For each point listed below, circle every inequality for wwhich it is a solution.   (0,0).   Y<=3/2x+3         2x+y<10         Y>-1

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Substitute 0 in for x and y into all inequalties.  Simplify and determine if the inequality is true or false.  If the inequality is true, then (0, 0) is part of the solution.  If the inequality is false, then (0, 0) is not part of the solution.

y `<=` 3/2...

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Substitute 0 in for x and y into all inequalties.  Simplify and determine if the inequality is true or false.  If the inequality is true, then (0, 0) is part of the solution.  If the inequality is false, then (0, 0) is not part of the solution.

y `<=` 3/2 x + 3

0 `<=` 3/2 * 0 + 3

0 `<=` 0 + 3

0 `<=` 0          true          (0, 0) is part of the solution

 

3x + y < 10

3 * 0 + 0 < 10

0 + 0 < 10

0 < 10          true          (0, 0) is part of the solution

 

y > -1

0 > -1          true          (0, 0) is part of the solution

 

Answer:

The point (0, 0) is part of the solution to all three given inequalities because when substituted in for x and y, the resulting inequalities were all true.

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The request of the problem is vague, hence, supposing that you need to check if the coordinates `(0,0)`  verify the given inequalities, you should substitute 0 for x and 0 for y such that:

`0 <= (3/2)*0 + 3 => 0 <= 3`

Notice that this inequality is invalid since 0`< 3 ` but `0 != 3` .

Substituting 0 for x and y in `2x+y<10`  yields:

`2*0 + 0 < 10 => 0 < 10`

Hence, the inequality holds for `(0,0).`

Substituting 0 for  y in `y>-1`  yields:

`0>-1`

Hence, the inequality holds for `y=0`  and it not depends on x, hence, x can have all real values.

Hence, checking if the coordinates of the point `(0,0)`  verify the given inequalities yields that the inequality `2x+y<10 ` holds for `(0,0).`

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