Question Video: Solving Systems of Inequalities by Graphing | Nagwa Question Video: Solving Systems of Inequalities by Graphing | Nagwa

# Question Video: Solving Systems of Inequalities by Graphing Mathematics • First Year of Secondary School

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Which of the following points belongs to the solution of the set of the inequalities 𝑥 > −2, 𝑦 < 4, 𝑥 + 𝑦 ≤ −1? [A] (1, 1) [B] (−1, −1) [C] (−2, 1) [D] (−2, −4)

04:56

### Video Transcript

Which of the following points belong to the solution of the set of the inequalities 𝑥 is greater than negative two, 𝑦 is less than four, 𝑥 plus 𝑦 is less than or equal to negative one? (A) one, one, (B) negative one, negative one, (C) negative two, one, or (D) negative two, negative four.

We’ve been given three different inequalities, and a point that belongs to the solution set of these three inequalities will make all three of these inequalities true. One way to solve this would be to simultaneously graph all three inequalities on a coordinate grid. First, we sketch a coordinate grid. And then we want to graph the inequality 𝑥 is greater than negative two. Because 𝑥 is greater than negative two but not equal to negative two, we’ll need to use a dotted line instead of a solid line. This vertical line is running all the places where 𝑥 would be equal to negative two. Since we’re only interested in the places where 𝑥 is greater than negative two, we’re going to eliminate everything to the left of this line. The values for 𝑥 is greater than negative two will all be to the right of this line.

And now moving to 𝑦 is less than four. Because 𝑦 is not equal to four, it’s only less than four, again we’ll use a dotted line and not a solid one. Here is our horizontal line that indicates 𝑦 being four. We want to eliminate all of the places where 𝑦 is greater than four. Our solution will not be above this pink line.

Now that we’ve graphed both 𝑥 is greater than negative two and 𝑦 is less than four, we know that both of these things have to be true. And therefore, our solution is going to be somewhere on the right side of the yellow line and below the pink line. However, we have one additional inequality to graph. In order to graph this, we want to rewrite it so the 𝑦-value is by itself. We can do that by subtracting 𝑥 from both sides of this inequality, which gives us 𝑦 is less than or equal to negative 𝑥 minus one.

Notice for this inequality, we do have less than or equal to, which means we’ll be dealing with a solid line. This value has a point at negative one, zero and a slope of negative one. This is what the solid line would look like. The 𝑦-value needs to be less than or equal to this line, and that means we want to exclude any value that falls above this line. We’re only interested in the parts below. Any values above this green line are excluded, and that means the area where our solution can fall will be here. If we try to graph one, one, it’s above our green line, and therefore (A) is not possible.

If we graph negative one, negative one, it does fall within our solution set, so we’ll check that off for option (B). We should be careful with the point negative two, one because it does fall on the green line. However, it also falls on the yellow dotted line. 𝑥 must be greater than negative two. It cannot be equal to negative two. And because of this, both option (C) and option (D) have to be eliminated. On our graph, only the point negative one, negative one falls in the solution set of these three inequalities. What we’ve just shown is a graphical method for solving.

However, you could have solved this problem algebraically. You could’ve said, “Okay, 𝑥 must be greater than negative two. Option (C) and (D) have 𝑥 being equal to negative two, which is not possible.” If you checked 𝑦 is less than four, you would see that all four of these options have 𝑦 being less than four. That doesn’t eliminate any. However, when you got two 𝑥 plus 𝑦 is less than or equal to negative one, you would’ve realized that one plus one is two, and that is not less than or equal to negative one, which would again eliminate option (A). Only option (B) satisfies all three of these inequalities and is therefore part of the solution set.

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