Solve the following system of inequalities graphically:
$$
x+y \geq 4, \quad 2 x-y<0
$$

Answer

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Algebra

NCERT Class 11 - Math

Chapter 6

Linear Inequalities

Section 3

Algebraic Solutions of Linear Inequalities in One Variable and their Graphical Representation

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Video Transcript

for this problem. We're solving a system of inequalities by graphing. So our inequalities are too. Why plus X is less than four and then why managed to X is greater than equal to four. So we have our equation for two x two wide plus X is less than four in the red, and then we have our equation for y minus two X is greater than equal to four in the blue. And so this area where the two shaded regions overlap is going to be our solution. Eso it's are feasible region and any points in this feasible region is a solution to the inequality. So just to prove that we're going to choose one point. So let's say this is negative too. And this is Hank before this is negative. Three So negative three comma one. So we have the coordinates negative three comma one. And then I'm gonna write out the inequalities because we're gonna have to plug in the coordinate to prove that it is indeed a solution to the system of inequalities. So our first inequalities to y plus X is listen for, then our second inequality is why minus two X is greater than equal to four. So we're gonna start with plugging in negative three as X and one as why so for our first inequality, we have two times one plus negative. Three is less than four. So two times one is two two plus negative three is a negative one. Then we have negative one is less than four. So that is true. That's correct. So now we're gonna plug these coordinates into the second inequality. Why minus two X is greater than equal before. So we have one minus two times the negative three greater than equal to four. So one negative two times negative three is positive. Six. So we have one plus six is greater than equal to 41 plus six is seven, so we have seven is greater than equal to four, which is also true. So that just proves that any point in our peaceful region is a solution to the system of inequalities. If we disappoint, that's in the blue shaded area. That means that it would only be a solution to y. Minus two X is greater than equal to four. If we choose a point in this red shaded area, that's outside of the feasible region. It would only be a solution to two y plus X is lesson for and if we choose a point, that's outside of, um, the shaded areas of both inequalities, that it wouldn't be a solution to any to the system.

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