You have the system of linear equations:

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

To solve this system, let's use the elimination method or substitution method. I will start by using the elimination method.

Step 1: Eliminate one variable

We want to eliminate one variable by making the coefficients of either $x$ or $y$ the same in both equations. Here, I'll multiply the first equation by 3 and the second equation by 2 to make the coefficients of $y$ match.

Equation 1: $(5x + 2y = 1) \times 3 \implies 15x + 6y = 3$

Equation 2: $(-3x + 3y = 5) \times 2 \implies -6x + 6y = 10$

Now we have the system:

$15x + 6y = 3$ $-6x + 6y = 10$

Step 2: Subtract the second equation from the first equation to eliminate $y$:

$(15x + 6y) - (-6x + 6y) = 3 - 10$ $15x + 6y + 6x - 6y = -7$ $21x = -7$

Step 3: Solve for $x$

$x = \frac{-7}{21} = -\frac{1}{3}$

Step 4: Substitute $x = -\frac{1}{3}$ into one of the original equations

Let’s substitute into the first equation $5x + 2y = 1$:

$5\left(-\frac{1}{3}\right) + 2y = 1$ $-\frac{5}{3} + 2y = 1$ Add $\frac{5}{3}$ to both sides: $2y = 1 + \frac{5}{3}$ Write 1 as $\frac{3}{3}$: $2y = \frac{3}{3} + \frac{5}{3} = \frac{8}{3}$ Now divide by 2: $y = \frac{8}{3} \div 2 = \frac{8}{6} = \frac{4}{3}$

Final solution:

$x = -\frac{1}{3}, \quad y = \frac{4}{3}$

So, the solution to the system of equations is:

$x = -\frac{1}{3}, \quad y = \frac{4}{3}$

Would you like me to explain any part of the solution in more detail or would you like to explore other methods of solving systems of equations?

Here are 5 related questions to further explore:

1. How do you solve a system of linear equations using substitution?
2. How can you check the solution to a system of equations?
3. What is the graphical interpretation of solving a system of linear equations?
4. Can you solve a system of equations with no solution or infinitely many solutions?
5. How do you solve systems of equations involving three variables?

Tip: Always try to eliminate one variable at a time to simplify the system!