



 
Hi James, (6*x^{2}*y^{3})(2*x*y^{4}) = 12*x^{3}*y^{7} We multiply the coefficients together (6*2 = 12) Now that we have a process to follow, let's look at your problem: (ax^{1}*y)(1/2xy^{b}) = 6y^{6} Let's multiply the lefthand side to see what it would look like. a*(1/2)*x^{1+1}*y^{1+b} = 6y^{6} Simplify what you can. a/2 * x^{0} *y^{1+b} = 6y^{6} But x^{0} = 1 so we don't need to write it. a/2*y^{1+b} = 6y^{6} Let's see what the "a" and "b" match up with on the other side. To determine the value of "a", we need to solve the following equation that equates the coefficients from both sides: a/2 = 6 solve this and we get a = 12 To determine the value of "b", we need to solve the following equation that equates the exponent on y from both sides: 1 + b = 6 solve this and we get b = 7 Try substituting these values into your original problem and see if it does yield the result you need. Hope this was clear enough for you. Leeanne  


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