Question 601393
Here are comments to your work:
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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.
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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:
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18nr - 6 = 10nr + 90
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Add 6 to both sides and subtract 10nr from both sides:
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8nr = 96
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Divide both sides by 8
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nr = 12
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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.
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Just as a matter of reference, once you got your two equations:
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3(nb)= 2(nr)+18 and
5(nb)= 6(nr)-2
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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:
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15(nb) = 10(nr) + 90 and
15(nb) = 18(nr) - 6
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Since the left sides are now equal, the right sides must also be equal, which leads to:
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10(nr) + 90 = 18(nr) - 6
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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.
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But what you did was fine and right on the mark. You did good, thoughtful work. Keep it up and good luck.
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