Simultaneous Equations

Date: 8/25/96 at 16:48:15
From: Anonymous
Subject:  Algebra Word Problem

Ten percent of the reds were added to twenty percent of the blues, 
and the total was 24. Yet the product of the number of reds and 3 
exceeded the number of blues by 20. How many were red, and how many 
were blue?

I don't know how to do it from here. 
I don't even know if I set this up right:

10%R x 20%B = 24  
 R x 3 = B+20

Date: 8/25/96 at 18:46:6
From: Doctor Paul
Subject: Algebra Word Problem

Let's call 'R' the number of reds and 'B' the number of blues. Now we 
need two equations:

     (0.10 * R) + (0.20 * B) = 24
                     (R * 3) = B + 20

Let's solve for B in the second equation:

B = 3R - 20

Now substitute this expression for B into the first equation:

(0.10*R) + (0.20*(3R - 20)) = 24

solve for R:

            0.1R + 0.6R - 4 = 24
                       0.7R = 28

R = 40

Now back to the original equation. Substitute R = 40 into either 
equation. For simplicity's sake, I will substitute it into the second 
equation:

                   (R * 3) = B + 20
                  (40 * 3) = B + 20
                         B = 100

(Note that which equation you choose to substitute r = 40 into does 
not matter. If you did the first part correctly, you should get the 
same answer)

So your final answer is that you have 100 Blues and 40 Reds.

I hope this clears things up.

Regards,
-Doctor Paul,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/