Expert: Josh - 2/23/2004

Question
Having trouble with these 3 problems -
1)  0.22y - 0.6 = 0.12y + 3 - 0.8y

2)  6(x-2) - 16 = 3 (2x - 5) + 11

3)  1/4x - 1/8x = 3 - 1/16x

Answer
The general idea is to separate the "knowns" from the "unknowns". Here, "y" is the unknown.

Remember what we are allowed to do to achieve this objective. We can add, subtract, multiply or divide both sides of the equation by the same number, without changing the validity of the equation.

OK, Q1.
0.22y-0.6 = 0.12y+3-0.8y
collect like terms on RHS first.
0.22y-0.6 =-0.68y +3
add 0.68y from both sides,
0.9y-0.6 = 3
add 0.6 to both sides,
0.9y=3.6
divide both sides by 0.9
y=4

Q2.
6(x-2) - 16 = 3 (2x - 5) + 11
if we expand the parentheses, we get
6x-12-16 = 6x-15+11
This doesn't make sense, since the x vanishes; which means there's nothing to solve for.

Did the question ask you to verify
6(x+2)-16=3(2x-5)+11 ?
You need to say what the question ask for.
Keywords like "evaluate", "prove", "solve for" etc are important directives.

Q3. Not sure whether you mean (1/4)*x, a quarter of x OR
1/(4x), as in one over "4x", with 4x in the denominator.

If what you meant was (1/4)x-(1/8)x=3-(1/16)x
Then, gather the unknowns onto one side of the equation, and add them just like fractions.
Add (1/16)x to both sides...
(1/4)x-(1/8)x+(1/16)x=3
(1/4-1/8+1/16)x=3
(1/8+1/16)x=3
(2/16+1/16)x=3
(3/16)x=3
x=(16/3)*3=16 OK?