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8.5: Building Rational Expressions and the LCD

8.5: Building Rational Expressions and the LCD



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The Process

Recall, from Section 8.2 the equality property of fractions.

Equality Property of Fractions

If (dfrac{a}{b} = dfrac{c}{d}), then (ad = bc).

Using the fact that (1 = dfrac{b}{b}, b ot = 0), and that (1) is the multiplicative identity, it follows that if (dfrac{P}{Q}) is a rational expression, then

(dfrac{P}{Q} cdot dfrac{b}{b} = dfrac{Pb}{Qb}, b ot = 0)

This equation asserts that a rational expression can be transformed into an equivalent rational expression by multiplying both the numerator and denominator by the same nonzero number.

Process of Building Rational Expressions

This process is known a(s the process of building rational expressions and it is exactly the opposite of reducing rational expressions. The process is shown in these examples:

Example (PageIndex{1})

(dfrac{3}{4}) can be built to (dfrac{12}{16}) since:

(dfrac{3}{4} cdot 1 = dfrac{3}{4} cdot dfrac{4}{4} = dfrac{3 cdot 4}{4 cdot 4} = dfrac{12}{16})

Example (PageIndex{2})

(dfrac{-4}{5}) can be built to (dfrac{-8}{10}) since

(dfrac{-4}{5} cdot 1 = dfrac{-4}{5} cdot dfrac{2}{2} = dfrac{-4 cdot 2}{5 cdot 2} = dfrac{-8}{10})

Example (PageIndex{3})

(dfrac{3}{7}) can be built to (dfrac{3xy}{7xy}) since

(dfrac{3}{7} cdot 1 = dfrac{3}{7} cdot dfrac{xy}{xy} = dfrac{3xy}{7xy})

Example (PageIndex{4})

(dfrac{4a}{3b}) can be built to (dfrac{4a^2(a+1)}{3ab(a+1)}) since:

(dfrac{4a}{3b} cdot 1 = dfrac{4a}{3b} cdot dfrac{a(a + 1)}{a(a+1)} = dfrac{4a^2(a + 1)}{3ab(a + 1)}).

Suppose we're given a rational expression (dfrac{P}{Q}) and wish to build it into a rational expression with denominator (Qb^2), that is,

(dfrac{P}{Q} ightarrow dfrac{?}{Qb^2})

Since we changed the denominator, we must certainly change the numerator in the same way. To determine how to change the numerator we need to know how the denominator was changed. Since one rational expression is built into another equivalent expression by multiplication by 1, the first denominator must have been multiplied by some quantity. Observation of

(dfrac{P}{Q} ightarrow dfrac{?}{Qb^2})

tells us that (Q) was multiplied by (b^2). Hence, we must multiply the numerator (P) by (b^2). Thus

(dfrac{P}{Q} ightarrow dfrac{Pb^2}{Qb^2})

Quite often a simple comparison of the original denominator with the new denominator will tell us the factor being used. However, there will be times when the factor is unclear by simple observation. We need a method for finding the factor.

Observe the following examples; then try to speculate on the method.

Example (PageIndex{5})

(dfrac{3}{4} = dfrac{?}{20})

The original denominator 4 was multiplied by 5 to yield 20. What arithmetic process will yield 5 using 4 and 20?

Example (PageIndex{6})

(dfrac{9}{10} = dfrac{?}{10y})

The original denominator 10 was multiplied by (y) to yield (10y).

Example (PageIndex{7})

(dfrac{-6xy}{2a^3b} = dfrac{?}{16a^5b^3})

The original denominator (2a^3b) was multiplied by (8a^2b^2) to yield (16a^5b^3)

Example (PageIndex{8})

(dfrac{5ax}{(a+1)^2} = dfrac{?}{4(a+1)^2(a-2)})

To determine the quantity that the original denominator was multiplied by to yield the new denominator, we ask, "What did I multiply the original denominator by to get the new denominator?" We find this factor by dividing the original denominator into the new denominator.

It is precisely this quantity that we multiply the numerator by to build the rational expression.

Sample Set A

Determine N in each of the following problems.

Example (PageIndex{9})

(dfrac{8}{3} = dfrac{N}{15})

The original denominator is (3) and the new denominator is (15). Divide the original denominator into the new denominator and multiply the numerator (8) by this result. (15÷3=5) Then, (8 cdot 5=40). So,

(dfrac{8}{3} = dfrac{40}{15}) and (N = 40).

Check by reducing (dfrac{40}{15})

Example (PageIndex{10})

(dfrac{2x}{5b^2y} = dfrac{N}{20b^5y^4})

The original denominator is (5b^2y) and the new denominator is (20b^5y^4). Divide the original denominator into the new denominator and multiply the numerator (2x) by this result.

