We are given the system of linear equations:

1. \( 3x + 5y = 1 \)
2. \( 2x - 3y = 7 \)

To solve this, we can use the **substitution method** or the **elimination method**. I'll use the **elimination method** to eliminate one of the variables.

### Step 1: Multiply the equations
We'll multiply the first equation by 2 and the second equation by 3, so that the coefficients of \(x\) will be the same:

\[
2(3x + 5y) = 2(1) \quad \text{(Multiply first equation by 2)}
\]
\[
3(2x - 3y) = 3(7) \quad \text{(Multiply second equation by 3)}
\]

This gives:
\[
6x + 10y = 2 \quad \text{(Equation 3)}
\]
\[
6x - 9y = 21 \quad \text{(Equation 4)}
\]

### Step 2: Subtract the equations
Now, subtract Equation 4 from Equation 3 to eliminate \(x\):

\[
(6x + 10y) - (6x - 9y) = 2 - 21
\]
\[
6x - 6x + 10y + 9y = -19
\]
\[
19y = -19
\]

### Step 3: Solve for \(y\)
\[
y = \frac{-19}{19} = -1
\]

### Step 4: Substitute \(y = -1\) into one of the original equations
We'll substitute \(y = -1\) into the first equation \( 3x + 5y = 1 \):

\[
3x + 5(-1) = 1
\]
\[
3x - 5 = 1
\]
\[
3x = 6
\]
\[
x = \frac{6}{3} = 2
\]

### Solution:
The solution to the system of equations is:
\[
x = 2, \quad y = -1
\]

Would you like more details or have any questions?

Here are 5 related questions:
1. How do you solve a system of equations using the substitution method?
2. Can you solve a system of equations using matrices?
3. What happens if a system of equations has no solution or infinitely many solutions?
4. How can you graphically interpret the solution of two linear equations?
5. How can you solve a system of nonlinear equations?

**Tip:** The elimination method works well when you want to quickly remove a variable by manipulating both equations.

Learn how to solve the system of linear equations 3x + 5y = 1 and 2x - 3y = 7 using the elimination method. This step-by-step guide covers algebraic concepts, key formulas, and is suitable for students in Grades 8-10.

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