To solve the equations \(3y + 5x = 0\) and \(5y + 2x = 0\) graphically, we will follow these steps:
### Step 1: Rearranging the equations
First, we need to rearrange both equations into the slope-intercept form \(y = mx + b\).
1. For the first equation \(3y + 5x = 0\):
\[
3y = -5x \implies y = -\frac{5}{3}x
\]
2. For the second equation \(5y + 2x = 0\):
\[
5y = -2x \implies y = -\frac{2}{5}x
\]
### Step 2: Finding points for the first equation
Next, we will find points for the first equation \(y = -\frac{5}{3}x\).
- For \(x = 0\):
\[
y = -\frac{5}{3}(0) = 0 \quad \text{(Point: (0, 0))}
\]
- For \(x = 3\):
\[
y = -\frac{5}{3}(3) = -5 \quad \text{(Point: (3, -5))}
\]
- For \(x = 1\):
\[
y = -\frac{5}{3}(1) \approx -1.67 \quad \text{(Point: (1, -1.67))}
\]
### Step 3: Finding points for the second equation
Now, we will find points for the second equation \(y = -\frac{2}{5}x\).
- For \(x = 0\):
\[
y = -\frac{2}{5}(0) = 0 \quad \text{(Point: (0, 0))}
\]
- For \(x = 5\):
\[
y = -\frac{2}{5}(5) = -2 \quad \text{(Point: (5, -2))}
\]
- For \(x = 3\):
\[
y = -\frac{2}{5}(3) \approx -1.2 \quad \text{(Point: (3, -1.2))}
\]
### Step 4: Plotting the points
Now, we will plot the points on a graph:
- For the first line \(y = -\frac{5}{3}x\), we plot the points (0, 0), (1, -1.67), and (3, -5).
- For the second line \(y = -\frac{2}{5}x\), we plot the points (0, 0), (3, -1.2), and (5, -2).
### Step 5: Drawing the lines
After plotting the points, we will draw straight lines through the points for both equations.
### Step 6: Finding the intersection point
The intersection point of the two lines is the solution to the system of equations. From our graph, we can see that the lines intersect at the point (0, 0).
### Final Solution
Thus, the solution to the system of equations \(3y + 5x = 0\) and \(5y + 2x = 0\) is:
\[
\boxed{(0, 0)}
\]
