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How do you graph the inequality #y < -2x + 6# and #y< x + 2#?

Answer 1

Eva Abramson

See the triangular shaded region, for answer.
The zenith is ( 4/3 10/3 ),. and the region is #darr#..

#y < -2x + 6 rArr 3 - y / 2 > x#,

#y < x + 2 rArr x > y - 2#. Combining,

#3 - y/2 > x > y - 2 rArr 3 - y / 2 > y -2 rArr y < 10 / 3 #.

The shaded triangular region below (4/3 10/3 ), sans the vertex, is

the answer. Note that the graph is the combined graph for

(y - x - 2)(y-6+2x)>0 that includes the graph for the the

reversed inequalities, as well.. The shaded triangular portion above

the vertex ( including the vertex ), has to be ignored.. graph{(y - x - 2)(y-6+2x)>0[-20/3 20/3 -10/3 10/3]}

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Answer 2

Eva Adamson

To graph the system of inequalities (y < -2x + 6) and (y < x + 2), you first graph the boundary lines for each inequality. Then, you shade the region below both boundary lines since the inequalities are less than. Finally, you indicate the intersection of the shaded regions as the solution area.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you graph the inequality #y &lt; -2x + 6# and #y&lt; x + 2#?

How do you graph the inequality #y < -2x + 6# and #y< x + 2#?