Problem

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

04:40

Problem 66 Hard Difficulty

Graph the solutions of each system.
$$
\left\{\begin{array}{l}
{3 x+y<-2} \\
{y>3(1-x)}
\end{array}\right.
$$

Answer

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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 the system off Inequality three x plus y has lost the negative to and why this greater than three times one minus X. And we are I lost a graft. The solutions of the system. Yeah, the first thing we can do about that IHS, look for the equation of the life. So here we want the y to be isolated and this will give us why, Yes, less than negative three X minus two. We just simply why is equal to negative three x minus dio And here we just want to distribute the E three inside the parentheses and that will give us why IHS were than three minus three x So hard line is simply going to be why is equal to negative three X plus three. And with this and for me so we can start to grab. Our first line is going to be negative three X minus two. So we know we have. And hey, on intercepted negative too in a slope of negative three. That's I agree. Is this our line? So this is our why's equal to I got a three x minus two and our next line is going to be having intercept at three. Girls are having the same Salo as negative three. So does this. R y is equal to negative three bucks plus the re and to know which part of the graph is going to be shaded. We can have 00 as our test point and plugging it into this and equality, and this will give us zero. It's less than, um zero minus two. So this will end up with zero is less than negative, too, which is not the case. So we know that little sheet off the area that that is not included, that that test point is not included. And we're going to do the same for the other line, which we will have 00 as their test boys and plugging it into our inequality. And this will have us with zero, is greater than zero plus three and will have zero Let's go here than three, which is not the case. So we know to shade off part of the graph that the test point is not included. So in this case, we don't have any solutions, since there is no overlap

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