1. ## graphs

I would like to rnsure that I am completing my graphs correctly:
1) To graph the solution set of x + 2y is less than or equal to 3 and x + 2y is less than or equal to 4,
I have y is less than or equal to -1/2 x + 2 and y is less than or equal to -1/2 x + 3/2
I have two solid diagonal lines relatively close together, going this direction: \ , and have shaded the lower half. Does this sound correct??

2) In what way should I plot my lines to graph the inequality: 3x is less than or equal to -4y - 4?

2. Originally Posted by jay1
I would like to rnsure that I am completing my graphs correctly:
1) To graph the solution set of x + 2y is less than or equal to 3 and x + 2y is less than or equal to 4,
I have y is less than or equal to -1/2 x + 2 and y is less than or equal to -1/2 x + 3/2
I have two solid diagonal lines relatively close together, going this direction: \ , and have shaded the lower half. Does this sound correct??

2) In what way should I plot my lines to graph the inequality: 3x is less than or equal to -4y - 4?
Usually you shade the half that you are DISregarding (this helps when you start linear programming).

What you have said looks right for Q.1, but since you want everything less than the y you have said, you will need to shade everything above these lines. And yes, you are right, they need to be solid.

For each of those inequalities, use a different colour.

2) $\displaystyle 3x \leq -4y - 4$

$\displaystyle -4y - 4 \geq 3x$

$\displaystyle -4y \geq 3x + 4$

$\displaystyle y \leq -\frac{3}{4}x - 1$.

Graph the straight line. It should be solid. Since you want everything that is less than this line as well, you shade everything above the line (what you are disregarding).

3. ## Graphing

x + y is less than or equal to 4
x is greater than or equal to 0
y is greater than or equal to 0

I came up with a right angled triangle in the middle, which I've shaded. Is this correct or am I way off?

4. Originally Posted by jay1
x + y is less than or equal to 4
x is greater than or equal to 0
y is greater than or equal to 0

I came up with a right angled triangle in the middle, which I've shaded. Is this correct or am I way off?
Yes you are correct. However, since you're supposed to shade everything you disregard, the triangle should be the only thing that is not shaded.

5. Originally Posted by jay1
I would like to rnsure that I am completing my graphs correctly:
1) To graph the solution set of x + 2y is less than or equal to 3 and x + 2y is less than or equal to 4,
I have y is less than or equal to -1/2 x + 2 and y is less than or equal to -1/2 x + 3/2
I have two solid diagonal lines relatively close together, going this direction: \ , and have shaded the lower half. Does this sound correct??`
You "two solid diagonal lines" should be the graphs of x+ 2y= 3 and x+ 2y= 4 which are parallel lines (slope -1/2), but it is not clear what you mean by the "lower half". Those two lines divide the plane into three regions. x= 1, y= 2 is on the first line. If x= 1, y= 0, that is below the first line. But you can check that 1+ 2(0)= 1< 3 so any point below that line satisfies the first inequality. Further, (1, 0) is clearly below the second line, since it is higher and 1< 4 so points below the second line satisfy that inequality. Points that are below the lower line are below both and so satify both inequalities. If that was what you mean by "the lower half", then you are correct. (I would have said "the lowest region".)

2) In what way should I plot my lines to graph the inequality: 3x is less than or equal to -4y - 4?
There is only a single inequality and so only a single line. You get it by changing the "$\displaystyle \le$" to "= ". Draw the line 3x= -4y- 4. One way to do that is to use the "intercepts", where the line crosses the axes: when x= 0, the equation becomes -4y- 4= 0 which is the same as 4y= 4 or y= 1. The graph crosses the y-axis at (0, 1). When y= 0, the equation becomes 3x= 4 or x= -4/3. The graph crosses the x-axis at (-4/3, 0). Mark those two points and draw the line through them.

If you were very clever, you could have thought "I can avoid fractions by making sure the right side (which is -4(y+1)) a multiple of 3 and the left side (3x) a multiple of 4. That is, take y= 2 so that y+1= 3 and -4(y+1)= -12. Then the equation becomes 3x= -12 so x= -4. Mark the point (-4, 2). Take x= 4 so that 3x= 12 and the equation becomes 12= -4(y+1) so y+ 1= -3 and y= -4. Mark the point (4, -4) and draw the line through (-4, 2) and (4, -4).

6. Thanks. My assumption is that I should shade the upper region, is that correct? (and just to confirm, I have plotted my single line in this direction: \ )

Also I am trying to ensure that I have the correct equation for y plotted at (0, 1) and x at (2, 0) and find the inequality for the shaded region, which is the upper region above my dashed line, going this direction: \ ) can you help me with this?