Question 1182457
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Check out this similar problem
<a href = "https://www.algebra.com/algebra/homework/expressions/expressions.faq.question.1182456.html">https://www.algebra.com/algebra/homework/expressions/expressions.faq.question.1182456.html</a>


Hopefully that's enough to tackle this current problem. If not, then I'll show you how to solve.


Let's draw the graph.


The boundary line is the equation -5x+2y = 0, which is the same as y = (5/2)x. This line goes through (0,0) and (2,5). We make the boundary line a dashed line to tell the reader "points on the boundary are <u>not</u> solutions".


If we plugged in the coordinates of choice A(8,-10), then we get
-5x+2y < 0
-5(8)+2(-10) < 0
-40-20 < 0
-60 < 0
This is a true statement since -60 is smaller than 0. On a number line, -60 is to the left of 0. The last inequality being true indicates the first one is true when (x,y) = (8,-10)


Because -5x+2y < 0 is true for choice A, this means we shade the entire region in which point A is located. This shades below the dashed boundary line as shown below


Graph of -5x+2y < 0
<img width="25%" src = "https://i.imgur.com/rTTPGgF.png">
Points A, B, C are in the shaded region. Therefore, they are solution points. In contrast, points D and E are not solutions as they are outside the shaded region.


A non-visual approach is to plug the coordinates of each point into the inequality to see which result in true statements or not. You should find that points A through C result in true statements while D and E lead to false statements. 


Here's an example of a false statement
We'll plug in the coordinates for choice D
-5x+2y < 0
-5(-4)+2(-2) < 0
20-4 < 0
16 < 0
we can see this is false because it should be 16 > 0

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