Simultaneous EquationsDate: 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/ |
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