You provided two linear equations:

1. \( 3x + 5y - 1 = 0 \)
2. \( 2x + 3y + 5 = 0 \)

It seems like you're asking for the solution to this system of equations. Let's solve it using either substitution or elimination. I will use the elimination method.

### Step 1: Align the system of equations
We have:
\[
3x + 5y = 1 \tag{1}
\]
\[
2x + 3y = -5 \tag{2}
\]

### Step 2: Eliminate one variable (let's eliminate \(x\))
To eliminate \(x\), multiply the first equation by 2 and the second equation by 3 to align the coefficients of \(x\):
\[
6x + 10y = 2 \tag{3}
\]
\[
6x + 9y = -15 \tag{4}
\]

### Step 3: Subtract Equation (4) from Equation (3)
\[
(6x + 10y) - (6x + 9y) = 2 - (-15)
\]
\[
6x + 10y - 6x - 9y = 2 + 15
\]
\[
y = 17
\]

### Step 4: Substitute \(y = 17\) into one of the original equations
Substitute \(y = 17\) into Equation (1):
\[
3x + 5(17) = 1
\]
\[
3x + 85 = 1
\]
\[
3x = 1 - 85
\]
\[
3x = -84
\]
\[
x = -28
\]

### Solution:
The solution to the system is:
\[
x = -28, \quad y = 17
\]

Would you like more details on any part of the process or have any questions?

### Related Questions:
1. How do you solve a system of equations using the substitution method?
2. What is the graphical interpretation of solving a system of linear equations?
3. How do you verify the solution of a system of equations?
4. Can systems of linear equations have no solutions or infinitely many solutions?
5. What are some applications of systems of equations in real-world problems?

**Tip**: Always double-check your solutions by plugging the values of \(x\) and \(y\) back into the original equations to ensure correctness.

Learn how to solve the system of linear equations 3x + 5y - 1 = 0 and 2x + 3y + 5 = 0 using the elimination method. This step-by-step guide demonstrates key algebraic techniques, providing a solution for x = -28 and y = 17. Suitable for students in Grades 9-12.

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