Question 617
<b><font face = "courier new" size = 3>{{{(3y+1)/2+(y+2)/3=1}}}
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Multiply each term by the LCD = 6
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{{{6*((3y+1)/2)+6*((y+2)/3)=6*1}}}
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Cancel the denominators into the 6's, and we have:
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3(3y+1) + 2(y+2) = 6
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9y + 3 + 2y + 4 = 6
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11y + 7 = 6
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11y = -1
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y = -1/11

Edwin <font face = "wingdings" size = 7 color = "red">J</font>