We are asked to solve the system of equations using the substitution method. The system of equations is:

$y = 3x + 5$ $x = 2y$

Step 1: Substitute one equation into the other

We already have $y = 3x + 5$, so we can substitute $y = 3x + 5$ into the second equation $x = 2y$.

Substitute $y = 3x + 5$ into $x = 2y$:

$x = 2(3x + 5)$

Step 2: Solve for $x$

Now, simplify the equation:

$x = 6x + 10$

Next, move all terms involving $x$ to one side of the equation:

$x - 6x = 10$ $-5x = 10$

Now solve for $x$:

$x = \frac{10}{-5} = -2$

Step 3: Substitute $x = -2$ back into one of the original equations

Now that we know $x = -2$, substitute this value into the first equation $y = 3x + 5$ to find $y$:

$y = 3(-2) + 5$ $y = -6 + 5$ $y = -1$

Final Solution:

The solution to the system of equations is:

$x = -2 \quad \text{and} \quad y = -1$

Conclusion:

The solution to the system of equations is $x = -2$ and $y = -1$.

Would you like more examples or further clarification on solving systems of equations? Here are some related questions:

1. What other methods can be used to solve systems of equations besides substitution?
2. How do you check your solution to a system of equations?
3. How does the substitution method compare to the elimination method for solving systems of equations?
4. What do you do when the system has no solution or infinitely many solutions?
5. How can you solve a system of equations involving more than two variables?

Tip: Always substitute one equation into the other carefully and double-check your algebra to avoid simple mistakes!