Ask your own question, for FREE!
Mathematics 22 Online
OpenStudy (anonymous):

Which inequality matches the graph? -2x + 3y > 7 2x - 3y < 7 -3x + 2y ≥ 7 3x - 2y ≤ 7

OpenStudy (anonymous):

OpenStudy (anonymous):

@texaschic101

OpenStudy (anonymous):

@AlexandervonHumboldt2

OpenStudy (anonymous):

@Michele_Laino

OpenStudy (kidrah69):

which ones can we eliinate? Look at the X axis

OpenStudy (anonymous):

The third option and the last option

OpenStudy (kidrah69):

are those the ones elimnated?

OpenStudy (anonymous):

yes

OpenStudy (kidrah69):

Shade above the line for a "greater than" (y> or y≥) or below the line for a "less than" (y< or y≤).

OpenStudy (kidrah69):

does that help you determine your answer?

OpenStudy (kidrah69):

There are three steps: -Rearrange the equation so "y" is on the left and everything else on the right. -Plot the "y=" line (make it a solid line for y≤ or y≥, and a dashed line for y< or y>) -Shade above the line for a "greater than" (y> or y≥) or below the line for a "less than" (y< or y≤). http://www.mathsisfun.com/algebra/graphing-linear-inequalities.html

OpenStudy (anonymous):

So it would be the first one right?

OpenStudy (kidrah69):

yep i think thats right

OpenStudy (anonymous):

Yeah so do I.

OpenStudy (anonymous):

Thanks!

OpenStudy (kidrah69):

:)

OpenStudy (texaschic101):

2x - 3y < 7 -3y < -2x + 7 y > 2/3x - 7/3 slope is 2/3....y int (0,-7/3)......-7/3 = - 2 1/3 x int can be found by subbing in 0 for y 0 > 2/3x - 7/3 -2/3x = -7/3 x = (-7/3) / (-2/3) x = -7/3 * -3/2 x = 21/6.....so x int (21/6,0).....21/6 = 3 1/2 and since it is a dashed line, there is no equal sign....and greater then means shaded above. I believe your graph is B

OpenStudy (texaschic101):

@FEARLESS_JOCEY

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!