algebra question: In printing an article of 48,000 words....

atwana_b

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In printing an article of 48,000 words, a printer decides to use two sizes of type. Using the larger type, a printed page contains 1,800 words. Using smaller type, a page contains 2,400 words. The article is allotted 21 full pages in a magazine. How
many pages must be in smaller type?
 
Re: algebra question

atwana_b said:
in printing an article of 48,000 words, a printer decides to use two sizes of type. using the larger type, a printed page contains 1,800 words. using smaller type, a page contains 2,400 words. the article is allotted 21 full pages in a magazine. how
many pages must be in smaller type.

Start by NAMING things.

I might do this:

let x = # of pages in large type
let y = # pages in small type

We knkow that there are to be 21 pages in all, so
x + y = 21

We also know that
words on small type pages + words on large type pages = 48000
since the article contains 48000 words.

Using large type, there are 1800 words per page, so x pages of large type will hold 1800x words.

Using small type, there are 2400 words per page, so y pages of small type will hold 2400 words.

Together, the two type sizes must hold 48000 words. So,
1800x + 2400y = 48000

Ok...now you have a system of two equations in two variables:

x + y = 21
1800x + 2400y = 48000

Solve this system....

If you have trouble solving the sysem, please repost showing your work.
 
algebra question

Ok, I am horrible at word problems --

1800x + 2400y = 48000

so would I do this:

1800x = 48000 (divide both sides by 1800) and then x=26.66

2400y = 48000 (divide both sides by 2400) and then y=20

If I did start off right, I am not sure what to do next........
 
Re: algebra question

atwana_b said:
Ok, I am horrible at word problems --
1800x + 2400y = 48000
so would I do this:
1800x = 48000 (divide both sides by 1800) and then x=26.66
2400y = 48000 (divide both sides by 2400) and then y=20
If I did start off right, I am not sure what to do next........
If that's what you did after being given this by Mrs. Pi:

Mrspi said:
x + y = 21
1800x + 2400y = 48000
Solve this system....
...then you need to see your teacher; we cannot teach here.
 
algebra question

I am sorry that you feel that way -- but thanks to Tchrwill & Eliz, I was taught here.

Is there anyone who would be willing to help me understand how to solve this equation? It would really mean a lot.
 
atwana_b said:
Is there anyone who would be willing to help me understand how to solve this equation?
As was mentioned earlier, this is not one equation; it is a system of (two) equations. Since we cannot teach courses here, please use the lessons already available online.

Thank you.

Eliz.
 
algebra question

Once again, you have helped me! Thank you so much

1800x + 2400y = 48000

1800 (-y+21) + 2400y = 48000
-1800y + 37800 + 2400y = 48000
600y + 37800= 48000
600y = 10200
y=17

x + 17 = 21
x = 4

Have a happy holiday!
 
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