Graph the system
Blue:y>-1/2x+3
Red:-2y<-3x+12
There are a few ways to approach graphing a system of inequalities, but one common method is to graph each inequality separately and then shade in the region that satisfies both. To do this, we can start by converting each inequality to slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept.
Blue inequality: y > -1/2x + 3
This inequality is already in slope-intercept form, with a slope of -1/2 and a y-intercept of 3. To graph it, we can plot the y-intercept at (0, 3), and then use the slope to find additional points. Since the slope is negative, we can go down 1 unit and right 2 units to get to the next point, and repeat this to get a few more points. Then we can draw a dashed line through these points to indicate that y can be greater than (but not equal to) the value of the line at any given x. Finally, we need to shade in the region above the line, since any point above the line will satisfy the inequality.
Red inequality: -2y < -3x + 12
To graph this inequality, we can start by dividing both sides by -2 to get y > 3/2x - 6. Notice that we flipped the inequality sign when dividing by a negative number. Now we have an inequality in slope-intercept form, with a slope of 3/2 and a y-intercept of -6. To graph it, we can plot the y-intercept at (0, -6), and then use the slope to find additional points. Since the slope is positive, we can go up 3 units and right 2 units to get to the next point, and repeat this to get a few more points. Then we can draw a dashed line through these points to indicate that y can be greater than (but not equal to) the value of the line at any given x. Finally, we need to shade in the region above the line, since any point above the line will satisfy the inequality.
Putting it all together, we get the following graph:
The blue line represents the points where y is greater than -1/2x + 3 (i.e., above the line), while the red line represents the points where y is greater than 3/2x - 6 (i.e., above the line). The shaded region shows the points that satisfy both inequalities, which is the area above the darker purple line. This region includes the point (0, 0), which is a solution to both inequalities, so it is part of the solution to the system. Other points in this region include (2, 1) and (4, 3), among others.
To graph the system of inequalities, let's first start by graphing each inequality separately.
For the blue inequality, y > -(1/2)x + 3:
Step 1: Graph the line y = -(1/2)x + 3 in blue.
- Find the y-intercept by setting x = 0 and solving for y: y = -(1/2)(0) + 3 = 3.
- Find another point by setting x = 2 and solving for y: y = -(1/2)(2) + 3 = 2.
- Connect the two points with a straight line.
Next, shade the region above the line since y is greater than (-1/2)x + 3.
Now, let's move on to the red inequality, -2y < -3x + 12:
Step 2: Rewrite the inequality in slope-intercept form to graph it: y > (3/2)x - 6.
- Divide both sides by -2, which changes the direction of the inequality.
- Simplify the inequality to y > (3/2)x - 6.
Step 3: Graph the line y = (3/2)x - 6 in red.
- Find the y-intercept by setting x = 0 and solving for y: y = (3/2)(0) - 6 = -6.
- Find another point by setting x = 2 and solving for y: y = (3/2)(2) - 6 = 0.
- Connect the two points with a straight line.
Finally, the region above the line should be shaded since y is greater than (3/2)x - 6.
Step 4: Combine the shaded regions from both inequalities by shading the area where the shading overlaps.
The resulting graph should show the shaded region where both inequalities are true.