You have the system of equations:

1. $x + y = 4$
2. $y = 3x + 1$

Solving the system:

You can substitute the expression for $y$ from the second equation into the first equation.

From the second equation, $y = 3x + 1$, substitute this into the first equation:

$x + (3x + 1) = 4$

Now simplify:

$x + 3x + 1 = 4$ $4x + 1 = 4$

Subtract 1 from both sides:

$4x = 3$

Now solve for $x$:

$x = \frac{3}{4}$

Find $y$:

Now substitute $x = \frac{3}{4}$ back into the second equation $y = 3x + 1$:

$y = 3\left(\frac{3}{4}\right) + 1$ $y = \frac{9}{4} + 1$ $y = \frac{9}{4} + \frac{4}{4}$ $y = \frac{13}{4}$

Final solution:

$x = \frac{3}{4}, \quad y = \frac{13}{4}$

Would you like more explanation on any part of this solution?

Related Questions:

1. How would you solve a system of equations with three variables?
2. How can you graph the system of equations $x + y = 4$ and $y = 3x + 1$?
3. What happens if one of the equations is a quadratic equation instead of linear?
4. How do you check if a solution to a system is correct?
5. How would you solve a system of equations using matrices?

Tip:

When solving systems of equations, substitution is helpful when one equation is already solved for one variable. If not, you can rearrange it into that form to make the process easier!