Draw the graph of the linear equations 2y-x=7 with the help of graph check whether x =3 and y=2 is the soluti9on of the equation

Answer

To solve the problem step by step, we will first draw the graph of the linear equation \(2y - x = 7\) and then check if the point \((3, 2)\) is a solution to this equation.

### Step 1: Rewrite the equation in slope-intercept form
We start with the equation:
\[ 2y - x = 7 \]

To express it in the form \(y = mx + b\) (where \(m\) is the slope and \(b\) is the y-intercept), we can rearrange it:
\[ 2y = x + 7 \]
\[ y = \frac{1}{2}x + \frac{7}{2} \]

### Step 2: Identify points to plot
To graph the equation, we need at least two points. We can find these by substituting different values for \(x\) and solving for \(y\).

**Point 1: Let \(x = 1\)**
\[ y = \frac{1}{2}(1) + \frac{7}{2} \]
\[ y = \frac{1}{2} + \frac{7}{2} = \frac{8}{2} = 4 \]
So, one point is \((1, 4)\).

**Point 2: Let \(y = 0\)**
\[ 0 = \frac{1}{2}x + \frac{7}{2} \]
Multiply through by 2 to eliminate the fraction:
\[ 0 = x + 7 \]
\[ x = -7 \]
So, another point is \((-7, 0)\).

### Step 3: Plot the points on the graph
Now we plot the points \((1, 4)\) and \((-7, 0)\) on the graph.

1. For the point \((1, 4)\):
- Move 1 unit to the right (positive x-direction).
- Move 4 units up (positive y-direction).

2. For the point \((-7, 0)\):
- Move 7 units to the left (negative x-direction).
- Stay at 0 on the y-axis.

### Step 4: Draw the line
Using a ruler, draw a straight line through the points \((1, 4)\) and \((-7, 0)\). This line represents the equation \(2y - x = 7\).

### Step 5: Check if \((3, 2)\) is a solution
To check if the point \((3, 2)\) lies on the line, we substitute \(x = 3\) and \(y = 2\) into the original equation:
\[ 2(2) - 3 = 7 \]
\[ 4 - 3 = 7 \]
\[ 1 \neq 7 \]

Since the left-hand side does not equal the right-hand side, the point \((3, 2)\) does not satisfy the equation.

### Conclusion
The point \((3, 2)\) is **not** a solution to the equation \(2y - x = 7\).

---