Can somebody help me figure out, how to graph the below equation? 3x is equal or greater than -y + 3 How can you graph it, when you only have the one point of 3, for the y intercept? How can you tell which way your line needs to go?
\[3x \ge -y+3\] You have to make it of the form \[y \le something\] You graph that line and it'll be that line and the area above that lIne
Sorry \[y \ge something \] and it's the line and the area below it
Let me attach a picture of the graph that the answer key has....it shows the area above the line shaded..
\[3x \ge -y +3\] \[3x+y \ge +3\] \[y \ge -3x+3\] and you can draw the line of \[y=-3x+3\]
And it'll be the area above it sorry I always get confuse
It's because y is greater than or equal to so above because greater numbers are above
but how did they come up with the second point, of 1, on the x axis?
Because if you plug in y=0 into y=-3x+3 you get x=1 right? Just pretending it's a normal equation not an equality D'you get it?
My textbook just said something about plugging in 0 for x and y both, to determine whether the origin was part of the equation...but anyway..so I just plug in 0 for y, to determine the second point of the line? Does that work like every time, is it a math rule?
Because I will be trying your method on other problems of this type possibly..
If you plug in y=0 you always get the point that it intersects the x-xis
ok thanks, I'll see if it works for my other problems here.
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