(dfrac{20b^5y^4}{5b^2y} = 4b^3y^3)

So, (2x cdot 4b^3y^3 = 8b^3xy^3). Thus,

(dfrac{2x}{5b^2y} = dfrac{8b^3xy^3}{20b^5y^4}) and (N = 8b^3xy^3).

Example (PageIndex{11})

(dfrac{-6a}{a+2} = dfrac{N}{(a+2)(a-7)})

The new denominator divided by the original denominator is

(dfrac{(a+2)(a-7)}{a+2} = a-7)

Multiply (-6a) by (a-7).

(-6a(a-7) = -6a^2 + 42a)

(dfrac{-6a}{a+2} = dfrac{-6a^2 + 42a}{(a+2)(a-7)}) and (N = -6a^2 + 42a)

Example (PageIndex{12})

(dfrac{-3(a-1)}{a-4} = dfrac{N}{a^2 - 16}).

The new denominator divided by the original denominator is

(dfrac{a^2-16}{a-4} = dfrac{(a+4)cancel{(a-4)}}{cancel{a-4}})

Multiply (-3(a-1)) by (a + 4)

(-3(a-1)(a+4) = -3(a^2 + 3a - 4))

( = -3a^2 - 9a + 12)

(dfrac{-3(a-1)}{a-4} = dfrac{-3a^2 - 9a + 12}{a^2 - 16}) and (N = -3a^2 - 9a + 12)

Example (PageIndex{13})

(egin{array}{flushleft}
7x&=dfrac{N}{x^2y^3}& ext{ Write } 7x ext{ as } dfrac{7x}{1}
dfrac{7x}{1}&=dfrac{N}{x^2y^3}& ext{ Now we can see clearly that the original denominator }
&& 1 ext{ was multiplied by } x^2y^3. ext{ We need to multiply the }
&& ext{ numerator } 7x ext{ by } x^2y^3
7x&=dfrac{7x cdot x^2y^3}{x^2y^3}
end{array})

(7x = dfrac{7x^3y^3}{x^2y^3} ext{ and } N = 7x^3y^3)

Example (PageIndex{14})

(egin{array}{flushleft}
dfrac{5x}{x+3}&=dfrac{5x^2-20x}{N} & ext{ The same process works in this case. Divide the original }
&& ext{ numerator } 5x ext{ into the new numerator } 5x^2 - 20x
dfrac{5x^2 - 20x}{5x} &= dfrac{cancel{5x}(x-4)}{cancel{5x}}
&=x-4
end{array})


((x+3)(x-4) ext{ Multiply the denominator by } x-4)

(dfrac{5x}{x+3} = dfrac{5x^2-20}{(x+3)(x-4)} ext{ and } N = 5x^2 - 20)

Example (PageIndex{15})

(egin{array}{flushleft}
dfrac{4x}{3-x} &= dfrac{N}{x-3} & ext{ The two denominators have nearly the same terms; each has }
&& ext{ The opposite sign. Factor } -1 ext{ from the original denominator}
3-x &= -1(-3+x)
&=-(x-3)
end{array})

(dfrac{4x}{3-x} = dfrac{4x}{-(x-3)} = dfrac{-4x}{x-3} ext{ and } N = -4x)

It is important to note that we factored (−1) from the original denominator. We did not multiply it by (−1). Had we multiplied only the denominator by (−1) we would have had to multiply the numerator by (−1) also.

Practice Set A

Determine N.

Practice Problem (PageIndex{1})

(dfrac{3}{8} = dfrac{N}{48})

Answer

(N=18)

Practice Problem (PageIndex{2})

(dfrac{9a}{5b} = dfrac{N}{35b^2x^3})

Answer

(N = 63abx^3)

Practice Problem (PageIndex{3})

(dfrac{-2y}{y-1} = dfrac{N}{y^2 - 1})

Answer

(N = -2y^2 - 2y)

Practice Problem (PageIndex{4})

(dfrac{a+7}{a-5} = dfrac{N}{a^2 - 3a - 10})

Answer

(N = a^2 + 9a + 14)

Practice Problem (PageIndex{5})

(4a = frac{N}{6a^3(a-1)})

Answer

(N = 24a^4(a-1))

Practice Problem (PageIndex{6})

(-2x = dfrac{N}{8x^3y^3z^5})

Answer

(N = -16x^4y^3z^5)

Practice Problem (PageIndex{7})

(dfrac{6ab}{b+3} = dfrac{N}{b^2 + 6b + 9})

Answer

(N = 6ab^2 + 18ab)

Practice Problem (PageIndex{8})

(dfrac{3m}{m+5} = dfrac{3m^2 - 18m}{N})

Answer

(N = m^2 - m - 30)

Practice Problem (PageIndex{9})

(dfrac{-2r^2}{r-3} = dfrac{-2r^3 + 8r^2}{N})

Answer

(N = r^2 - 7r + 12)

Practice Problem (PageIndex{10})

(dfrac{-8ab^2}{a-4} = dfrac{N}{4-a})

Answer

(N = 8ab^2)

The Reason For Building Rational Expressions

Building Rational Expressions

Normally, when we write a rational expression, we write it in reduced form. The reason for building rational expressions is to make addition and subtraction of rational expressions convenient (simpler).

