Question 870471
If the ordered pair, or point, (2,-2) was a solution to the system of inequalities, then it has to make ALL of the inequalities true inequalities (ie true statements). 
Something like 1 < 5 is true because 1 is indeed less than 5. 
On the other hand, 1 > 5 is false because 1 is smaller than 5 (not the other way around).



To determine if (2,-2) is a solution to the system, we simply plug in (x,y) = (2,-2) into each inequality individually. 



Let's plug in (x,y) = (2,-2) into y < x-3



y < x-3


-2 < 2-3 .... plug in (x,y) = (2,-2), ie plug in x = 2, y = -2


-2 < -1



The last inequality -2 < -1 is true since -2 is less than -1. 


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Let's plug in (x,y) = (2,-2) into y > -x+3



y > -x+3


-2 > -(2)+3 .... plug in (x,y) = (2,-2), ie plug in x = 2, y = -2


-2 > -2 + 3


-2 > 1



The last inequality -2 > 1 is false since -2 is really smaller than -1 (not larger).


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So (x,y) = (2,-2) makes y < x-3 true, but it makes y > -x+3 false.



To be a solution to the system, it has to make ALL of the inequalities of that system true. Because at least one is false (ie not all are true), (x,y) = (2,-2) is NOT a solution to the system.



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Final Answer: (2,-2) is NOT a solution to the system.