How do you solve #4y^2 - 3(y-4)=4y(y-1)+3y#?

1 Answer
Apr 23, 2018

Answer:

When only one variable is there, an equation can be solved by using simplification and #y=6# in this equation. You can verify it also by putting #y=6# on both sides and the results will be same.

Explanation:

#4y^2-3(y-4)=4y(y-1)+3y#

#4y^2 -3y+12=4y^2-4y+3y " " " " [-3 * (-4)=12, 4y * y=4y^2]#

or

#4y^2-4y^2 -3y +12=-y#

#-3y+12=-y#

So

#12=3y-y#

#2y=12#

#y=12/2#

# y=6#

Hope it helps...