My Math Forum Inverse Proportion Problem

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 August 11th, 2012, 01:09 AM #1 Member   Joined: Jul 2010 Posts: 69 Thanks: 0 Inverse Proportion Problem Hello, I have a problem that I would not have been able to figure out if it were not for tutors. The problem is that after trying so hard to figure this math problem out on my own, I cannot see how it is done. Here we have the word problem of: If y is inversely proportional to 4x + 5 and y = 4 when x = 8 1/2, find x when y = 12. If someone could help me solve this problem I would appreciate it. I also will ask, if you get the problem correct, how you did it. Thanks.
 August 11th, 2012, 01:41 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs Re: Inverse Proportion Problem We are told: y is inversely proportional to 4x + 5 hence, we may state: (1) $y(4x+5)=k$ We are given: y = 4 when x = 8 1/2 = 17/2 Substituting these values for x and y into (1), we obtain: $4$$4\(\frac{17}{2}$$+5\)=k$ $4$$2\cdot17+5$$=k$ $4(40-1)=k$ $k=160-4$ $k=156$ When $y=12$ we have by substitution into (1) and using the value we found for k: $12(4x+5)=156$ Solve for x. First divide both sides by 12: $4x+5=13$ Subtract through by 5: $4x=8$ Divide through by 4: $x=2$
 August 11th, 2012, 11:39 AM #3 Member   Joined: Jul 2010 Posts: 69 Thanks: 0 Re: Inverse Proportion Problem MarkFL, You got the answer correct. Where did the 40-1 come from? Also, how are you using those fraction symbols?
 August 11th, 2012, 01:18 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs Re: Inverse Proportion Problem $2\cdot17+5=34+5=39=40-1$ I used 40 - 1 to make the multiplication easier. I am using $\LaTeX$ to write the mathematical expressions.

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