Solve graphically each of the following systems of inear equations. Also, find the coordinates of the points where the lines meet the axis of x.
`x + 2y =5`
` 2x - 3y =-4`

Answer

To solve the given system of linear equations graphically, we will follow these steps:

### Step 1: Write down the equations
The equations we need to solve are:
1. \( x + 2y = 5 \)
2. \( 2x - 3y = -4 \)

### Step 2: Find the x-intercepts of each equation
To find the x-intercept of a line, we set \( y = 0 \) and solve for \( x \).

**For the first equation \( x + 2y = 5 \):**
- Set \( y = 0 \):
\[ x + 2(0) = 5 \implies x = 5 \]
- So, the x-intercept is \( (5, 0) \).

**For the second equation \( 2x - 3y = -4 \):**
- Set \( y = 0 \):
\[ 2x - 3(0) = -4 \implies 2x = -4 \implies x = -2 \]
- So, the x-intercept is \( (-2, 0) \).

### Step 3: Find the y-intercepts of each equation
To find the y-intercept of a line, we set \( x = 0 \) and solve for \( y \).

**For the first equation \( x + 2y = 5 \):**
- Set \( x = 0 \):
\[ 0 + 2y = 5 \implies 2y = 5 \implies y = \frac{5}{2} \]
- So, the y-intercept is \( \left(0, \frac{5}{2}\right) \).

**For the second equation \( 2x - 3y = -4 \):**
- Set \( x = 0 \):
\[ 2(0) - 3y = -4 \implies -3y = -4 \implies y = \frac{4}{3} \]
- So, the y-intercept is \( \left(0, \frac{4}{3}\right) \).

### Step 4: Plot the points and draw the lines
Now we can plot the points we found:
- For the first equation \( x + 2y = 5 \): Points \( (5, 0) \) and \( \left(0, \frac{5}{2}\right) \).
- For the second equation \( 2x - 3y = -4 \): Points \( (-2, 0) \) and \( \left(0, \frac{4}{3}\right) \).

Draw the lines through these points on a graph.

### Step 5: Find the intersection point
To find the intersection point of the two lines, we can solve the equations simultaneously.

We can multiply the first equation by 2:
\[ 2(x + 2y) = 2(5) \implies 2x + 4y = 10 \]

Now we have:
1. \( 2x + 4y = 10 \)
2. \( 2x - 3y = -4 \)

Now, subtract the second equation from the first:
\[ (2x + 4y) - (2x - 3y) = 10 - (-4) \]
This simplifies to:
\[ 4y + 3y = 10 + 4 \implies 7y = 14 \implies y = 2 \]

Now substitute \( y = 2 \) back into one of the original equations to find \( x \):
Using \( x + 2y = 5 \):
\[ x + 2(2) = 5 \implies x + 4 = 5 \implies x = 1 \]

Thus, the intersection point is \( (1, 2) \).

### Final Result
The solution to the system of equations is \( (1, 2) \).

### Points where the lines meet the x-axis
1. For \( x + 2y = 5 \): \( (5, 0) \)
2. For \( 2x - 3y = -4 \): \( (-2, 0) \)