On this page we hope to clear up any problems you might have with
fractional expressions and equations. Fractions are used
continually in math, and even with the advent of powerful calculators, are
used extensively in advanced math courses. Click any of the links
below to start understanding fractions better!
Lowest common multiple
Addition and subtraction
Division of polynomials
Variation (direct and indirect)
Quiz on Fractional Expressions
The multiplication of fractions is rather straightforward. The Fraction Multiplication Theorem says for any fractional expressions, (a/b) and (c/d) where b and d do not equal 0, (a/b)(c/d) = (a * c)/(b * d). Example:
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)
The division of fractions is also not very complicated. The Fraction Division Theorem says for any fractional expressions, (a/b) and (c/d) where b, c, and d do not equal 0, (a/b)/(c/d) = (a/b)(d/c). In other words, you divide by multiplying by a reciprocal. Example:
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
Finding the LCM, or Lowest Common Multiple is a necessary skill if you want to be able to add to and subtract fractions from other numbers. However, finding the LCM is usually covered in most elementary algebra (Algebra I) courses. This custom has been followed on this site, so you can click here to get a better understanding of Lowest Common Multiples.
Adding and subtracting fractions from other numbers is a very useful skill since you encounter fractions so often. This skill is usually covered in elementary algebra (Algebra I) courses. We have followed that custom on this site, so click here to get a better understanding of addition and subtraction when fractions are involved.
Complex fractions, or fractions that have a fraction for its numerator, denominator, or both, are encountered less often than normal fractions, but are just as easy to solve because they, like normal fractions, are division problems. The most common way to simplify complex fractions is to simplify the numerator and denominator, treat it as a division problem, and then simplify as usual. Example:
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
Always keep in mind that a fraction bar is the same as a division sign. Division by a monomial can be done by rewriting the problem as a fraction. Example:
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.
A fractional equation is an equation that contains one or more fractional expressions. To solve a fractional equation you multiply each side of the equation by the LCM of all the denominators. Example:
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
A worker earns $10.00/hr. In one hour, $10.00 is earned. In two hours, $20.00 is earned, and so forth. This gives us a set of ordered pairs which all have the same ratio — (1, 10), (2, 20), (3, 30), . . . When we encounter this situation, where there are pairs of numbers in which the ratio is constant, we have direct variation. A situation such as the one described above gives rise to the function f(x) = kx, where k is a positive constant, there is direct variation. Example:
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 k. 34 = k(7) (34/7) = k The constant of variation is (34/7).
The problem above 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.
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. For example, if you were driving 100 km/hr., you would go 6.4 kilometers in 3.84 minutes. (This is figured by plugging in the speed for y and solving for x. Try any speed you like, it works!)
Take the Quiz on fractional expressions. (Very useful to review or to see if you've really got this topic down.) Do it!