To add or subtract two or more rational expressions they must have the same denominator.

Building rational expressions allows us to transform fractions into fractions with the same denominators (which we can then add or subtract). The most convenient new denominator is the least common denominator (LCD) of the given fractions.

The Least Common Denominator (LCD)

In arithmetic, the least common denominator is the smallest (least) quantity that each of the given denominators will divide into without a remainder. For algebraic expressions, the LCD is the polynomial of least degree divisible by each denominator. Some examples are shown below.

Example (PageIndex{16})

(dfrac{3}{4}, dfrac{1}{6}, dfrac{5}{12})

The LCD is 12 since 12 is the smallest number that 4, 6, and 12 will divide into without a remainder.

Example (PageIndex{17})

(dfrac{1}{3}, dfrac{5}{6}, dfrac{5}{8}, dfrac{7}{12})

The LCD is 24 since 24 is the smallest number that 3, 6, 8, and 12 will divide into without a remainder.

Example (PageIndex{18})

(dfrac{2}{x}, dfrac{3}{x^2})

The LCD is (x^2) since (x^2) is the smallest quantity that (x) and (x^2) will divide into without a remainder.

Example (PageIndex{19})

(dfrac{5a}{6a^2b}, dfrac{3a}{8ab^3})

The LCD is (24a^2b^3) since (24a^2b^3) is the smallest quantity that (6a^2b) and (8ab^3) will divide into without a remainder.

Example (PageIndex{20})

(dfrac{2y}{y-6}, dfrac{4y^2}{(y-6)^3}, dfrac{y}{y-1})

The LCD is ((y-6)^3(y-1)) since ((y-6)^3 cdot (y-1)) is the smallest quantity that (y-6, (y-6)^3), and (y-1) will divide into without a remainder

We’ll now propose and demonstrate a method for obtaining the LCD.

Method for Obtaining the LCD

  1. Factor each denominator. Use exponents for repeated factors. It is usually not necessary to factor numerical quantities.
  2. Write down each different factor that appears. If a factor appears more than once, use only the factor with the highest exponent.
  3. The LCD is the product of the factors written in step 2.

Sample Set B

Find the LCD

Example (PageIndex{21})

(dfrac{1}{x}, dfrac{3}{x^3}, dfrac{2}{4y})

1. The denominators are already factored.

2. Note that (x) appears as (x) and (x^3). Use only the (x) with the higher exponenent, (x^3). The term (4y) appears, so we must also use (4y).

3. THe LCD is (4x^3y).

Example (PageIndex{22})

(dfrac{5}{(x-1)^2}, dfrac{2x}{(x-1)(x-4)}, dfrac{-5x}{x^2 -4x + 2})

1. Only the third denominator needs to be factored:

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

Now the three denoinators are ((x-1)^2, (x-1)(x-4)), and ((x-2)(x-1)).

2. Note that (x-1) appears as ((x-1)^2, x-1), and (x-1). Use only the (x-1) with the highest exponent, ((x-1)^2). Also appearing are (x-4) and (x-2).

3. The LCD is ((x-1)^2(x-4)(x-2)).

Example (PageIndex{23})

(dfrac{-1}{6a^4}, dfrac{3}{4a^3b}, dfrac{1}{3a^3(b+5)})

1. We can see that the LCD of the numbers (6, 4), and (3) is (12). We also need (a^4, b), and (b + 5).

3. The LCD is (12a^4b(b+5)).

Example (PageIndex{24})

(dfrac{9}{x}, dfrac{4}{8y})

1. (x, 8y).

3. The LCD is (8xy).

Practice Set B

Find the LCD.

Practice Problem (PageIndex{11})

(dfrac{3}{x^2}, dfrac{4}{x^5}, dfrac{-6}{xy})

Answer

(x^5y)

Practice Problem (PageIndex{12})

(dfrac{x+1}{x-4}, dfrac{x-7}{(x-4)^2}, dfrac{-6}{x+1})

Answer

((x-4)^2(x+1))

Practice Problem (PageIndex{13})

(dfrac{2}{m-6}, dfrac{-5n}{(m+1)^2(m-2)}, dfrac{-3x}{x^2 - 6x + 9})

Answer

((m-6)(m+1)^2(m-2)^3)

Practice Problem (PageIndex{14})

(dfrac{1}{x^2 - 1}, dfrac{2}{x^2 - 2x - 3}, dfrac{-3x}{x^2-6x+9})

Answer

((x+1)(x-1)(x-3)^2)

Practice Problem (PageIndex{15})

(dfrac{3}{4y^2-8y}, dfrac{8}{y^2-4y+4}, dfrac{10y-1}{3y^3-6y^2})

Answer

(12y^2(y-2)^2)

Sample Set C

Change the given rational expressions into rational expressions having the same denominator.

