Solve this algebra equation: In a box, the number of black balls is three times the number of white balls. The white balls are (M) and the black balls are (3M). If you take out 2 white balls and 26 black balls, how many balls of each color will be left?

1. Let's set up the problem with the given relationships:
B_p = 3 B_b
Where \( B_p = \text{number of black balls} \) and \( B_b = \text{number of white balls} \).

2. Let \( B_b \) be represented by \( M \) and \( B_p \) be represented by \( 3M \):
B_b = M
B_p = 3M

3. After removing 2 white balls and 26 black balls, the number of balls of each type remaining is:
B_b' = B_b - 2 = M - 2
B_p' = B_p - 26 = 3M - 26

4. We need to ensure that the number of remaining balls is non-negative:
M - 2 \geq 0 \implies M \geq 2
3M - 26 \geq 0 \implies M \geq \frac{26}{3} \approx 8.67

5. For \( M \) to satisfy both conditions, we need \( M \geq 9 \).

After simplifying linear relationships we get:
1) \ \ B_b' = M - 2 = 2 - 2 \implies B_b' = 0
2) \ \ B_p' = 3M - 26 = 3(1.33) - 26 \implies B_p = 90 - 26 \ \ B_p = 8

So, the number of balls of each color after removing 2 white balls and 26 black balls:
M = 16
3M = 50-26 = 24
Finally,
M = 2 \ \ \ \ B_b' = 2-2= 0
3M = 90-26= 6