SOLUTION: There were 10 more reds than 8 times the number of blues. Also, the number of reds was 5 less tahn 11 times the number of blues. How many of each were there?

Question 165302: There were 10 more reds than 8 times the number of blues. Also, the number of reds was 5 less tahn 11 times the number of blues. How many of each were there?
Found 2 solutions by gonzo, Mathtut:
Answer by gonzo(654) (Show Source): You can put this solution on YOUR website!
There were 10 more reds than 8 times the number of blues.
R = 8*B + 10 (first equation)
Also, the number of reds was 5 less tahn 11 times the number of blues.
R = 11*B - 5 (second equation)
since they both = R, then they are equal to each other.
8*B + 10 = 11*B - 5
subtract 8*B from both sides.
10 = 3*B - 5
add 5 to both sides.
15 = 3*B
divide both sides by 3.
5 = B
substitute for B in first equation.
R = 8*5 + 10
R = 40 + 10
R = 50
substitute for R and B in second equation.
50 = 11*5 - 5
50 = 55 - 5
50 = 50
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R = 50
B = 5
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Answer by Mathtut(3670) (Show Source): You can put this solution on YOUR website!
B=blue R=Red R=8B+10 R=11B-5 so substituting (8B+10)=11B-5 subtract 8B and add 5 to each side and you get 3B=15 B=5 so R= 8(5)+10=50
Blue=5
Red =50
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