Example (PageIndex{25})

(egin{array}{flushleft}
dfrac{3}{x^2}, dfrac{4}{x} & ext{ The LCD, by inspection, is } x^2 ext{. Rewrite each expression with }
&x^2 ext{ as the new denominator. }
dfrac{?}{x^2}, dfrac{?}{x^2} & ext{ Determine the numerators. In } dfrac{3}{x^2} ext{, the denominator was not }
& ext{ changed so we need not change the numerator}
& ext{ In the second fraction, the original denominator was } x
dfrac{3}{x^2}, dfrac{?}{x^2} & ext{ We can see that } x ext{ must be multiplied by } x ext{ to build it to } x^2
& ext{ So we must also multiply the numerator } 4 ext{ by } x ext{. Thus, } 4 cdot x = 4x.
dfrac{3}{x^2}, dfrac{4x}{x^2}
end{array})

Example (PageIndex{26})

(egin{array}{flushleft}
dfrac{4b}{b-1}, dfrac{-2b}{b+3} & ext{ By inspection, the LCD is } (b-1)(b+3)
dfrac{?}{(b-1)(b+3)}, dfrac{?}{(b-1)(b+3)} & ext{ The denominator of the first rational expression has been multiplied }
& ext{ by } b + 3 ext{, so the numerator } 4b ext{ must be multiplied by } b + 3.
4b(b + 3) = 4b^2 + 12b
dfrac{4b^2 + 12b}{(b-1)(b+3)}, dfrac{?}{(b-1)(b+3)} & ext{ The denominator of the second rational expression has been multiplied }
& ext{ by } b-1 ext{, so the numerator } -2b ext{ must be multiplied by } b-1.
-2b(b-1) = -2b^2 + 2b
dfrac{4b^2 + 12b}{(b-1)(b+3)}, dfrac{-2b^2 + 2b}{(b-1)(b+3)}
end{array})

Example (PageIndex{27})

(egin{array}{flushleft}
dfrac{6x}{x^2 - 8x + 15}, dfrac{-2x^2}{x^2 - 7x + 12}& ext{ We first find the LCD. Factor.}
dfrac{6x}{(x-3)(x-5)}, dfrac{-2x^2}{(x-3)(x-4)} & ext{The LCD is } (x-3)(x-5)(x-4) ext{. Rewrite each of these}
& ext{ fractions with new denominator } (x-3)(x-5)(x-4)
dfrac{?}{(x-3)(x-5)(x-4)}, dfrac{?}{(x-3)(x-5)(x-4)} & ext{ By comparing the denominator of the first fraction with the LCD }
& ext{ we see that we must multiply the numerator } 6x ext{ by } x-4
6x(x-4) = 6x^2-24x
dfrac{6x^2 - 24x}{(x-3)(x-5)(x-4)}, dfrac{?}{(x-3)(x-5)(x-4)} & ext{ By comparing the denominator of the second fraction with the LCD },
& ext{ we see that we must multiply the numerator } -2x^2 ext{ by } x - 5
-2x^2(x-5) = -2x^3 + 10x^2
dfrac{6x^2 - 24x}{(x-3)(x-5)(x-4)}, dfrac{-2x^3 + 10x^2}{(x-3)(x-5)(x-4)}
end{array})

These examples have been done step-by-step and include explanations. This makes the process seem fairly long. In practice, however, the process is much quicker.

Example (PageIndex{28})

(egin{array}{flushleft}
dfrac{6ab}{a^2-5a+4}, dfrac{a+b}{a^2-8a+16}
dfrac{6ab}{(a-1)(a-4)}, dfrac{a+b}{(a-4)^2} & ext{ LCD } = (a-1)(a-4)^2
dfrac{6ab(a-4)}{(a-1)(a-4)^2}, dfrac{(a+b)(a-1)}{(a-1)(a-4)^2}
end{array})

Example (PageIndex{29})

(egin{array}{flushleft}
dfrac{x+1}{x^3 + 3x^2}, dfrac{2x}{x^3 - 4x}, dfrac{x-4}{x^2 - 4x + 4}
dfrac{x+1}{x^2(x + 3)}, dfrac{2x}{x(x+2)(x-2)}, dfrac{x-4}{(x-2)^2} & ext{ LCD } = x^2(x+3)(x+2)(x-2)^2
dfrac{(x+1)(x+2)(x-2)^2}{x^2(x+3)(x+2)(x-2)^2}, dfrac{2x^2(x+3)(x-2)}{x^2(x+3)(x+2)(x-2)^2}, dfrac{x^2(x+3)(x+2)(x-4)}{x^2(x+3)(x+2)(x-2)^2}
end{array})

Practice Set C

Change the given rational expressions into rational expressions with the same denominators.

