Solve :
` 3x- 5y + 1=0 `
` 2x - y + 3=0 `

Answer

To solve the system of equations given by:

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

we will use the elimination method. Here’s the step-by-step solution:

### Step 1: Rearranging the equations
First, we can rearrange both equations to express them in a standard form (if needed):

1) \( 3x - 5y = -1 \)
2) \( 2x - y = -3 \)

### Step 2: Make the coefficients of \(x\) the same
To eliminate \(x\), we can multiply the first equation by 2 and the second equation by 3 so that the coefficients of \(x\) in both equations become equal.

Multiplying the first equation by 2:
\[ 2(3x - 5y) = 2(-1) \implies 6x - 10y = -2 \]

Multiplying the second equation by 3:
\[ 3(2x - y) = 3(-3) \implies 6x - 3y = -9 \]

Now we have the new system of equations:
1) \( 6x - 10y = -2 \)
2) \( 6x - 3y = -9 \)

### Step 3: Subtract the equations
Now, we will subtract the first equation from the second equation:
\[ (6x - 3y) - (6x - 10y) = -9 - (-2) \]
This simplifies to:
\[ -3y + 10y = -9 + 2 \]
\[ 7y = -7 \]

### Step 4: Solve for \(y\)
Now we can solve for \(y\):
\[ y = \frac{-7}{7} = -1 \]

### Step 5: Substitute \(y\) back into one of the original equations
Now that we have \(y\), we can substitute it back into one of the original equations to find \(x\). Let's use the first equation:
\[ 3x - 5(-1) + 1 = 0 \]
This simplifies to:
\[ 3x + 5 + 1 = 0 \]
\[ 3x + 6 = 0 \]
\[ 3x = -6 \]
\[ x = \frac{-6}{3} = -2 \]

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