Problem

For each nonlinear inequality in Exercises $33-40…

Problem 39 Easy Difficulty

For each nonlinear inequality in Exercises $33-40,$ a restriction is placed on one or both variables. For example, the inequality
$$x^{2}+y^{2} \leq 4, \quad x \geq 0$$
is graphed in the figure. Only the right half of the interior of the circle and its boundary is shaded, because of the restriction that $x$ must be nonnegative. Graph each nonlinear inequality with the given restrictions.
$$
x^{2}+4 y^{2} \geq 1, \quad x \geq 0, y \geq 0
$$

Answer

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Precalculus

Algebra

Beginning and Intermediate Algebra

Chapter 13

Nonlinear Functions, Conic Sections, and Nonlinear Systems

Section 5

Second-Degree Inequalities and Systems of Inequalities

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Video Transcript

Hey, what's up, guys? Eric? And today I'm gonna be walking us through how to solve this inequality over here graphically and AJ directly. So let's get started. I've graphed here absolutely, for close one minus three in red. And you y equals two in blue here with a horizontal line here. So, um, in our inequality were asked, defying the domain essentially grass if on the domain for X that satisfies this inequality that makes this inequality troop. So if we're asked when the left hand side is greater than the right hand side when the red side is greater than the right on the blue side were asked to find where on which vise of X is the red higher up than the blue. So you see, the two lines intersect, er the two grabs intersect. A X equals negative Three halves and X equals one. And we clearly see that the red is greater than the blue from negative infinity up to ah, negative three halves. And from one, all the two positive infinity. And I mean that means our demand of X that makes us about inequality is from negative infinity up to, but not including negative. Three halves union from one, not including one up to positive infinity. You have worked out. Ah, question directly here I've combined light terms and said, Isolated the absolute value. Set it up so that we're comparing the magnitude of the absolute value here. For one for X plus, one is greater than five and a four plus one is less than negative. Five. Ah, and those are equivalent expressions for this absolute value. And then I subtracting one from both sides and dividing four from both sides gets ourselves these two inequalities. X is greater than one and X is less than negative, perhaps. And that's the exam. Exact answer as what we determined. A graphically Yeah, that's pretty much it. Thanks for watching.

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