Given that `y=2[x]+3` and `y=3[x-2]+5` then find the value of `[x+y]`

Answer

To solve the problem step by step, we start with the given equations:

1. **Given Equations:**
\[ y = 2[x] + 3 \]
\[ y = 3[x - 2] + 5 \]

2. **Equating the Two Expressions for \( y \):**
Since both expressions represent \( y \), we can set them equal to each other:
\[ 2[x] + 3 = 3[x - 2] + 5 \]

3. **Expanding the Right Side:**
We expand the right side:
\[ 3[x - 2] + 5 = 3[x] - 6 + 5 = 3[x] - 1 \]
So, we have:
\[ 2[x] + 3 = 3[x] - 1 \]

4. **Rearranging the Equation:**
Now, we rearrange the equation to isolate terms involving \([x]\):
\[ 2[x] + 3 + 1 = 3[x] \]
\[ 2[x] + 4 = 3[x] \]
Subtract \( 2[x] \) from both sides:
\[ 4 = 3[x] - 2[x] \]
\[ 4 = [x] \]

5. **Finding the Range of \( x \):**
Since \([x] = 4\), this means \( x \) can take values in the range:
\[ 4 \leq x < 5 \]

6. **Finding \( y \):**
Now, we substitute \([x] = 4\) back into the equation for \( y \):
\[ y = 2[4] + 3 = 2 \cdot 4 + 3 = 8 + 3 = 11 \]

7. **Finding \( [x + y] \):**
Now we need to find \([x + y]\):
\[ x + y = x + 11 \]
Since \( 4 \leq x < 5 \), we can write:
\[ 4 + 11 \leq x + 11 < 5 + 11 \]
\[ 15 \leq x + 11 < 16 \]
Therefore, the greatest integer function gives us:
\[ [x + y] = 15 \]

8. **Final Answer:**
Thus, the value of \([x + y]\) is:
\[ \boxed{15} \]