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## Digital SAT Math

### Course: Digital SAT Math>Unit 2

Lesson 7: Linear inequality word problems: foundations

# Systems of linear inequalities word problems — Harder example

Watch Sal work through a harder Systems of linear inequalities word problem.

## Video transcript

- [Instructor] We're told, "Luis is cooking meals for at least 20 people. He estimates that the cost of each vegetarian meal is \$3, and the cost of each meal with meat is \$4.50. His budget for the meals is no more than \$100, and he wants to cook at least six of each type of meal. Which of the following systems of inequalities represents the conditions described if x is the number of vegetarian meals and y is the number of meals with meat Luis cooks?" So pause this video and have a go at this yourself before we work through it together. And I know this is a long question and these systems feel complicated, but trust me, if you do it step-by-step you'll actually find that it all falls into place. All right, now let's work through it together, and it's important to emphasize that they've already defined the two key variables for us. x is the number of vegetarian meals and y is the number of meals with meat. So let's look at each of the constraints they give us and each of these can set up a different inequality. So the first one is, they say, "Luis is cooking meals for at least 20 people." So that tells us that the total number of meals, which is going to be the number of vegetarian meals, that's x, plus the number of meat meals, that has to be at least equal to 20. So that has to be greater than or equal to 20. That's what that first sentence tells us. And if I wasn't doing this as a multiple choice, I would just keep adding more and more constraints here, but they give us some choices. And so we can see that x plus y is greater than or equal to 20, that's in choice A. It's actually not in choice B. So we can already rule out choice B. They have less than or equal to 20 here. Same thing for choice C. So we can rule that out. And then choice D does have that. So we are still in the running. The next constraint they tell us, "He estimates that the cost of each vegetarian meal is \$3, and the cost of each meal with meat is 4.50. His budget for the meals is no more than \$100." So how much is he gonna spend in total? Well, on the vegetarian meals, he's going to spend the number of vegetarian meals times \$3 per meal. So that's how much he's going to spend on vegetarian meals. And what about meat meals? Well, it's going to be y meals times 4.50 per meal. So it's 4.5 times y. The amount that he's spending on vegetarian meals, the amount that he's spending on non-vegetarian meals, that's the total he's spending on meals. And they say it is no more than \$100. So this has to be less than or equal to 100. And so let's see, we actually over here, we have 3x plus 4.5y is greater than or equal to 100. So we can rule choice A out as well and just by deductive reasoning, we see choice D does have that in there. But this must be the answer, but let's keep going to make sure that these other constraints work. We are also told he wants to cook at least six type of each meal. So that means that x, the number of vegetarian meals, has to be greater than or equal to six. And y, the number of non-vegetarian meals, also has to be greater than or equal to six. And we see both of these down here, so we can feel pretty good about choice D.