Practice Problem (PageIndex{16})

(dfrac{4}{x^3}, dfrac{7}{x^5})

Answer

(dfrac{4x^2}{x^5}, dfrac{7}{x^5})

Practice Problem (PageIndex{17})

(dfrac{2x}{x+6}, dfrac{x}{x-1})

Answer

(dfrac{2x(x-1)}{(x+6)(x-1)}, dfrac{x(x+6)}{(x+6)(x-1)})

Practice Problem (PageIndex{18})

(dfrac{-3}{b^2-b}, dfrac{4b}{b^2-1})

Answer

(dfrac{-3(b+1)}{b(b-1)(b+1)}, dfrac{4b^2}{b(b-1)(b+1)})

Practice Problem (PageIndex{19})

(dfrac{8}{x^2-x-6}, dfrac{-1}{x^2 + x - 2})

Answer

(dfrac{8(x-1)}{(x-3)(x+2)(x-1)}, dfrac{-1(x-3)}{(x-3)(x+2)(x-1)})

Practice Problem (PageIndex{20})

(dfrac{10x}{x^2 + 8x + 16}, dfrac{5x}{x^2 - 16})

Answer

(dfrac{10x(x-4)}{(x+4)^2(x-4)}, dfrac{5x(x+4)}{(x+4)^2(x-4)})

Practice Problem (PageIndex{21})

(dfrac{-2ab^2}{a^3-6a^2}, dfrac{6b}{a^4-2a^3}, dfrac{-2a}{a^2 - 4a + 4})

Answer

(dfrac{2^2b^2(a-2)^2}{a^3(a-6)(a-2)^2}, dfrac{6b(a-6)(a-2)}{a^3(a-6)(a-2)^2}, dfrac{-2a^4(a-6)}{a^3(a-6)(a-2)^2})

Exercises

For the following problems, replace N with the proper quantity.

Exercise (PageIndex{1})

(dfrac{3}{x} = dfrac{N}{x^3})

Answer

(3x^2)

Exercise (PageIndex{2})

(dfrac{4}{a} = dfrac{N}{a^2})

Exercise (PageIndex{3})

(dfrac{-2}{x} = dfrac{N}{xy})

Answer

(−2y)

Exercise (PageIndex{4})

(dfrac{-7}{m} = dfrac{N}{ms})

Exercise (PageIndex{5})

(dfrac{6a}{5} = dfrac{N}{10b})

Answer

(12ab)

Exercise (PageIndex{6})

(dfrac{a}{3z} = dfrac{N}{12z})

Exercise (PageIndex{7})

(dfrac{x^2}{4y^2} = dfrac{N}{20y^4})

Answer

(5x^2y^2)

Exercise (PageIndex{8})

(dfrac{b^3}{6a} = dfrac{N}{18a^5})

Exercise (PageIndex{9})

(dfrac{-4a}{5x^2y} = dfrac{N}{15x^3y^3})

Answer

(-12axy^2)

Exercise (PageIndex{10})

(dfrac{-10z}{7a^3b} = dfrac{N}{21a^4b^5})

Exercise (PageIndex{11})

(dfrac{8x^2y}{5a^3} = dfrac{N}{25a^3x^2})

Answer

(40x^4y)

Exercise (PageIndex{12})

(dfrac{2}{a^2} = dfrac{N}{a^2(a-1)})

Exercise (PageIndex{13})

(dfrac{5}{x^3} = dfrac{N}{x^3(x-2)})

Answer

(5(x−2))

Exercise (PageIndex{14})

(dfrac{2a}{b^2} = dfrac{N}{b^3-b})

Exercise (PageIndex{15})

(dfrac{4x}{a} = dfrac{N}{a^4-4a^2})

Answer

(4ax(a+2)(a−2) )

Exercise (PageIndex{16})

(dfrac{6b^3}{5a} = dfrac{N}{10a^2-30a})

Exercise (PageIndex{17})

(dfrac{4x}{3b} = dfrac{N}{3b^5 - 15b})

Answer

(4x(b^4 - 5))

Exercise (PageIndex{18})

(dfrac{2m}{m-1} = dfrac{N}{(m-1)(m+2)})

Exercise (PageIndex{19})

(dfrac{3s}{s + 12} = dfrac{N}{(s + 12)(s-7)})

Answer

(3s(s-7))

Exercise (PageIndex{20})

(dfrac{a+1}{a-3} = dfrac{N}{(a-3)(a-4)})

Exercise (PageIndex{21})

(dfrac{a+2}{a-2} = dfrac{N}{(a-2)(a-4)})

