# SOLUTION: here is the problem: three times the number of blue marbles exceeded twice the number of red marbles by 18 Also 5 times the number of blue marbles was 2 less then 6 times the numbe

Algebra ->  Algebra  -> Human-and-algebraic-language -> SOLUTION: here is the problem: three times the number of blue marbles exceeded twice the number of red marbles by 18 Also 5 times the number of blue marbles was 2 less then 6 times the numbe      Log On

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 Click here to see ALL problems on Human-and-algebraic-language Question 601393: here is the problem: three times the number of blue marbles exceeded twice the number of red marbles by 18 Also 5 times the number of blue marbles was 2 less then 6 times the number of red marbles. here is how far I got 3(NB)= 2(nr)+18 5(NB)= 6(nr)-2 nb= 6/5(nr)-2/5 3(6/5nr-2/5) = 2(nr)+18 18/5NR-6/5 = 2NR =18Answer by bucky(2189)   (Show Source): You can put this solution on YOUR website!Here are comments to your work: . 3(nb)= 2(nr)+18 <---- correct (I put NB into lower case) 5(nb)= 6(nr)-2 <---- also correct (I put NB into lower case) nb= 6/5(nr)-2/5 <---- correct (solving second equation for nb) 3((6/5)nr-2/5) = 2(nr)+18 <---- correct (substituting for nb in first equation. Note that I put the 6/5 in parentheses just to ensure the understanding that nr is not in the denominator. (18/5)nr-6/5 = 2nr + 18 <---- basically correct. Changed the equal sign in front of the 18 to a + sign. Also lower cased the NR for consistency. And put the 18/5 in parentheses. . Looks good so far. You can now get rid of the denominator 5 on the left side by multiplying 5 times all terms on both sides to get: . 18nr - 6 = 10nr + 90 . Add 6 to both sides and subtract 10nr from both sides: . 8nr = 96 . Divide both sides by 8 . nr = 12 . Then return to either of the original equations and substitute 12 for nr. Then solve that equation for nb. You will get that nb = 14. . Just as a matter of reference, once you got your two equations: . 3(nb)= 2(nr)+18 and 5(nb)= 6(nr)-2 . you could have multiplied the top equation (both sides, all terms) by 5 and the bottom equation (both sides, all terms) by 3. This would have given you: . 15(nb) = 10(nr) + 90 and 15(nb) = 18(nr) - 6 . Since the left sides are now equal, the right sides must also be equal, which leads to: . 10(nr) + 90 = 18(nr) - 6 . Solving this leads to the same equation you got with your method, but it saves messing around with fractions during the solution process. It still gives the same answers. . But what you did was fine and right on the mark. You did good, thoughtful work. Keep it up and good luck. .