find three different solution of the following equation                                                                                                      1. 2x - 3y = 0                                                                                                                                                                      2. 2x + y =0                                                                                                                                                                                      3. x + 2y = 6                                                                                                                                                                                    4. 5x - y = 1

To find three different solutions for each of the given equations, we can choose various values for \( x \) or \( y \) and solve for the other variable. Here are the solutions for each equation:

### 1. \( 2x - 3y = 0 \)
Let's solve for \( x \):
\[ x = \frac{3y}{2} \]

#### Solution 1:
Let \( y = 2 \):
\[ x = \frac{3(2)}{2} = 3 \]
So, one solution is \( (3, 2) \).

#### Solution 2:
Let \( y = -2 \):
\[ x = \frac{3(-2)}{2} = -3 \]
So, another solution is \( (-3, -2) \).

#### Solution 3:
Let \( y = 0 \):
\[ x = \frac{3(0)}{2} = 0 \]
So, another solution is \( (0, 0) \).

### 2. \( 2x + y = 0 \)
Let's solve for \( y \):
\[ y = -2x \]

#### Solution 1:
Let \( x = 1 \):
\[ y = -2(1) = -2 \]
So, one solution is \( (1, -2) \).

#### Solution 2:
Let \( x = -1 \):
\[ y = -2(-1) = 2 \]
So, another solution is \( (-1, 2) \).

#### Solution 3:
Let \( x = 0 \):
\[ y = -2(0) = 0 \]
So, another solution is \( (0, 0) \).

### 3. \( x + 2y = 6 \)
Let's solve for \( x \):
\[ x = 6 - 2y \]

#### Solution 1:
Let \( y = 0 \):
\[ x = 6 - 2(0) = 6 \]
So, one solution is \( (6, 0) \).

#### Solution 2:
Let \( y = 1 \):
\[ x = 6 - 2(1) = 4 \]
So, another solution is \( (4, 1) \).

#### Solution 3:
Let \( y = -1 \):
\[ x = 6 - 2(-1) = 8 \]
So, another solution is \( (8, -1) \).

### 4. \( 5x - y = 1 \)
Let's solve for \( y \):
\[ y = 5x - 1 \]

#### Solution 1:
Let \( x = 0 \):
\[ y = 5(0) - 1 = -1 \]
So, one solution is \( (0, -1) \).

#### Solution 2:
Let \( x = 1 \):
\[ y = 5(1) - 1 = 4 \]
So, another solution is \( (1, 4) \).

#### Solution 3:
Let \( x = -1 \):
\[ y = 5(-1) - 1 = -6 \]
So, another solution is \( (-1, -6) \).

Would you like more details or have any questions about these solutions?

Here are some additional questions you might be interested in:
1. How to find the intersection of two lines?
2. What is the graphical representation of these linear equations?
3. How to solve a system of linear equations using substitution or elimination?
4. What is the method to find the determinant of a matrix?
5. How to solve quadratic equations?

**Tip:** When solving linear equations, it's often helpful to choose values that simplify the calculations.

Explore three different solutions for each of the given linear equations: 2x - 3y = 0, 2x + y = 0, x + 2y = 6, 5x - y = 1. Understand how to find solutions by choosing values for x or y and solving for the other variable. Suitable for Grades 9-12.

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