2y)+(-3*-5)color(orange)(+(-15*y)+(-15*1))#
#->25-[2+5ycolor(red)(-3y-6)]=color(blue)(-6y+15)color(orange)(-15y-15)-3y+3#

There's still a pesky negative symbol in front of the brackets. Let's distribute that within the bracket.

#25-color(red)((1))[2+5y-3y-6]=-6y+15-15y-15-3y+3#
#->25color(red)(+(-1*2)+(-1*5y)+(-1*-3y)+(-1*-6))=-6y+15-15y-15-3y+3#
#->25color(red)(-2-5y+3y+6)=-6y+15-15y-15-3y+3#

Now everything is expanded thoroughly, combine like terms.

#25-2-5y+3y+6=-6y+15-15y-15-3y+3#
#->25-2+6-5y+3y=-6y-15y-3ycancel(+15)cancel(-15)+3#
#->29-2y=-24y+3#

Isolate the #y#.

#29-2y=-24y+3#
#->cancel(29)cancel(color(red)(-29))-2ycolor(blue)(+24y)=cancel(-24y)cancel(color(blue)(+24y))+3color(red)(-29)#
#->22ycolor(lime)(/22)=-26color(lime)(/22)#
#->y=-13/11#