Question

6(z+2)-2(2z+5)=18 show step by step

Answer 1

`6(z+2)-2(2z+5)=18\\\\6\cdot z+6\cdot2-2\cdot2z-2\cdot5=18\\\\6z+12-4z-10=18\\\\\underbrace{6z-4z}_(2z)+\underbrace{12-10}_(2)=18\\\\2z+2=18\ \ \ |\text{subtract 2 from both sides}\\\\2z=16\ \ \ |\text{divide both sides by 2}\\\\\boxed{z=8}`

Answer 2

Follow PEMDAS.

Note:
Parenthesis, Exponents (and roots), Multiplication, Division, Addition, Subtraction)

Because you are trying to isolate the z (variable), you cannot do parenthesis. Instead, distribute 6 to z and 2, and -2 to 2z and 5

6(z + 2) = 6z + 12
-2(2z + 5) = -4z - 10


6z + 12 - 4z - 10 = 18

Next, combine like terms. Combine terms with common variables together, as well as constants.

6z - 4z = 2z
12 - 10 = 2

2z + 2 = 18

Next, isolate the z, do the opposite of PEMDAS. Note that there is an equal sign. What you do to one side, you have to do to the other.

2z + 2 = 18

subtract 2 from both sdies

2z + 2 (-2) = 18 (-2)
2z = 18 - 2
2z = 16

isolate the z, divide 2 from both sides

2z = 16
2z/2 = 16/2
z = 16/2
z = 8

8 is your answer for z


hope this helps