The Answer Page
Fractions

UNH Mathematics Center

1  Multiplying and Dividing Fractions

1. 1. Notice that we will disallow x = 0, x = 2, and y = 0.
      

      x y · -3y x2 · y+4 x-2 = x x2 · -3y y · y+4 x-2

      which can be rewritten
      

      x x · 1 x ·(-3)· y y · y+4 x-2

      When we take into account the two fractions whose values are 1, what's left is
      

      -3(y+4) x(x-2)

   2. 
      

      (-6)·(x-2)· 3-x x+2 ·4· 1 (2-x)2

      which is, when rearranged,
      

      -6 1 · x-2 2-x · 3-x x+2 ·4· 1 2-x ·

      or,
      

      (-6)(-1) 3-x x+2 ·(4)· 1 2-x

      which might be written
      

      24 3-x (x+2)(2-x) or 24 x-3 (x+2)(x-2)

      As usual, we will leave the denominator factored.

   3. 
      

      x-3 y -2x+6 y3(3-x) = x-3 y · y3(3-x) -2x+6 · = x-3 -2(x-3) · y3 y · 3-x 1

      which might be rewritten
      

      -1 2 · x-3 x-3 ·y2· y y ·(3-x)

      or, taking into account fractions with value 1,
      

      -y2 2 ·(3-x) = y2(x-3) 2

2. You did notice that -1 and -2 are roots of the denominator? We will not allow x to assume either of those values.
   

   x5-x3 x2+3x+2 = x3(x2-1) (x+1)(x+2) = x+1 x+1 · x3(x-1) x+2

   which would be more simply written as
   

   x3(x-1) x+2

   Now we can look at the individual pieces of the expression. Because x is negative, x³ will be negative also; and because x < -1, the number 1 less than x is less than -2. That is, x-1 < -2 ... which of course makes x-1 negative. Thus, the simplified expression's numerator is always positive.

   The denominator of the simplified expression, x+2, is also positive. Because -2 < x, the number x ''sits to the right of'' -2 on a number line. Adding 2 to x moves it 2 units to the right; so that x+2 will be located to the right of zero on the number line. That is, x+2 > 0.

   With a positive numerator and a positive denominator, the simplified expression's value is positive. And so also is the value of the original expression.

3. 1. We notice first the numerator's common factor -5:
      

      -5x2-5x+10 (x4+x)-(x3+1) = (-5) x2+x-2 (x4+x)-(x3+1)

      and the denominator's common factor x³+1.
      

      (-5) x2+x-2 (x4+x)-(x3+1) = -5 x3+1 · x2+x-2 x-1

      When we notice that the quadratic x²+x-2 has linear factors, we can finish the example:
      

      -5 x3+1 · x2+x-2 x-1 = -5 x3+1 · (x+2)(x-1) x-1 = -5 x3+1 ·(x+2)· x-1 x-1

      so that the simplified version of the expression is
      

      -5 x+2 x3+1

   2. This example has the same denominator, and we recognize the ''difference of squares'' pattern in its numerator:
      

      x2-1 (x4+x)-(x3+1) = (x+1)(x-1) (x3+1)(x-1) = x+1 x3+1 · x-1 x-1

      Now, the cubic expression x³+1 factors. Sums and differences of cubes are a factoring pattern you should recognize. If you don't know them already, it's never too late to learn it them. Here are the patterns:
      

      A3+B3 = (A+B)(A2-AB+B2)

      and
      

      A3-B3 = (A-B)(A2+AB+B2)

      In case of emergencies it's nice to remember that the sum and difference of the roots is one of the factors. You can always use long division to find the other factor, in a pinch.

      In the present example, x³+1 = (x+1)(x²-x+1). (Notice that the second factor is not the square of x-1, which would have a different middle term.)

      Back to our example:
      

      x+1 x3+1 · x-1 x-1 = x+1 x3+1 = x+1 (x+1)(x2-x+1) = x+1 x+1 · 1 x2-x+1

      so that the final, simplified version is
      

      1 x2-x+1

2  Adding and Subtracting Fractions

1. In the first example, we add (or subtract) the expressions.

   1. The simplest common denominator will be (x-2)(x-1).
      

      4 x-2 · x-1 x-1 - 3 x-1 · x-2 x-2 - 5 x+1 (x-1)(x-2)

      which is
      

      4(x-1) -3(x-2)-5(x+1) (x-1)(x-2) = 4x-4 -3x+6 -5x-5 (x-1)(x-2)

      
      

      = -4x-3 (x-1)(x-2) = - 4x+3 (x-1)(x-2)

      Notice that we have left the denominator written as a product of its factors. This way, it's easy for us to see that 1 and 2 are not possible values of x, because either of them would make the denominator's value zero.

