Solve for all real values of y: |3y+7|=|2y-1|.

this can be reduced to two cases, actually 4, but 2 are duplicates

+ -- +
+ -- -
- -- +
- -- -

3y+7 = 2y-1 or 3y+7 = -2y + 1
y = -8 or 5y = -6

y = -8 or y = -6/5

To solve the equation |3y+7|=|2y-1| for all real values of y, we need to consider the two cases when the expressions inside the absolute values are positive and when they are negative.

When 3y + 7 is positive and 2y - 1 is positive, the equation becomes:
3y + 7 = 2y - 1

Simplifying this equation, we get:
y = -8

When 3y + 7 is negative and 2y - 1 is negative, the equation becomes:
-(3y + 7) = -(2y - 1)

Expanding and simplifying this equation, we get:
-3y - 7 = -2y + 1

Combining like terms, we have:
-y = 8

Multiplying through by -1, we get:
y = -8

Therefore, the solution to the equation |3y+7|=|2y-1| is y = -8.

To solve for all real values of y in the equation |3y+7|=|2y-1|, we can approach it by considering two cases.

Case 1: (3y+7) = (2y-1)
Start by removing the absolute value signs on both sides. In this case, the expression inside the absolute value on the left side (3y + 7) is equal to the expression inside the absolute value on the right side (2y - 1).

Simplifying this case:
3y + 7 = 2y - 1

Next, we can solve for y:
3y - 2y = -1 - 7
y = -8

So, for this case, the solution is y = -8.

Case 2: -(3y+7) = (2y-1)
In this case, we have the expression inside the absolute value on the left side (3y + 7) multiplied by -1, and it is equal to the expression inside the absolute value on the right side (2y - 1).

Simplifying this case:
-3y - 7 = 2y - 1

Now, solving for y:
-3y - 2y = -1 + 7
-5y = 6
y = -6/5

So, for this case, the solution is y = -6/5.

Therefore, the real values of y that satisfy the equation |3y+7|=|2y-1| are y = -8 and y = -6/5.