1. Problem: x + 3 x3
---- * -----
y - 4 y + 5
Solution: Using the Fraction Multiplication
Theorem, multiply the numerator
of the first fraction by the numer-
ator of the second fraction and the
denominator of the first fraction
by the denominator of the second
fraction.
(x + 3)x3
--------------
(y - 4)(y + 5)
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1. Problem: x - 2 x + 5
----- / -----
x + 1 x - 3
Solution: Utilizing the Fraction Division
Theorem, we take the reciprocal of the
divisor and multiply.
x - 2 x - 3
----- * -----
x + 1 x + 5
Use the Fraction Multiplication
Theorem to multiply the problem out.
x2 - 5x + 6
-----------
x2 + 6x + 5
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Complex Fractions
1. Simplify: 1
1 + -
x
-----
1
1 - -
x2
Solution: Combine the numerator into one
fraction and then do the same to the
denominator. (If you do not know how
to add and subtract fractions,
click here.)
x 1 x + 1
- + - -----
x x x
----- = ------
x2 - 1 x2 - 1
- - - ------
x2 x2 x2
Now you have a normal division problem
with fractional expressions.
x + 1 x2 - 1
----- / ------
x x2
Using the Fractional Division
Theorem, take the reciprocal of the
divisor and multiply.
x + 1 x2
----- * ------
x x2 - 1
After multiplication, you have the
following expression:
(x + 1)x2
---------------
x(x + 1)(x - 1)
Now, cancel out any factors that are
in both the denominator and the
numerator to simplify.
(They both have (x + 1) and
x as a factor. Cancel them
out.)
You are left with the following:
x
-----
x - 1
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1. Divide: 12x3 + 8x2 + x + 4 by 4x
Solution: Rewrite the problem as a fraction.
12x3 + 8x2 + x + 4
-----------------
4x
This expression shows that each
term in the numerator is divided
by 4x. Rewrite the prob-
lem to show that.
12x3 8x2 x 4
---- + --- + -- + --
4x 4x 4x 4x
Now do the four divisions in-
dicated by each fraction.
3x2 + 2x + (1/4) + (1/x)
When the divisor is not a monomial, you have to use a
procedure that resembles long division as you learned
way back in 5th grade arithmetic! Example:
2. Divide: x2 + 5x + 6 by x + 3
Solution: Write the problem as a long
division problem.
___________
x + 3 )x2 + 5x + 6
Divide first term by first term —
(x2/x) = x.
x__________
x + 3 )x2 + 5x + 6
x2 + 3x Multiply x by divisor.
-------
2x Subtract.
Bring down the next term and repeat
the process.
x_+_2______
x + 3 )x2 + 5x + 6
x2 + 3x
-----------
2x + 6
2x + 6
------
0
The quotient is x + 2.
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1. Solve: 2 5 1
- - - = -
3 6 x
Solution: The LCM of all denominators is
6x. Multiply each side of the
equation by 6x.
6x((2/3) - (5/6)) = 6x(1/x)
Use the distributive law of multi-
plication, which says a(b + c) =
ab + ac, to rewrite the
equation.
6x(2/3) - 6x(5/6) = 6x(1/x)
Multiply each group of terms
together.
(12x/3) - 30x/6 = 6x/x
Perform each of the indicated
divisions.
4x - 5x = 6
Solve for x.
Combine like terms.
-x = 6
Multiply each side by -1.
x = -6
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1. Problem: Find the variation constant k
and an equation of variation where y
varies directly as x, and where
y = 34 and x = 7.
Solution: The problem tells us (7, 34)
is a solution of the direct variation
equation, y = kx.
Plug in the givens for y and x
and solve for l.
34 = k(7)
(34/7) = k
The constant of variation is (34/7).
The above problem could be used in real life if someone
made $4.857/hr. and they wanted to know how much they
would make if they worked for 7 hours. Using 7 for x
would tell us they would make $34.00.
Sometimes, things do not vary directly. An example of this is a bus that travels 20 kilometers in one hour at a speed of 20 km/hr. At 40 km/hr. it would only take half of an hour to go 20 kilometers. This gives us a pair of ordered pairs — (20, 1), (40, .5), . . . These numbers, whose product is constant, vary indirectly. Whenever a situation such as the one described above gives rise to the function f(x) = (k/x), where k is a positive constant, there is inverse variation. Example:
2. Problem: Find the variation constant and
then an equation of variation where
y varies inversely as x, and
y = 32 when x = (1/5).
Solution: We know that (.2, 32) is
a solution of the inverse variation
equation.
Plug in the given information and
solve for k.
32 = (k/.2)
6.4 = k
The equation of variation is the
following:
y = (6.4/x)
This can be applied to real life, too. If you
were driving 100 km/hr., you would go 6.4 kilometers
in 3.84 minutes.
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Take the quiz on fractional expressions. The quiz is very useful for either review or to see if you've really got the topic down.