In solving systems of equations, one effective method is to manipulate the equations to eliminate a variable, allowing for easier resolution of the remaining variable. Consider the two equations: 2x + 3y = 1 and x - y = 3. The goal is to determine what to multiply one or both equations by to achieve cancellation of either the x or y coefficients.

To start, we can analyze the coefficients of x. The first equation has a coefficient of 2 for x, while the second has an implied coefficient of 1. Since these coefficients are factors of each other, we can multiply the second equation by -2 to create opposite signs. This results in:

2x + 3y = 1

-2(x - y) = -2(3)

which simplifies to:

-2x + 2y = -6

Now, when we add these two equations together, the x terms cancel out:

(2x - 2x) + (3y + 2y) = 1 - 6

This simplifies to:

5y = -5

From here, we can solve for y, yielding y = -1. To find x, we would substitute y back into one of the original equations, although this step is not necessary for our current focus.

Alternatively, we could choose to eliminate y instead. Observing the coefficients of y, we see that they are already set up with opposite signs (3 and -1). We can multiply the second equation by 3, leading to:

2x + 3y = 1

3(x - y) = 3(3)

which simplifies to:

3x - 3y = 9

Adding these equations results in:

(2x + 3x) + (3y - 3y) = 1 + 9

This simplifies to:

5x = 10

Thus, we find x = 2. Again, substituting back into one of the original equations would yield the corresponding value for y, confirming that y = -1.

This approach illustrates that there are multiple valid methods to solve a system of equations, emphasizing the flexibility in choosing which variable to eliminate based on the coefficients present. Understanding these strategies enhances problem-solving skills in algebra.