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How do you solve #y>x+1# and #y<2/3x + 3#?

Answer 1

Solve the system:
(1) y > x + 1
(2) # y < (2x)/3 + 3#

First graph Line (1): y = x + 1 by its 2 intercepts. Make x = 0 --> y = 1. Make y = 0 --> x = - 1. The solution set of the inequality (1) is the area above the Line (1). Color or shade it.

Next, graph the Line (2): #y = (2x)/3 + 3# Make x = 0 --> y = 3. Make y = 0 --> #x = - 9/2# The solution set of the inequality (2) is the area below this Line (2). Shade or color it. The compound solution set of the system is the commonly shared area.

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Answer 2

Julia Adams

To solve the system of inequalities (y>x+1) and (y<\frac{2}{3}x + 3), you first graph each inequality on the coordinate plane. Then, find the region where the shaded areas overlap, representing the solution to the system. Finally, determine the values of (x) and (y) within this overlapping region that satisfy both inequalities.

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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 solve #y&gt;x+1# and #y&lt;2/3x + 3#?

How do you solve #y>x+1# and #y<2/3x + 3#?