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Rearrange inequality
Now we can find the boundary line. We will do that by temporarily considering the related equation instead. We find a related equation by changing the inequality sign to an equals sign.
Notice that the inequality is strict. This means the boundary line is not part of the solution set which is why we dash the line. Finally, we must shade the correct side of the line. To do that, we will test a point that does not fall on the line in the original inequality. In this case, the easiest point we can choose is the origin.
x= 0, y= 0
Since the inequality holds true, we should shade the side of the boundary line that contains the origin.
Next, we will consider the related equation which will allow us to write the boundary line.
Notice that y<3 tells us that every y-value that is less than 3 is part of the solution set. Therefore, we know that we must shade the region below the boundary line.
LHS+2y≤RHS+2y
LHS-6≤RHS-6
Rearrange inequality
.LHS /2.≥.RHS /2.
a* b/c=a/c* b
Now we can find the boundary line. We will do that by temporarily considering the related equation instead. We find a related equation by changing the inequality sign to an equals sign.
Notice that the inequality is non-strict which is why we keep the boundary line solid. Finally, we must shade the correct side of the boundary line. To do that, we will test a point in the original inequality that is not on the boundary line. In this case, the easiest point we can choose is the origin.
Since the inequality holds true, we should shade the side of the boundary line that contains the origin.
To graph the boundary line, we will consider the related equation.
Like in Part C, we have a non-strict inequality which means the boundary curve is part of the solution set. Therefore, we keep the curve solid. To shade the correct side of the boundary curve, we will substitute a point that is not on the curve into the original inequality. We will choose the origin.
Since the inequality is true when substituting the test point, we should shade the region that contains the origin.