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### Course: Digital SAT Math > Unit 2

Lesson 6: Systems of linear equations word problems: foundations# Systems of linear equations word problems — Harder example

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

## Want to join the conversation?

- you can solve it with less time and effort. you can substract 4 children and 1 adult from each choice and see if the remaining children are double the adults since each of the remaining adults brought 2 children. think smarter not harder haha(84 votes)
- Agnes has 23 collectible stones, all of which are labradorite crystals or galena crystals. Labradorite crystals are worth $20 each, while galena crystals are worth $13 each. Agnes earns $439 by selling her entire collection. How many stones of each type did she sell?

I'm still stuck on this one problem....please help!(21 votes)- 20 labradorite and 3 galena

because 20*20=400 and 3*13=39 so the total will be 439 which is exactly how much Agnes made(25 votes)

- good concept but a little bit confusing(35 votes)
- That was a tricky one!(31 votes)
- Right do you need help>(1 vote)

- My solution:

Consider the number of adults to be x

Subsequently, the number of children can be written in terms of x as "2x + 2". (Each adult brings 2 children and one of the adults brought 2 extra children)

Now multiply the number of adults and children with their corresponding prices and equate that to 60.

That is "2(2x + 2) + 4x = 60"

Upon solving, you shall get x = 7 which implies that there are 7 adults. Thereafter, you can also find out the number of children which is "2x + 2" and that is 16. Thus, Option C is correct.(23 votes) - Why are we given harder practice problems then what the example shows? This one looks more simple, but the other practice problems are so complicated sometimes. The SAT is coming soon, I just feel so under prepared..(19 votes)
- exactly. this is why i find these videos ineffective sometimes. i wish the exercises had the same difficulty level as the videos(2 votes)

- Here's my solution:

4+2(a-1)=c

4+2a-2=c

2a+2=c

2(2a+2)+4a=60

4a+4+4a=60

8a=56

a=7

*Note: a-1 = number of children who brought 2 children

subtract the one who brought 4(11 votes) - A lil confusing(10 votes)
- I did it in a different way

Since each adult brought 2 children so the number of children is double the number of adults except one adult brought extra two

So the number of children is double plus two the number of adults(7 votes) - Better use the calculator and check de options. 7(4)+16(2)=60(6 votes)

## Video transcript

- [Instructor] Tickets for a
play were $2 for each child and $4 for each adult. At one showing of the play,
one adult brought four children and the remaining adults
brought two children each. The total ticket sales from the
children and adults was $60. How many children and
adults attended the play? Alright, this is an interesting one. Okay, so let's just think about how much we've spent at the play and we know it has to add up to $60. Let's think about it in
terms of the children and the adults and their admissions. So you have this one adult right over here that brought four children. So how much is that adult,
how much is this family? Let's just assume it's a family. How much are they going to spend? Well, that one adult is going to spend $4 for their own ticket and
then four children at $2 each. So plus four children times $2 per child, this is going to be $8
for the children's tickets plus $4 on theirs. They're going to spend $12. So that adult is going to spend $12. And then there's some
remaining number of adults that brought two children. So let's just say r is I could say the remaining number of adults or the number of adults with two children. Adults with two children, that's r. So each of these adults with two children, how much are they going to spend? Well, they're each going to
spend $4 on their own ticket for the adult and then they're
gonna have two children at $2 each, so they're gonna spend $4 on the children's tickets. So they're gonna spend $8 in total. So each of these adults with two children is gonna spend $8 at the
play and there's r of them. So they're going to spend
$8 for each of these adults with two children and there are r of them. So this is the total
amount of ticket sales from the adults with two children. And we add that to the ticket sales from this one adult with the four children and they're gonna have to add up to $60. So this is gonna have to add up to $60. Let's see, we can subtract
12 from both sides and so on the left, we'll be left with 8r is equal to 60 minus 12 is 48. Divide both sides by eight
and you get r is equal to six. So we wanna be very careful, you might say, okay there was six adults these are just the
adults with two children. There's six adults with two children, but there's another adult. There's another adult who
brought four children. So there's a total of seven adults, seven adults total. Pardon my handwriting. Seven adults total. So we could just look at these choices, only one of these choices
have seven adults. And we could verify that this would also amount to 16 children because this person up here, in magenta, they would bring four children, so you would have four children plus six adults brought two children, so six adults bringing two children each, that would amount to 12 children. And that indeed, that
indeed does add up to be 16. And if you're under time pressure, you can see there is only one
choice that has seven adults, so you could just pick that one.