the problem is to graph the equation below. 3x - y < 2 and x + y > 2
my solution is y = 3x - 2, y = -1x + 2
I,m lost at where my lines ar drawn. In y = 3x - 2 (y line is 3)
and y = -1x +2 (y line is -1)
or am I totaly lost.
After continued effort I beleive
1st equation
m = 3 and y-intercept is -2
2nd question
m = -1 and y- intercept is 2
However my graph shows 2 parallel lines shading becomes an issue or am I still lost
You did the lines right, however....
The regions are:
y>3x-2 Notice that the allowed region is ABOVE the line y=3x-2
y>-x+2 The allowed region is above that line.
Now on the two allowed regions, where do they have common area?
http://www.purplemath.com/modules/syslneq.htm
To graph the inequalities, y > 3x - 2 and y > -x + 2, follow these steps:
1. Graph the line y = 3x - 2:
- Start by plotting the y-intercept at (0, -2).
- Use the slope, which is 3 (since the coefficient of x is 3), and the y-intercept to find another point.
- Connect the two points to draw the line.
2. Graph the line y = -x + 2:
- Start by plotting the y-intercept at (0, 2).
- Use the slope, which is -1 (since the coefficient of x is -1), and the y-intercept to find another point.
- Connect the two points to draw the line.
3. Determine the shaded region:
- Since both inequalities are "greater than" (y >), the shaded region should be above both lines.
- Shade the region above the line y = 3x - 2.
- Shade the region above the line y = -x + 2.
- The region where the two shaded regions overlap is the solution.
Remember to be careful with the shading, as the overlap of the regions is what represents the solution to the system of inequalities.
If you are having difficulty visualizing the solution, you can refer to the link provided in the previous response for a visual representation of the solution.
To graph the equations 3x - y < 2 and x + y > 2, you first need to find the equations for the lines.
For the first equation, 3x - y < 2, you can rewrite it as y > 3x - 2. The line for this equation has a slope of 3 and a y-intercept of -2.
For the second equation, x + y > 2, you can rewrite it as y > -x + 2. The line for this equation has a slope of -1 and a y-intercept of 2.
To graph these lines, you can choose some x-values and calculate the corresponding y-values. For example, for the first line, you can choose x = 0 and find the y-intercept, which is -2. This gives you one point on the line. You can choose additional x-values, calculate the respective y-values, and plot the points to draw the line. Doing the same for the second line will allow you to draw the second line on the graph.
However, the shaded region of the graph represents the solutions to both inequalities. The solution to the first inequality, y > 3x - 2, is above the line y = 3x - 2. The solution to the second inequality, y > -x + 2, is also above the line y = -x + 2.
To find the common area between these two regions, you need to determine where they overlap. You can refer to the link provided (http://www.purplemath.com/modules/syslneq.htm) for a visual representation of the overlapping regions. The common area is the region that satisfies both inequalities and is shaded in the graph.