Answer

((a+2)(a−4))

Exercise (PageIndex{22})

(dfrac{b+7}{b-6} = dfrac{N}{(b-6)(b+6)})

Exercise (PageIndex{23})

(dfrac{5m}{2m + 1} = dfrac{N}{(2m+1)(m-2)})

Answer

(5m(m−2))

Exercise (PageIndex{24})

(dfrac{4}{a+6} = dfrac{N}{a^2 + 5a - 6})

Exercise (PageIndex{25})

(dfrac{9}{b-2} = dfrac{N}{b^2 - 6b + 8})

Answer

(9(b−4))

Exercise (PageIndex{26})

(dfrac{3b}{b-3} = dfrac{N}{b^2 - 11b + 24})

Exercise (PageIndex{27})

(dfrac{-2x}{x-7} = dfrac{N}{x^2 - 4x - 21})

Answer

(−2x(x+3))

Exercise (PageIndex{28})

(dfrac{-6m}{m+6} = dfrac{N}{m^2 + 10m + 24})

Exercise (PageIndex{29})

(dfrac{4y}{y+1} = dfrac{N}{y^2 + 9y + 8})

Answer

(4y(y+8))

Exercise (PageIndex{30})

(dfrac{x + 2}{x - 2} = dfrac{N}{x^2 - 4})

Exercise (PageIndex{31})

(dfrac{y-3}{y + 3} = dfrac{N}{y^2 - 9})

Answer

((y-3)^2)

Exercise (PageIndex{32})

(dfrac{a+5}{a-5} = dfrac{N}{a^2 - 25})

Answer

((z - 4)^2)

Exercise (PageIndex{33})

(dfrac{4}{2a + 1} = dfrac{N}{2a^2 - 5a - 3})

Exercise (PageIndex{34})

(dfrac{1}{3b - 1} = dfrac{N}{3b^2 + 11b - 4})

Answer

(b+4)

Exercise (PageIndex{35})

(dfrac{a+2}{2a - 1} = dfrac{N}{2a^2 + 9a - 5})

Exercise (PageIndex{36})

(dfrac{-3}{4x + 3} = dfrac{N}{4x^2 - 13x - 12})

Answer

(−3(x−4))

Exercise (PageIndex{37})

(dfrac{b+2}{3b - 1} = dfrac{N}{6b^2 + 7b - 3})

Exercise (PageIndex{38})

(dfrac{x - 1}{4x - 5} = dfrac{N}{12x^2 - 11x - 5})

Answer

((x−1)(3x+1))

Exercise (PageIndex{39})

(dfrac{3}{x + 2} = dfrac{3x - 21}{N})

Exercise (PageIndex{40})

(dfrac{4}{y + 6} = dfrac{4y + 8}{N})

Answer

((y+6)(y+2))

Exercise (PageIndex{41})

(dfrac{-6}{a - 1} = dfrac{-6a - 18}{N})

Exercise (PageIndex{42})

(dfrac{-8a}{a+3} = dfrac{-8a^2 - 40a}{N})

Answer

((a+3)(a+5))

Exercise (PageIndex{43})

(dfrac{y+1}{y-8} = dfrac{y^2 - 2y - 3}{N})

Exercise (PageIndex{44})

(dfrac{x - 4}{x + 9} = dfrac{x^2 + x - 20}{N})

Answer

((x+9)(x+5))

Exercise (PageIndex{45})

(dfrac{3x}{2-x} = dfrac{N}{x-2})

Exercise (PageIndex{46})

(dfrac{7a}{5-a} = dfrac{N}{a-5})

Answer

(-7a)

Exercise (PageIndex{47})

(dfrac{-m + 1}{3 - m} = dfrac{N}{m-3})

Exercise (PageIndex{48})

(dfrac{k + 6}{10 - k} = dfrac{N}{k- 10})

Answer

(−k−6)

Exercise (PageIndex{49})

(dfrac{2}{a^2} = dfrac{N}{a^2(a-1)})

For the following problems, convert the given rational expressions to rational expressions having the same denominators.

Exercise (PageIndex{50})

(dfrac{2}{a}, dfrac{3}{a^4})

Exercise (PageIndex{51})

(dfrac{5}{b^2}, dfrac{4}{b^3})

Answer

(dfrac{5b}{b^3}, dfrac{4}{b^3})

Exercise (PageIndex{52})

(dfrac{8}{z}, dfrac{3}{4z^3})

Exercise (PageIndex{53})

(dfrac{9}{x^2}, dfrac{1}{4x})

Answer

(dfrac{36}{4x^2}, dfrac{x}{4x^2})

Exercise (PageIndex{54})

(dfrac{2}{a+3}, dfrac{4}{a+1})

Exercise (PageIndex{55})

(dfrac{2}{x + 5}, dfrac{4}{x-5})

