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First. we need this equation in y = mx+b form Next, pick a test point.. say (0,0) letting x = 0 and y = 0 and plug it into the inequality . If the inequality holds, then it's true and (0,0) is shaded. If the inequality doesn't hold, then it's false and (0,0) won't be shaded Third, let x = 1,2,3, and solve for y to gain more points to plot Finally, compare the graph to the picture
ok. let's have x = 0 and y = 0 plug those values into the choices you are given which of these inequalities are true?

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@UsukiDoll I'm sorry i just went full retard. I'm reading everything youre telling me, but nothing is making sense I'm so sorry
I'll do an example picking a test point of (0,0) and the inequality x-3y>5 we have 0-3(0) > 5 0-0>5 0>5 and that doesn't make sense at all. so throw that inequality out
there's one more inequality that needs to go. which one is it?
x - 3y < 5
no that inequality holds true 0 is less than 5
yes
dude . x-3y < 5 is a true inequality when the test point is (0,0) so that stays. It's just that there are two of them out there and we need to know which inequality is the same as the picture
so we either have x-3y <5 or x-2y <5 we know that the test point (0,0) is true so that test point is shaded.
wait wouldn't the answer be what i just mentioned? x - 3y < 5
prove that it's correct.... I'm not even done explaining things and this is out of the blue.
oh. no i didn't do anything. i just have strong feelings for that choice. somethings just telling me thats it.
so when you have this on a test, what are you going to do? You can't just guess.
@UsukiDoll thats true. I'm sorry for interrupting you. can you help me figure it out??
anyway changing it to y = mx+b form x-3y =5 \[-3y=5-x\] \[y = \frac{5}{-3}+\frac{-x}{-3} \] \[y = \frac{x}{3} +(\frac{-5}{3})\]
and for x-2y < 5 just change signs to = and then we need y =mx+b form again x-2y=5 -2y=5-x \[y=\frac{-5}{2}+\frac{x}{2}\]
so which colored line is above the -2 in the y-axis?
it was x-3y<5 lucky guess
are you serious?! wow thanks. i had a feeling lol
yeah but I've outlined the process for manual solving (except for graphing because there is no way that I was doing that by hand)

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