To graph the system of equations given by y = 2x + 3 and 2x + 4y = 8, we will first manipulate both equations so we can easily plot them.
Starting with the first equation, y = 2x + 3, this is already in slope-intercept form (y = mx + b), where the slope (m) is 2 and the y-intercept (b) is 3. This means we can begin by plotting the point (0, 3) on the graph, and from there, we will rise 2 units and run 1 unit to the right to find another point, which will be (1, 5). Connecting these points will give us the line of the first equation.
Next, we’ll rearrange the second equation, 2x + 4y = 8, into slope-intercept form as well. We can start by isolating y:
- 4y = 8 – 2x
- y = (8 – 2x)/4
- y = 2 – (1/2)x
This shows that the slope is -1/2 and the y-intercept is 2. We can plot the y-intercept (0, 2) first, and then from here, we’ll move down 1 unit and run 2 units to the right to find another point, which is (2, 1).
Now we have two lines plotted on the same coordinate system. The intersection of these two lines represents the solution to the system of equations. If we calculate it algebraically:
Substituting y from the first equation into the second:
2x + 4(2x + 3) = 8
2x + 8x + 12 = 8
10x + 12 = 8
10x = 8 - 12
10x = -4
x = -2/5
Now substitute back to find y:
y = 2(-2/5) + 3 = -4/5 + 3 = 11/5
The intersection point is
(-2/5, 11/5). Therefore, this point will be where the two lines cross on the graph.
In brief, to summarize:
- Plot the line for y = 2x + 3.
- Plot the line for y = 2 – (1/2)x.
- Identify the intersection point (-2/5, 11/5), which demonstrates the solution to the system.