Answer

(dfrac{2(x-5)}{(x+5)(x-5)}, dfrac{4(x+5)}{(x+5)(x-5)})

Exercise (PageIndex{56})

(dfrac{1}{x-7}, dfrac{4}{x-1})

Exercise (PageIndex{57})

(dfrac{10}{y+2}, dfrac{1}{y+8})

Answer

(dfrac{10(y+8)}{(y+2)(y+8)}, dfrac{y+2}{(y+2)(y+8)})

Exercise (PageIndex{58})

(dfrac{4}{a^2}, dfrac{a}{a+4})

Exercise (PageIndex{59})

(dfrac{-3}{b^2}, dfrac{b^2}{b+5})

Answer

(dfrac{-3(b+5)}{b^2(b+5)}, dfrac{b^4}{b^2(b+5)})

Exercise (PageIndex{60})

(dfrac{-6}{b-1}, dfrac{5b}{4b})

Exercise (PageIndex{61})

(dfrac{10a}{a-6}, dfrac{2}{a^2 - 6a})

Answer

(dfrac{10a^2}{a(a-6)}, dfrac{2}{a(a-6)})

Exercise (PageIndex{62})

(dfrac{4}{x^2 + 2x}, dfrac{1}{x^2 - 4})

Exercise (PageIndex{63})

(dfrac{x+1}{x^2 - x - 6}, dfrac{x+4}{x^2 + x - 2})

Answer

(dfrac{(x+1)(x-1)}{(x-1)(x+2)(x-3)}, dfrac{(x+4)(x-3)}{(x-1)(x+2)(x-3)})

Exercise (PageIndex{64})

(dfrac{x-5}{x^2 - 9x + 20}, dfrac{4}{x^2 - 3x - 10})

Exercise (PageIndex{65})

(dfrac{-4}{b^2 + 5b - 6}, dfrac{b+6}{b^2 - 1})

Answer

(dfrac{-4(b +1)}{(b+1)(b-1)(b + 6)}, dfrac{(b+6)^2}{(b+1)(b-1)(b+6)})

Exercise (PageIndex{66})

(dfrac{b+2}{b^2 + 6b + 8}, dfrac{b-1}{b^2 + 8b + 12})

Exercise (PageIndex{67})

(dfrac{x+7}{x^2 - 2x - 3}, dfrac{x+3}{x^2 - 6x - 7})

Answer

(dfrac{(x+7)(x-7)}{(x+1)(x-3)(x-7)}, dfrac{(x+3)(x-3)}{(x+1)(x-3)(x-7)})

Exercise (PageIndex{68})

(dfrac{2}{a^2 + a}, dfrac{a+3}{a^2 - 1})

Exercise (PageIndex{69})

(dfrac{x-2}{x^2 + 7x + 6}, dfrac{2x}{x^2 + 4x - 12})

Answer

(dfrac{(x-2)^2}{(x+1)(x-2)(x+6)}, dfrac{2x(x+1)}{(x+1)(x-2)(x+6)})

Exercise (PageIndex{70})

(dfrac{x-2}{2x^2 + 5x - 3}, dfrac{x-1}{5x^2 + 16x + 3})

Exercise (PageIndex{71})

(dfrac{2}{x-5}, dfrac{-3}{5-x})

Answer

(dfrac{2}{x-5}, dfrac{3}{x-5})

Exercise (PageIndex{72})

(dfrac{4}{a-6}, dfrac{-5}{6-a})

Exercise (PageIndex{73})

(dfrac{6}{2-x}, dfrac{5}{x-2})

Answer

(dfrac{-6}{x-2}, dfrac{5}{x-2})

Exercise (PageIndex{74})

(dfrac{k}{5-k}, dfrac{3k}{k-5})

Exercise (PageIndex{75})

(dfrac{2m}{m-8}, dfrac{7}{8-m})

Answer

(dfrac{2m}{m-8}, dfrac{-7}{m-8})

Exercises For Review

Exercise (PageIndex{76})

Factor (m^2x^3 + mx^2 + mx)

Exercise (PageIndex{77})

Factor (y^2 - 10y + 21)

Answer

((y−7)(y−3))

Exercise (PageIndex{78})

Write the equation of the line that passes through the points (1, 1) and (4, −2). Express the equation in slope-intercept form.

Exercise (PageIndex{79})

Reduce (dfrac{y^2 - y - 6}{y-3})

Answer

(y+2)

Exercise (PageIndex{80})

Find the quotient. (dfrac{x^2 - 6x + 9}{x^2 - x - 6} div dfrac{x^2 + 2x - 15}{x^2 + 2x})


8.7 Solving Rational Equations

When solving equations that are made up of rational expressions, we use the same strategy we used to solve complex fractions, where the easiest solution involved multiplying by the LCD to remove the fractions. Consider the following examples.

