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

Answer

To solve the equations graphically, we will follow these steps:

### Step 1: Write down the equations
We have the following equations:
1. \( 2x + 2y - 3 = 0 \)
2. \( x + 2y + 1 = 0 \)

### Step 2: Rearrange the equations to slope-intercept form
To plot these equations, we can rearrange them into the form \( y = mx + b \).

1. For the first equation:
\[ 2y = 3 - 2x \\ y = \frac{3}{2} - x \]

2. For the second equation:
\[ 2y = -x - 1 \\ y = -\frac{1}{2}x - \frac{1}{2} \]

### Step 3: Find points for each equation
We will find two points for each equation to plot.

**For the first equation \( y = \frac{3}{2} - x \):**
- Let \( x = 0 \):
\[ y = \frac{3}{2} - 0 = \frac{3}{2} \quad \text{(Point: (0, 1.5))} \]
- Let \( y = 0 \):
\[ 0 = \frac{3}{2} - x \\ x = \frac{3}{2} \quad \text{(Point: (1.5, 0))} \]

**For the second equation \( y = -\frac{1}{2}x - \frac{1}{2} \):**
- Let \( x = 0 \):
\[ y = -\frac{1}{2}(0) - \frac{1}{2} = -\frac{1}{2} \quad \text{(Point: (0, -0.5))} \]
- Let \( y = 0 \):
\[ 0 = -\frac{1}{2}x - \frac{1}{2} \\ \frac{1}{2} = -\frac{1}{2}x \\ x = -1 \quad \text{(Point: (-1, 0))} \]

### Step 4: Plot the points on a graph
- For the first equation, plot the points (0, 1.5) and (1.5, 0).
- For the second equation, plot the points (0, -0.5) and (-1, 0).

### Step 5: Draw the lines
Draw a straight line through the points for each equation.

### Step 6: Find the intersection point
The intersection point of the two lines is the solution to the system of equations. By observing the graph, we can find the coordinates of the intersection point.

### Step 7: State the solution
From the graph, we find that the intersection point is:
\[ (4, -2.5) \quad \text{or} \quad (4, -\frac{5}{2}) \]

### Final Answer:
The solution to the equations \( 2x + 2y - 3 = 0 \) and \( x + 2y + 1 = 0 \) is:
\[ (4, -\frac{5}{2}) \]

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