Hi! My teacher wanted me to sketch the graph of solution set of each system. For the first question, she wants me to use the slope intercept form and the second question using a table of values, but, I'm a little lost. Can someone help? Thanks!
1.) 3x + 2y < 6
x > 0
y > 0
2.) 2x2 + y ≥ 2
x ≤ 2
y ≤ 1
3x + 2y < 6
2y < -3x + 6
y < -(3/2)x + 3
so the slope is -3/2 and the y-intercept is 3
so start with the y-intercept of 3 , or the point (0,3)
slope is rise/run = -3/2
so from (0,3) , drop down 3 , then 2 units to the right. This should bring you to the point (-3,5)
Joint the points with a dotted line, shade in everthing in the first quadrant below the dotted line
see: http://www.wolframalpha.com/input/?i=plot+3x+%2B+2y+%3C+6+%2C+y+%3E+0+%2C+x+%3E+0
Lines should be dotted
2x^2 + y ≥2
y ≥ -2x^2 + 2
shade in everything "above" the graph of
y = -2x^2 + 2 , making the boundary a solid line
Restrict your graph to all values of x ≤ 2 , and y values ≤ 1
here are some points of the parabola boundary
(0,2)
(1,0), (-1,0)
(2,-2) , (-2,-2)
etc.
Of course, I'd be happy to help you with graphing the solution sets for these systems of inequalities! I'll explain the process step by step for each question.
1.) Using the slope-intercept form:
To graph the solution set of the first system, which consists of the inequalities 3x + 2y < 6, x > 0, and y > 0, we can start by graphing each inequality separately and then finding the overlapping region.
To graph 3x + 2y < 6, we can rewrite it in slope-intercept form, which is y = mx + b. In this case, let's solve the inequality for y:
3x + 2y < 6
2y < -3x + 6
y < (-3/2)x + 3
Now, we can treat y = (-3/2)x + 3 as an equation and graph the corresponding line. Since it's a strict inequality (<), we need to use a dashed line instead of a solid line.
Next, we consider the inequality x > 0. This means that x should be greater than 0. We can represent this on the graph by shading the half-plane to the right of the y-axis.
Finally, for the inequality y > 0, we need to shade the half-plane above the x-axis.
The solution set would be the region where all the shaded areas overlap. This region represents the values of x and y that satisfy all three inequalities simultaneously.
2.) Using a table of values:
To graph the solution set of the second system, which consists of the inequalities 2x^2 + y ≥ 2, x ≤ 2, and y ≤ 1, we can generate a table of values and plot the points that satisfy each inequality.
Starting with the inequality 2x^2 + y ≥ 2, we can rewrite it in the form y ≥ -2x^2 + 2. We can then create a table of x and y values that satisfy this inequality. For example, when x = 0, y should be greater than or equal to 2. You can choose a few more x-values and find the corresponding y-values to create a table.
Next, we consider the inequality x ≤ 2, which means x should be less than or equal to 2. We can add this condition to our table by checking if each x-value is less than or equal to 2.
Similarly, for the inequality y ≤ 1, we need to ensure that each y-value is less than or equal to 1.
Now, plot the points from your table on a graph, using x-values for the horizontal axis and y-values for the vertical axis. Shade the region that satisfies all three inequalities simultaneously.
The solution set would be the shaded region on the graph, which represents the values of x and y that satisfy all three inequalities.
Remember to label your axes and indicate the boundaries of any dashed lines accurately.
I hope this explanation helps! If you have any further questions, feel free to ask.