To solve the system of equations:

$x + 3y = 1$ $2x - 5y = -3$

we can use either the substitution or elimination method. Here, I’ll use the elimination method.

Step 1: Make the Coefficients of $x$ Equal

To eliminate $x$, let's multiply the first equation by 2:

$2(x + 3y) = 2 \cdot 1$ $2x + 6y = 2$

Now we have the following system:

$2x + 6y = 2$ $2x - 5y = -3$

Step 2: Subtract the Equations

Subtract the second equation from the first to eliminate $x$:

$(2x + 6y) - (2x - 5y) = 2 - (-3)$ $2x + 6y - 2x + 5y = 5$ $11y = 5$

Step 3: Solve for $y$

Divide both sides by 11:

$y = \frac{5}{11}$

Step 4: Substitute $y$ into One of the Original Equations

Now, substitute $y = \frac{5}{11}$ into the first equation to solve for $x$:

$x + 3\left(\frac{5}{11}\right) = 1$ $x + \frac{15}{11} = 1$

Convert 1 to $\frac{11}{11}$ to combine terms:

$x = \frac{11}{11} - \frac{15}{11}$ $x = \frac{-4}{11}$

Solution

The solution to the system is:

$x = -\frac{4}{11}, \quad y = \frac{5}{11}$

Would you like a detailed explanation of any part of this solution, or do you have questions about the methods used?

Here are 5 related questions you might explore:

1. How would the solution change if both equations were multiplied by different values?
2. What would happen if we used the substitution method instead?
3. How can we verify the solution by plugging the values back into the original equations?
4. What if the system had no solution or infinitely many solutions?
5. How does the elimination method compare to matrix methods for solving systems?

Tip: Always check your solution by substituting the values of $x$ and $y$ back into the original equations to ensure they satisfy both equations.