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)Back to top.
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 + 5Back to top.
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1. Simplify: 1
1 + -
1 - -
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,
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
----- / ------
Using the Fractional Division
Theorem, take the reciprocal of the
divisor and multiply.
x + 1 x2
----- * ------
x x2 - 1
After multiplication, you have the
(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
You are left with the following:
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.Back to top.
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 = -6Back to top.
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.