How do you solve #(2(2b+1))/3=3(b-2)#?

Question

Avery Alexander · Accepted Answer

#b=4# #\frac{2(2b+1)}{3}=3(b-2)# Let's remove the denominator from the left multiplying both side with the same quantity, in this case #3# so we can simplify #\color(blue)\cancel{3}*\frac{2(2b+1)}{\cancel{3}}=3(b-2)\color(blue){*3}# #2(2b+1)=9(b-2)# Now let's do the multiplication #4b+2=9b-18# And move all the terms with #b# to the left and all the constants (terms without letter) to the right. Every time we move something to the other side let's change the sign. #4b-9b=-2-18# #-5b = -20# Multiply with #-1# to get a positive #b# #\color(blue){(-1)}-5b = -20 \color(blue){(-1)}# #5b=20# #\frac{\cancel{5}}{\cancel{5}}b=\frac{20}{5}# #b=4#

Jace Anderson · Answer

undefined To solve the equation (2(2b+1))/3=3(b-2), follow these steps:

1. Distribute the 2 on the left side of the equation: (4b + 2)/3 = 3(b - 2).
2. Multiply both sides of the equation by 3 to clear the fraction: 4b + 2 = 9(b - 2).
3. Distribute the 9 on the right side of the equation: 4b + 2 = 9b - 18.
4. Subtract 4b from both sides: 2 = 5b - 18.
5. Add 18 to both sides: 20 = 5b.
6. Divide both sides by 5: b = 4.

So the solution to the equation is b = 4.