   2. The simplest common denominator is t²u². So we would convert each term first to have t²u² as a denominator:
      

      t+1 tu · tu tu + 5 t2u · u u - u-1 u2 · t2 t2

      which would be
      

      (t+1)tu t2u2 + 5u t2u2 - (u-1)t2 t2u2 = t2u+tu+5u-t2u+t2 t2u2 .

      We would combine the terms in the numerator that add up to zero, to get:
      

      tu + 5u + t2 t2u2

   3. First we notice that the simplest denominator is not the product of the two individual denominators, because they both have the factor y-2. The first term's denominator is y(y-2) and the denominator in the second term is (y+2)(y-2). So we multiply each term by 1 in a productive manner:
      

      5 y(y-2) · y+2 y+2 - y (y+2)(y-2) · y y

      to get
      

      5(y+2) (y)(y+2)(y-2) - y2 (y)(y+2)(y-2) = -y2+5y+10 (y)(y+2)(y-2) = - y2-5y-10 (y)(y+2)(y-2)

      The denominator is left in its factored form, making it easy to see which values of y would make it zero.

2. In the second example, we first combine the expressions as indicated, and then check to see where the combined expression's value is positive.
   1. 
      

      6 - 2x2+1 x2 = 6x2-2x2-1 x2 = 4x2-1 x2

      Now, think about this expression's possible values. We will certainly disallow x = 0, because we can't have zero as a denominator.

      Provided the denominator isn't zero, its sign will be positive because it's a square. So, the entire expression will have the same sign as does the numerator.

      The numerator is 4x²-1, so it will be positive if 4x² > 1. That can happen for both positive and negative values of x, provided x² > 1/4. What is needed (graph the parabola y = 4x²-1 if it helps you see this) is for the |x| > 1/2. That is, either x > 1/2, or x < -1/2.

      Do you remember the definition of absolute value, by the way? The absolute value of a number, say |A|, is either A or -A ... whichever isn't negative. In this problem, the absolute value |x| is either x or -x ... whichever is positive. Make sure you understand how the definition applies to this problem's solution, in the case where it's -x that is positive!

   2. We will use (y+5)³ as a common denominator, so that
      

      y (y+5)2 · y+5 y+5 - y2 (y+5)3 = y2+5y-y2 (y+5)3 = 5y (y+5)3

      This expression will be positive whenever the numerator and denominator have the same sign. Here are the possibilities:
      • Both y and (y+5)³ are positive. Now, cubes of numbers always have the same sign as the numbers themselves (because a number's cube is the product of the number itself and its square - which isn't negative). So our request comes down to this: y, and y+5, should both be positive. This will happen whenever y > 0; because y+5 will then be even larger.
      • Both y and (y+5)³ are negative - which means, as noted above, that both y and y+5 are negative. This will happen whenever y+5 < 0, or y < -5.

      We conclude: this expression is positive provided either y > 0, or y < -5. (You will notice that these conditions also preclude the disastrous possibility that y = 5.)

   3. Here we use the entire square root denominator from our second term as the common denominator. This has the pleasant result of removing square roots entirely from the numerator:
      

      2r   ___ Ör+3    +  r2 2   ___ Ör+3   = 2r   ___ Ör+3   ·2   ___ Ör+3    + r2 2   ___ Ör+3   = 4r(r+3)+r2 2   ___ Ör+3   = r(5r+12) 2   ___ Ör+3

      To see where this sum is positive, we need only look at each piece of the expression in turn:
      • The denominator will never be negative. That's because the square root function is defined so that its values are never negative! It's part of the bargain we make with the world when we use the square root as a function, which must not produce ambiguous values. Of course we will exclude r = -3 because that would make the denominator zero, and if r < -3 the square root won't even make sense. So we begin by restricting r to numbers r > -3.
      • Because the denominator is positive, the numerator must be positive also. That could happen if both r > 0, and 5r+12 > 0. But r is positive, then surely 5r+12 is, too; so the one condition is all we need here.
      • The numerator would also be positive if r < 0 and 5r+12 < 0. Notice that if the second of these conditions holds, the first one does too: if 5r+12 < 0 then r < -(12/5) so r will be negative.

      Here are our conclusions. The expression will have a positive value if either

      • r is positive (r > 0); or
      • r is bigger than -3 but smaller than -12/5 (-3 < r < -12/5).

