OpenStudy (anonymous):

A delivery truck is transporting boxes of two sizes: large and small. The large boxes weigh 40 pounds each, and the small boxes weigh 15 pounds each. There are 135 boxes in all. If the truck is carrying a total of 3650 pounds in boxes, how many of each type of box is it carrying?

2 years ago
OpenStudy (anonymous):

2 years ago
OpenStudy (anonymous):

@MattAlva98

2 years ago
OpenStudy (anonymous):

i am doing my math

2 years ago
OpenStudy (anonymous):

okay.

2 years ago
OpenStudy (anonymous):

@mathmate

2 years ago
OpenStudy (mathmate):

Have you already done systems of two equations?

2 years ago
OpenStudy (anonymous):

this is hard

2 years ago
OpenStudy (anonymous):

no... not sure how to set that up

2 years ago
OpenStudy (anonymous):

40x +15y=135??

2 years ago
OpenStudy (mathmate):

So you have already learned system of two equations, right?

2 years ago
OpenStudy (anonymous):

yes

2 years ago
OpenStudy (mathmate):

Ok, do you want the normal way, or the easy way?

2 years ago
OpenStudy (anonymous):

idc i just really neeed to get this question answered ,

2 years ago
OpenStudy (mathmate):

I will set up both, will let you choose which one to use, but I will not solve it for you. You can choose to give me your answer to check if you want. Do you want me to go ahead?

2 years ago
OpenStudy (anonymous):

okay yes

2 years ago
OpenStudy (mathmate):

Method 1: Let x=number of heavy boxes at 40 lbs ea. y=number of small boxes at 15 lbs ea. Total number of boxes = x+y = 135 .....(1) Total weight = 40x+15y = 3650 ......(2) So you can solve for x and y in the above system of equations. method 2: Let x=number of heavy boxes at 40 lbs each (135-x) = number of small boxes at 15 lbs each. Total weight = 40x + 15(135-x) = 3650 Distribute and solve for x. Do you have any question on the setups? I will give you a third method in a post that follows.

2 years ago
OpenStudy (anonymous):

x=65 y=70

2 years ago
OpenStudy (anonymous):

so the small box is 65 and the large is 70?

2 years ago
OpenStudy (mathmate):

Yes, try checking using 65*40+70*15=?

2 years ago
OpenStudy (mathmate):

Good job!

2 years ago
OpenStudy (anonymous):

thanks

2 years ago
OpenStudy (mathmate):

You're welcome! :)

2 years ago
OpenStudy (anonymous):

i have one more if you dont mind

2 years ago
OpenStudy (anonymous):

@mathmate

2 years ago
OpenStudy (anonymous):

Two mechanics worked on a car. The first mechanic worked for 10 hours, and the second mechanic worked for 5 hours. Together they charged a total of 1350 . What was the rate charged per hour by each mechanic if the sum of the two rates was 195 per hour?

2 years ago
OpenStudy (mathmate):

Why don't you try your setup on this one, it's almost the same as the previous!

2 years ago
OpenStudy (anonymous):

x+y = 1350 and 10x +5y=195?

2 years ago
OpenStudy (anonymous):

is that right?

2 years ago
OpenStudy (mathmate):

Almost, just two things. Always define the variables used, so when you set up the equations you will not get confused.

2 years ago
OpenStudy (mathmate):

So what do x and y stand for?

2 years ago
OpenStudy (mathmate):

Define them before you set up equations!

2 years ago
OpenStudy (anonymous):

x is for the numberof hours the first mechinic worked

2 years ago
OpenStudy (anonymous):

y is for the second mechianic

2 years ago
OpenStudy (mathmate):

If you reread the question and compare with your definitions of x and y, you will see a little problem, can you spot that?

2 years ago
OpenStudy (anonymous):

uhh not really

2 years ago
OpenStudy (mathmate):

"The first mechanic worked for 10 hours, and the second mechanic worked for 5 hours" so you don't really need x=10 and y=5. Think of x and y as numbers you need to find.

2 years ago
OpenStudy (anonymous):

so is my first equation right just not the second?

2 years ago
OpenStudy (mathmate):

Equations depend on what the variables x and y represent. I cannot say which one is right or wrong before we know what x and y stand for. That is why we need to define x and y before setting up equations, agree?

2 years ago
OpenStudy (anonymous):

okay

2 years ago
OpenStudy (mathmate):

remember that charges are proportional to the number of hours worked and the rate. So if the first one charges $x per hour and worked for 10 hours, his charge will be 10x dollars. 2 years ago OpenStudy (anonymous): so if the mechinaic that worked 10 hours charged 10 a hour? 2 years ago OpenStudy (mathmate): We don't know if he charged$10 an hour, that's why we need to solve the problem.

2 years ago
OpenStudy (anonymous):

okay, i just need help setting up the equations i have it from there

2 years ago
OpenStudy (mathmate):

Can you define your variables x and y, knowing the number of hours worked, and the total bill?

2 years ago
OpenStudy (anonymous):

please this thing is timed i really need to get it answered ::(

2 years ago
OpenStudy (anonymous):

im not asking for the answer just show me how to set up the equations again please

2 years ago
OpenStudy (mathmate):

I am not even supposed to help you if this is timed, meaning this is a test. But I will give you a last hint: The only thing you don't know is the rate in $/hour. So Let x=rate for first person in$/hour, and y=rate for second person in \$/hour. Proceed as in the first question. You should have no problems with that.

2 years ago
OpenStudy (anonymous):

not a test, it will log me out after a few minutes of not using it.

2 years ago
OpenStudy (anonymous):

i got it it was 75 and 120

2 years ago
OpenStudy (anonymous):

thanks

2 years ago
OpenStudy (mathmate):

Well, if you define the variables as above, you can give it a try with the setup. Remember that defining the right variables is the most important step in the setup.

2 years ago
OpenStudy (mathmate):