Use a separate sheet of paper to solve the system of inequalities by graphing. Use the graph to decide if the point (1, –1) is in the solution set of the system. Explain your answer in the box provided below. y≤-3x+3 3x+8y≥3

The answer is no though, the point (1,-1) is below both.

Step-by-step explanation:

So the two inequalities are y≤-3x+3 and 3x+8y≥3

With inequalities you have to know a few things.  First, when you graph you may have to use non solid lines.  for instance, if you had x<5, 5 would not be part of the answer because it is only answers less than 5.  so you represent this by making a dashed line that says "you cannot include values on this line".  Second, with algebra, if you multiply both sides by a negative it flips the inequality.  for instance -y > 54 you would multiply (or divide) by -1 and it would change to y < -54.  Third, when graphing an inequality, treat it like an equation and graph the line as if it were said equation(with solid or dashed lines as needed)  then, if it is y > f(x) you shade in the section above the line and y < f(x) shades in below.  The shaded part is your answer basically.  Now lets get into yours.

y≤-3x+3

It is already in the form y = f(x)  so grapht he line 3x + 3 with a solid line since it is y is less than or equal to,  then shade below that line.  

3x+8y≥3

First get it so y is by itself

3x+8y≥3

8y≥3 - 3x

y ≥ 3/8 - 3/8 x

Now grapht his line  with a solid line and shade the area above it.

Now, the shaded area below between y ≥ 3/8 - 3/8 x and  y≤-3x+3 and the lines are your answer, so you need to check if the point (1, -1) is in this shared shaded area or on one of the lines.

The answer is no though, the point (1,-1) is below both.

idk

step-by-step explanation: i just don't

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.