Find the original fraction

Question

The numerator of a fraction is 1 less than the denominator. If the numerator and the denominator are both increased by 4, the new fraction will be 1/8 more than the original fraction. What is the original fraction. (I haven't used the quadratic formula because my teacher doesn't want us to use it yet, so try not to use it in the method for finding the solution)

Answer 1

Let numerator of original fraction = x - 1

Let denominator of original fraction = x

 

We set up the fractional equation as

 

[(x - 1 + 4 ) / (x + 4)] = [(x - 1) / x) + (1/8)]

 

Note that the original fraction is coded in orange.

 

 

Using some algebra, we obtain

 

[(x + 3) / (x + 4)] = [(8x - 8 + x) / 8x]

 

[(x + 3) / (x + 4)] = [(9x - 8) / 8x]

 

 

Now we cross multiply.

 

8x(x + 3) = (x + 4)(9x - 8)

 

8x² + 24x = 9x² + 28x - 32

 

 

Move the all the terms to one side of equation to make the other side equal to zero.  Since we want to have a positive x² term, we move all the terms to the right side of equation.

 

0 = x² + 4x - 32

0 = x² + 4x - 32

0 = (x + 8)(x - 4)

 

x = -8     or     x = 4

 

Lets take the positive value of x:    x = 4

 

Substituting this value into the original numerator and original denominator, the original fraction is

 

3/4

Answer 2

We'll call the numerator n.

 

Since the numerator is 1 less than the denominator, the fraction can be written as

 

n/(n+1).

 

If we increase both the numerator and the denominator by 4, the fraction that we get is

 

(n+4)/(n+1+4)=(n+4)/(n+5).

 

This is 1/8 more than the original fraction:

 

(n+4)/(n+5)-(n/(n+1))=1/8

 

We have one equation and one unknown, so we can solve it. The first thing we need to do is get rid of all of the fractions. In order to cancel all three denominators, we multiply all three terms by 8(n+5)(n+1). This gives

 

8(n+5)(n+1)(n+4)/(n+5)-8(n+5)(n+1)n/(n+1)=8(n+5)(n+1)/8.

 

Note how we chose this factor in order to cancel all three denominators. After canceling them, we have

 

8(n+1)(n+4)-8n(n+5)=(n+5)(n+1)

 

Now we use the FOIL method to expand all of these terms:

 

8(n^2+5n+4)-8(n^2+5n)=n^2+6n+5

 

Now I'll distribute the 8's through the parentheses on the left hand side of the equation:

 

8n^2+40n+32-8n^2-40n=n^2+6n+5

 

Combining like terms on the left hand side, we get

 

n^2(8-8)+n(40-40)+32=n^2+6n+5

32=n^2+6n+5 (The first two terms on the LHS came to 0.)

 

I'll subtract 32 from both sides to get everything on the same side:

 

0=n^2+6n-27

 

This is where we could apply the quadratic formula if we wanted. Factoring also works, though. We need two numbers that multiply to -27 and add to 6.

 

27 is divisible by 3. When we divide by it, we get 9. This is also divisible by 3, (9=3*3). Since we are working with -27, one of the numbers has to be positive and the other is negative.

 

So the possible pairs of numbers that multiply to -27 are

 

1   -27

-1  27

3   -9

-3  9

 

The pair that works from these choices is 3 and -9. (3*(-9)=-27, 3+(-9)=-6). So we factor the quadratic function to

(n-3)(n-(-9)=(n-3)(n+9)

 

This is equal to 0, so we need to solve each of these two terms equal to 0:

 

n-3=0

n=3

 

n+9=0

n=-9

 

Plugging these two possibilities into the expression for the fraction, we get

 

n/(n+1)=3/(3+1)=3/4.

or

n/(n+1)=-9/(-9+1)=-9/-8=9/8.

The second choice doesn't work after canceling the negatives, so our fraction is 3/4.

 

To check, let's add 4 to the numerator and denominator:

 

(3+4)/(4+4)=7/8,

 

and

 

7/8-3/4=7/8-6/8=1/8, as required.