By drawing a graph for each of the equations `3x+y+5=0,3y-x=5 and 2x+5y=1` on the same graph paper. Show that the lines given by these equations are concurrent (i.e., they pass through the same point). Take 2cm=1 unit the both the axes.

Answer

To show that the lines given by the equations \(3x + y + 5 = 0\), \(3y - x = 5\), and \(2x + 5y = 1\) are concurrent, we will graph each equation and find their intersection point.

### Step 1: Rewrite the equations in slope-intercept form
1. **Equation 1:** \(3x + y + 5 = 0\)
- Rearranging gives: \(y = -3x - 5\)

2. **Equation 2:** \(3y - x = 5\)
- Rearranging gives: \(3y = x + 5\) or \(y = \frac{1}{3}x + \frac{5}{3}\)

3. **Equation 3:** \(2x + 5y = 1\)
- Rearranging gives: \(5y = -2x + 1\) or \(y = -\frac{2}{5}x + \frac{1}{5}\)

### Step 2: Create a table of values for each equation
For each equation, we will select values for \(x\) and calculate the corresponding \(y\) values.

#### For \(y = -3x - 5\):
| \(x\) | \(y\) |
|-------|----------------|
| -1 | -2 |
| 0 | -5 |
| 1 | -8 |

#### For \(y = \frac{1}{3}x + \frac{5}{3}\):
| \(x\) | \(y\) |
|-------|----------------|
| 0 | \(\frac{5}{3}\) |
| 3 | 4 |
| -3 | 0 |

#### For \(y = -\frac{2}{5}x + \frac{1}{5}\):
| \(x\) | \(y\) |
|-------|----------------|
| 0 | \(\frac{1}{5}\) |
| 5 | -1 |
| -5 | 3 |

### Step 3: Plot the points on a graph
1. **Plot points for the first equation** \(y = -3x - 5\):
- Points: \((-1, -2)\), \((0, -5)\), \((1, -8)\)

2. **Plot points for the second equation** \(y = \frac{1}{3}x + \frac{5}{3}\):
- Points: \((0, \frac{5}{3})\), \((3, 4)\), \((-3, 0)\)

3. **Plot points for the third equation** \(y = -\frac{2}{5}x + \frac{1}{5}\):
- Points: \((0, \frac{1}{5})\), \((5, -1)\), \((-5, 3)\)

### Step 4: Draw the lines
- Connect the points for each equation with a straight line.

### Step 5: Identify the point of intersection
- Observe where all three lines intersect. The intersection point is \((-2, 1)\).

### Conclusion
Since all three lines intersect at the point \((-2, 1)\), we conclude that the lines are concurrent.

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