How do you solve #2(-y + 5) = -3y – 3#?

1 Answer
Jul 27, 2015

Answer:

You isolate #y# on one side of the equation.

Explanation:

Your equation contains one variable, #y#, which means that, in order to solve for #y#, you must isolate it on one side of the equation.

To do that, get rid of the paranthesis first

#2(-y + 5) = -3y - 3#

#-2y + 10 = -3y -3#

Add #3y# to both sides of the equation to get

#-2y + 3y + 10 = color(red)(cancel(color(black)(3y))) - color(red)(cancel(color(black)(3y))) - 3#

#y + 10 = -3#

Finally, add #-10# to both sides of the equation

#y - color(red)(cancel(color(black)(10))) + color(red)(cancel(color(black)(10))) = -3 - 10#

#y = color(green)(-13)#

Check to see if the calculations are correct

#2 * (-(color(green)(-13)) + 5) = -3* (-13) - 3#

#2 * 18 = 39 -3#

#36 = 36# #-># the solution is correct!