3  Fractions that describe proportionality

1. The proportionality can be expressed as
   

   36 hotdogs 21 burgers .

   We multiply the quantity x burgers by this fraction to obtain
   

   x burgers · 36 hotdogs 21 burgers = 36x 21 hotdogs.

   Naturally, if this quantity isn't a whole number, we round up. Even at the deli counter, they won't sell us fractional parts of hot dogs.

   It's good to notice that the reciprocal of the proportionality fraction,
   

   21 burgers 36 hotdogs

   would have expressed the proportionality just as well. Of course, to keep track of the units properly, we would have divided the number of burgers by this fraction to find the corresponding number of hotdogs to purchase.

2. There's more information in the problem than we use at present (perhaps next time we'll want an estimate of the Keeshonds). But right now, we have a number S of German shepherd dogs and we want to convert it to a number of Goldens. So we can use a conversion factor whose units are ''Goldens/Shepherds.'' The necessary conversion factor is a fraction which we get from the problem's data:
   

   5 Goldens 3 Shepherds

   We estimate the Goldens in our next class by multiplying the quantity S Shepherds by this converting fraction:
   

   S Shepherds · 5 Goldens 3 Shepherds = 5S 3 Goldens .

   And, as in the first problem, fractions of Goldens just won't do! We will round any fractional number up, and plan for a whole number of Golden retrievers in the class.

3. We have N eggs on hand, and we want to find from that the appropriate quantity of flour. We can go to the problem's data and construct a proportionality constant with units ''cups of flour per egg'' or one with units ''eggs per cup of flour.'' Let's do the latter, just for a change. The fraction will be
   

   5 eggs 6.5 cups of flour

   and to make the units sensible, we would divide a number of eggs by this fraction:
   

   N eggs 5 eggs 6.5 cups of flour

   Dividing by a fraction is the same as multiplying by its reciprocal, so this calculation is the same as
   

   N eggs · 6.5 cups of flour 5 eggs = 6.5N 5 cups of flour .

   Notice that we will not round the answer in this instance. Although eggs are counted one by one, as integers, flour is measured! We are really good at fractions: just hand over that measuring cup and we'll get it right.

4. Let's take as our starting point that two tablespoons of supplement S₁ will provide u units of vitamin A. We'll compare that to each supplement in turn:
   • There are u units of vitamin A in 2 tablespoons of supplement S₁. The conversion fraction is
     

     u units A 2 Tbsp S1

     If we multiply a number of tablespoons of S₁ by this fraction we convert it to a number of units of vitamin A.

   • There are u units of vitamin A in 3 tablespoons of supplement S₂. The conversion fraction is
     

     u units A 3 Tbsp S2

     Multiplication by this fraction converts a number of tablespoons of S₂ to a number of units of vitamin A.

   • And, there are u units of vitamin A in 4 tablespoons of supplment S₃. That produces the conversion fraction
     

     u units A 4 Tbsp S3

     Same here: multiplication by this fraction converts tablespoons of S₃ to units of the vitamin.

   Our first task is to decide how much vitamin A we will get from 4 tablespoons of supplement S₂ and 1 tablespoon of supplement S₃.

   • From 4 tablespoons of supplement S₂ we will get
     

     4 Tbsp S2· u units A 3 Tbsp S2 = 4u 3 units A

     while ...
   • from 1 tablespoon of supplement S₃ we will get
     

     1 Tbsp S3· u units A 4 Tbsp S3 = u 4 units A

   Notice that these quantities have the same units! Both are quantities of units of vitamin A. The algebra reflects our common sense of the problem's narrative: quantities with the same units can be added. From 4 tablespoons of supplement S₂ and 1 tablespoon of supplement S₃ we would get
   

   4u 3 units A + u 4 units A = ( 16u 12 + 3u 12 ) units A = 19u 12 units A

   Finally we are ready to finish solving the problem: we know how much vitamin A is in the combination, and we need only convert it to the corresponding amount of supplement S₁. To do that we would either divide:
   

   19u 12 units A ¸ u units A 2 Tbsp S1

   or multiply by the reciprocal fraction:
   

   19u 12 units A · 2 Tbsp S1 u units A = 19 6 Tbsp S1

   Notice that rounding is not involved here either. We do not need to measure out a whole number of tablespoons: 19/6 tablespoons is 1/6 of a tablespoon more than 3 tablespoons. It makes sense to measure out 1/6 of a tablespoon.

   (Besides, we already know that there are 3 teaspoons in a tablespoon. If we want to be really clever about the measurement, we can always find a half-teaspoon measure in the kitchen cupboard.)