To graph the system of equations 3x - 2y = 4 and 3x + 2y = 8, we first need to solve for the value of x and y.
To solve for x and y, we can use the first equation 3x - 2y = 4,
3x - 2y = 4
y = (3x/2) - 2
we can substitute this value of y into the second equation 3x + 2y = 8
3x + 2(3x/2 - 2) = 8
3x + 3x - 4 = 8
6x = 12
x = 2
Now we can substitute the value of x into the first equation to find the value of y
3(2) - 2y = 4
6 - 2y = 4
-2y = -2
y = 1
so the point of intersection is (2,1)
In order to graph the system of equations, we can plot the point of intersection (2,1) and then use the slope-intercept form of the equations to graph the lines.
The slope-intercept form of the first equation is y = (3/2)x - 2
The slope-intercept form of the second equation is y = (1/2)x + 4
So, the lines have a slope of 3/2 and 1/2 respectively.
We can now plot these lines on the coordinate plane, with the point of intersection (2,1) being the point where the lines cross.
The graph of the system of equations is the two lines crossing at the point (2,1)