Problem

Graph the solutions of each system. $$ \left\{\be…

03:46

Problem 32 Medium Difficulty

Graph the solutions of each system.
$$
\left\{\begin{array}{l}
{y>2 x-4} \\
{y \geq-x-1}
\end{array}\right.
$$

Answer

$y>2 x-4 \mathrm{i}$

Related Courses

Algebra

Elementary and Intermediate Algebra

Chapter 4

Systems of Linear Equations and Inequalities

Section 5

Solving Systems of Linear Inequalities

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Video Transcript

So here we have our system of inequalities. Starting with why, ISS Greer, Than to, uh, X minus four. And why is greater than or equal to? Negative one Negative X minus one on the first thing we will do is look for our lines, a question of lines and this Wilson. Think of us. Why is equal to two X minus four? And here are lion is going to be. Why is equal to negative X minus one. And with these information, we can start graphic. Our first line is going to be. Why did you go to two X minus four? So we have ah, intercept that Negative 41234 And our slope is a positive too. So that's what we want to. I am too. Okay, So here is our life. So this is our why is equal to two looks when this four, our next line is going to be. Why is it go to negative X minus one? So we have a y intense up that negative one and a negative one slope. So that is simply going to bid this line. Why is equal to negative X minus one now to check which fitted well which areas are going to be shaded. We can choose a point and our faces and choose this 0.0 and plug it into this inequality. And this will give us zero greater than zero minus four. And this will give us zero greater than negative for which satisfies our condition. And actually, our line is going to be dotted here since our inequality is restricted now we can shade the area where the point ISS contained next is going to be nous and equality. And we can choose the same point at 00 and plug it into this inequality. And this will simply give us zero is greater equal to zero minus one. Just skip us. Zero is great and are equal to negative one which satisfies our condition again. So we can she the area where the point is contained, which is this area? Nowhere do these graphs overlap, and that is going to be here. And this is going to be our solution

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