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Mathematics 14 Online
OpenStudy (anonymous):

Part a.) Please provide only one ordered pair (x, y) that will work for this system of inequalities. Part b.) prove your answer by substituting the (x, y) order pair into both inequalities. Show all work for each inequality y > 3x + 1 y is less than or equal to -2/3x + 4

Directrix (directrix):

y > 3x + 1 y ≤ -(2/3) x + 4 -------------- @Danky We can try to solve this system of inequalities or test points to find one that satisfies both inequalities. The instructions do not state that the inequalities must be solved by algebra.

OpenStudy (anonymous):

its algebra

Directrix (directrix):

@Danky Are there options for this problem or are you just to name a point and show that it works.

OpenStudy (anonymous):

no options

OpenStudy (anonymous):

these are the complete instructions

Directrix (directrix):

Part a.) Please provide only one ordered pair (x, y) that will work for this system of inequalities. I am going to demonstrate how to determine if a point works for this or does not work for this.

Directrix (directrix):

I choose (0,0) as a test point. I will test it in the first inequality: y > 3x + 1 x = 0 and y = 0 Is 0 > 3*0 + 1? Is 0 > 0 + 1? Is 0 > 1? NO. So the test point (0,0) is NOT an answer.

OpenStudy (anonymous):

i think u would need to graph it then find ordered pairs

Directrix (directrix):

What I am saying is that the instructions did not direct us to graph or to solve the inequalities with algebra. What I read is to find a point and show that it works. So, that is what I am doing.

Directrix (directrix):

Now, you test the point (0, 10). Does it satisfy both inequalities. Begin with y > 3x + 1 Is 10 > 3*0 + 1 ? Yes or No?

OpenStudy (anonymous):

no ?

Directrix (directrix):

10 > 0 --> I say yes because ten is greater than zero.

Directrix (directrix):

Now, test (0,10) here: y ≤ -(2/3) x + 4 Is 10 ≤ -(2/3)*0 + 4 ?

OpenStudy (anonymous):

no?

Directrix (directrix):

A new test point --> (0,3) Begin with y > 3x + 1 Is 3 > 3*0 + 1 ? yes or no?

Directrix (directrix):

3 >0 + 1 3>1 so yes. The test point (0,3) works for the first inequality.

Directrix (directrix):

Let's test for the second inequality. y ≤ -(2/3) x + 4 test point --> (0,3) Is 3 ≤ -(2/3)*0 + 4 ? Is 3 ≤ 0 + 4? Yes. So, an answer for part a is the point with coordinates (0,3).

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