To solve the given system of linear equations:
[tex]\[
4x - 2y = 8 \quad \text{and} \quad y = \frac{3}{2}x - 2
\][/tex]
we can follow the steps provided.
### Step 1: Plot the [tex]\( x \)[/tex]-intercept of the first equation.
To find the [tex]\( x \)[/tex]-intercept of the equation [tex]\( 4x - 2y = 8 \)[/tex], set [tex]\( y = 0 \)[/tex]:
[tex]\[
4x - 2(0) = 8 \\
4x = 8 \\
x = 2
\][/tex]
So the [tex]\( x \)[/tex]-intercept is at the point [tex]\( (2, 0) \)[/tex].
### Step 2: Plot the [tex]\( y \)[/tex]-intercept of the first equation.
To find the [tex]\( y \)[/tex]-intercept of the first equation, set [tex]\( x = 0 \)[/tex]:
[tex]\[
4(0) - 2y = 8 \\
-2y = 8 \\
y = -4
\][/tex]
So the [tex]\( y \)[/tex]-intercept is at the point [tex]\( (0, -4) \)[/tex].
### Step 3: Plot the [tex]\( y \)[/tex]-intercept of the second equation.
For the second equation [tex]\( y = \frac{3}{2}x - 2 \)[/tex], set [tex]\( x = 0 \)[/tex]:
[tex]\[
y = \frac{3}{2}(0) - 2 \\
y = -2
\][/tex]
So the [tex]\( y \)[/tex]-intercept is at the point [tex]\( (0, -2) \)[/tex].
### Step 4: Use the slope of the second equation to plot one more point.
The slope of the second equation, [tex]\( y = \frac{3}{2}x - 2 \)[/tex], is [tex]\( \frac{3}{2} \)[/tex]. This means for every increase in [tex]\( x \)[/tex] by 1, [tex]\( y \)[/tex] increases by [tex]\( \frac{3}{2} \)[/tex].
Starting from the [tex]\( y \)[/tex]-intercept [tex]\( (0, -2) \)[/tex], if [tex]\( x \)[/tex] increases by 1:
[tex]\[
x = 1 \\
y = \frac{3}{2}(1) - 2 \\
y = \frac{3}{2} - 2 \\
y = -0.5
\][/tex]
So, another point is [tex]\( (1, -0.5) \)[/tex].
### Conclusion:
From the above steps, we have identified the following points:
- [tex]\( (0, -4) \)[/tex] from the first equation's [tex]\( y \)[/tex]-intercept.
- [tex]\( (0, -2) \)[/tex] from the second equation's [tex]\( y \)[/tex]-intercept.
- [tex]\( (2, 0) \)[/tex] from the first equation's [tex]\( x \)[/tex]-intercept.
- [tex]\( (1, -0.5) \)[/tex] from the slope of the second equation.
Putting these points into the table, we get:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
0 & -4 \\
0 & -2 \\
2 & 0 \\
1 & -0.5 \\
\hline
\end{array}
\][/tex]
So, we have plotted the intercepts and a few other key points which would help in graphically solving or verifying the system of equations.
