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

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

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?

2. ## Re: algebra question

Originally Posted by atwana_b
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.

3. ## 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........

4. ## Re: algebra question

Originally Posted by atwana_b
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:

Originally Posted by Mrspi
x + y = 21
1800x + 2400y = 48000
Solve this system....
...then you need to see your teacher; we cannot teach here.

5. ## 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.

6. Originally Posted by atwana_b
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.

7. ## 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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