Question: Solve for $y$, where $y$ is a real number.
$$\sqrt{25} - 3y = y + 1$$
If there is more than one solution, separate them with commas.
If there is no solution, click on "No solution".
y =
1. **State the problem:** Solve for $y$ in the equation $$\sqrt{25} - 3y = y + 1$$ where $y$ is a real number.
2. **Recall the value of the square root:** $$\sqrt{25} = 5$$ because $5^2 = 25$.
3. **Substitute the value:** The equation becomes $$5 - 3y = y + 1$$.
4. **Group like terms:** Add $3y$ to both sides to get rid of $-3y$ on the left:
$$5 - \cancel{3y} + 3y = y + 1 + 3y$$
which simplifies to
$$5 = 4y + 1$$.
5. **Isolate $y$:** Subtract $1$ from both sides:
$$5 - 1 = 4y + \cancel{1} - 1$$
which simplifies to
$$4 = 4y$$.
6. **Divide both sides by $4$ to solve for $y$:**
$$\frac{4}{\cancel{4}} = \frac{4y}{\cancel{4}}$$
which simplifies to
$$1 = y$$.
7. **Check the solution:** Substitute $y=1$ back into the original equation:
Left side: $$\sqrt{25} - 3(1) = 5 - 3 = 2$$
Right side: $$1 + 1 = 2$$
Both sides equal $2$, so $y=1$ is a valid solution.
**Final answer:**
$$y = 1$$