x + 2y = 1 1/3

+x - y =+1/3
3y = ?

not just and answer

A.0
B.- 1/3
C.1/3
D.1
E.1 2/3

No ideas at all? From the problems you have posted, I would think that an algebra I problem like this would be no challenge.

solve for x in the 1st, and substitute it into the 2nd. Then solve for 3y.

Just subtract them as they are to get

3y = ....

To find the value of 3y in the given equations, we'll need to solve the system of equations first. Let's go step by step:

Step 1: Write down the given equations:
x + 2y = 1 1/3 -- (Equation 1)
x - y = 1/3 -- (Equation 2)

Step 2: We can solve this system of equations by the elimination or substitution method. Let's use the elimination method.

Multiply Equation 2 by 2 to make the coefficients of 'y' the same and obtain a system of equations:
2(x - y) = 2(1/3)
2x - 2y = 2/3 -- (Equation 3)

Now we have two equations:
x + 2y = 1 1/3 -- (Equation 1)
2x - 2y = 2/3 -- (Equation 3)

Step 3: Add both equations together to eliminate 'y':
(x + 2y) + (2x - 2y) = 1 1/3 + 2/3
3x = 3

Simplifying:
3x = 3

Step 4: Divide both sides of the equation by 3 to solve for 'x':
3x/3 = 3/3
x = 1

Step 5: Now that we have the value of 'x', substitute it back into either Equation 1 or Equation 2 to find 'y'. Let's use Equation 2:
x - y = 1/3
1 - y = 1/3

Solving for 'y':
-y = 1/3 - 1
-y = 1/3 - 3/3
-y = -2/3

Multiplying both sides by -1 to solve for 'y':
y = 2/3

Step 6: Now, we have the values of 'x' and 'y', so we can find 3y:
3y = 3 * (2/3)
3y = 2

Therefore, the value of 3y is 2.