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

Question

Isaiah Anthony · Accepted Answer

Less thought, more mechanical:  first remove the parentheses on both sides by distributing the multiplication.  Then collect all the terms involving #y# on one side (say the left) and the terms without #y# on the right.  Finally, divide both sides by the coefficient of #y#.  (the number in front of #y#). This is how it appears: #4(2y-1)=4(y+3)# #8y-4=4y+12# #8y-4y=12+4#  (subtract #4y# from both sides and add #4# to both sides) #4y=16# #(4y)/4=16/4# #y=4# Verify your response. Further consideration, examining the equation: Instead of distributing the #4#'s on both sides, you could start by dividing both sides by #4# (multiplying by #1/4#.  This keeps the numbers smaller.  It looks like this:
#4(2y-1)=4(y+3)# #1/4 [4(2y-1)] = 1/4 [4(y+3])# 1/ 4 * 4 = 1/ 4 * 4 (y + 3) #2y-1=y+3# #2y-y=3+1#  (subtract #y# and add #1# on both sides) #y=4# Verify your response.

Isaiah Anthony · Answer

First multiply #4# to get rid of the brackets:
#8y-4=4y+12#
Now, isolate the #y#s on one side and numbers on the other of the #=# sign (remembering to change sign in crossing the #=# sign);
#8y-4y=+4+12#
Notice the change of sign of #4y# and #-4#;
now we can add and subtract:
#4y=16#
The #4# is multiplying the #x# and can go to the right dividing to get:
#x=16/4=4# I hope it's useful.

Hazel Anglin · Answer

undefined To solve 4(2y - 1) = 4(y + 3) for y, you can follow these steps:

1. Distribute the 4 on both sides of the equation:
   8y - 4 = 4y + 12

2. Move the 4y term to the left side of the equation by subtracting 4y from both sides:
   8y - 4y - 4 = 12

3. Combine like terms on the left side:
   4y - 4 = 12

4. Move the constant term to the right side of the equation by adding 4 to both sides:
   4y = 12 + 4

5. Simplify the right side:
   4y = 16

6. Finally, solve for y by dividing both sides by 4:
   y = 16 / 4
   y = 4