Point (8,10) is not included in the solution area of the inequalities 2x+y<=8 and x+y>4.
Given two inequallities 2x+y<=8 and x+y>4.
We are required to graph the inqualities and find whether (8,10) is included in the solution of these inequalities.
To graph the inequalities we have to assume both the inequalities as equalities.
2x+y=8 x+y=4
x 0 4 x 0 4
y 8 0 y 4 0
Points of 2x+y<=8 are (0,8),(4,0).
Points of x+y>4 are (0,4),(4,0).
Now we have to graph these points to find the line.
Put (0,0) in both the inequalities to find how to shade the graph.
2x+y<=8
0<=8
True means shade will be towards origin.
x+y>4
0>4
False means shade will be away from origin.
From the graph we can say that (8,10) is not included in common shaded region of both the inequalities so it is not part of inequalities.
We have chosen (1,2) , it means the number of cupcakes is 1 and the number of siblings is 2.
Hence (8,10) is not a part of the solution of the inequalities 2x+y<=8,x+y>4.
Learn more about inequalities at https://brainly.com/question/24372553
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