URGENT: Word Problems involving rational equations

Question

the numerator of a fraction is 3 less than its denominator.  If 7 is added to the numerator and denominator, the resulting fraction equals 3/4. What was the original fraction?

Answer 1

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Answer 2

Assume that original fraction is n/d (numerator over denominator).

n = d - 3

 

(n+7)/(d+7) = 3/4.

 

These are two equations in two variables, and you should solve--I suggest substituting for n in the latter equation, so that you'll have one equation in only the one variable d

Answer 3

   Let x = numerator

 

Since the numerator is 3 less than the denominator,

 

   Let x + 3 = denominator

 

Adding 7 to both numerator and denominator and setting result to 3/4,

 

   (x + 7) / (x + 3 + 7) = 3/4

   (x + 7)/(x + 10) = 3/4

 

Cross multiply

 

   4(x+7) = 3(x+10)

   4x + 28 = 3x + 30

 

Subtract 3x from both sides;  subtract 28 from both sides

 

    4x - 3x = 30 - 28

    x = 2

 

The original fraction:

 

    x/(x + 3)

  = 2/(2+3)

  = 2/5

 

 

Answer 4

The fraction's denominator = x

   the numerator therefore = x-3,

   so the initial fraction would look like:

     x-3

    ------    now, add 7 to both, and you have

       x

 

      (x-3) +7

    ------------  =  3/4 → begin solving

       x+7

 

         x+4

        ------ = 3/4

         x+7

 

I would first get rid of the

x+7 denominator by multiplying

both sides by x+7, then the result

would be:

 

       x+4 = 3x+21

                ---------  

                     4

 

Multiply both sides by 4 to

get rid of the last denominator.

The result would be:

    4x+16 = 3x+21

 

Solve for x:

      4x+16 = 3x+21

     -3x         -3x

------------------------

        x+16 =       21

          -16         -16

------------------------

        x        =5

 

 The fraction is 2/5

 

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