For these three fractions, the LCD is 12. Therefore, multiply all parts of the equation by 12:

This reduces the rational equation to:

Multiplying this out yields:

Now isolate and solve for :

For these three fractions, the LCD is />. Therefore, multiply all parts of the equation by />:

This reduces the rational equation to:

Multiplying this out yields:

Now subtract 5 from both sides of the equation to turn this into an equation that can be easily factored:


Adding and Subtracting Rational Expressions with Unlike Denominators

There are a few steps to follow when you add or subtract rational expressions with unlike denominators.

  1. To add or subtract rational expressions with unlike denominators, first find the LCM of the denominator. The LCM of the denominators of fraction or rational expressions is also called least common denominator , or LCD.
  2. Write each expression using the LCD. Make sure each term has the LCD as its denominator.
  3. Add or subtract the numerators.
  4. Simplify as needed.

Since the denominators are not the same, find the LCD.

Since 3 a and 4 b have no common factors, the LCM is simply their product: 3 a &sdot 4 b .

That is, the LCD of the fractions is 12 a b .

Rewrite the fractions using the LCD.

Since the denominators are not the same, find the LCD.

Here, the GCF of 4 x 2 and 6 x y 2 is 2 x . So, the LCM is the product divided by 2 x :

Rewrite the fractions using the LCD.

Since the denominators are not the same, find the LCD.

The LCM of a and a &minus 5 is a ( a &minus 5 ) .

That is, the LCD of the fractions is a ( a &minus 5 ) .

Rewrite the fraction using the LCD.

2 a &minus 3 a &minus 5 = 2 ( a &minus 5 ) a ( a &minus 5 ) &minus 3 a a ( a &minus 5 )

= 2 a &minus 10 a ( a &minus 5 ) &minus 3 a a ( a &minus 5 )

Since the denominators are not the same, find the LCD.

The LCM of c + 2 and c &minus 3 is ( c + 2 ) ( c &minus 3 ) .

That is, the LCD of the fractions is ( c + 2 ) ( c &minus 3 ) .

Rewrite the fraction using the LCD.

5 c + 2 + 6 c &minus 3 = 5 ( c &minus 3 ) ( c + 2 ) ( c &minus 3 ) + 6 ( c + 2 ) ( c + 2 ) ( c &minus 3 )

= 5 c &minus 15 ( c + 2 ) ( c &minus 3 ) + 6 c + 12 ( c + 2 ) ( c &minus 3 )

= 5 c &minus 15 + 6 c + 12 ( c + 2 ) ( c &minus 3 )

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8.5: Building Rational Expressions and the LCD

After studying this section, you will be able to:

1. Simplify an algebraic fraction by factoring.

Recall that a rational number is a number that can be written as one integer divided by another integer such as 3 ÷ 4 or 3/4 . We usually use the word fraction to mean 3/4 We can extend this idea to algebraic expressions. A rational expression is an algebraic expression divided by another algebraic expression such as

The last fraction is sometimes called a fractional algebraic expression. There is a special restriction for all fractions, including fractional algebraic expressions. The denominator of the fraction cannot be 0 . For example, in the expression (3x+2)/(x+4) the denominator cannot be 0 . Therefore, the value of x cannot be -4 . The following important restriction will apply throughout this lesson. We state it here to avoid having to mention it repeatedly throughout this lesson.

We have discovered that fractions can be simplified (or reduced) in the following way.

This is sometimes referred to as the basic rule of fractions and can be stated as follows:

We will examine several examples where a , b , c are real numbers, as well as more involved examples where a, b, and c are polynomials. In either case we shall make extensive use of our factoring skills in this section.

There is one essential property that is revealed by the basic rule of fractions. It is this: If the numerator and denominator of a given fraction are multiplied by the same quantity, an equivalent fraction is obtained. The rule can be used two ways. You can start with (ac)/(bc) and end with the equivalent fraction a/b . Or, you can start with a/b and end with the equivalent fraction (ac)/(bc) .

(a) Write a fraction equivalent to 3/5 with a denominator of 10 .

(a) 3/5=(39*2)/(5*2)=6/10 &emsp&emspUse the rule a/b=(ac)/(bc) . Let c = 2 since 5*2 = 10 .

(b) 21/39 = (7*3)/(13*3)=7/13 &emsp&emspUse the rule (ac)/(bc)=a/b . Let c = 3 because 3 is the greatest common factor of 21 and 39 .

Simplifying Algebraic Fractions

The process of reducing the fraction shown above is sometimes called dividing out common factors. Remember, only factors of both the numerator and the denominator can be divided out. Now, to apply this rule, it is usually necessary that the numerator and denominator of the fraction be completely factored. You will need to use your factoring skills from earlier tutorial to accomplish this step. When you are applying this rule you are simplifying the fraction .

EXAMPLE 2. Simplify, (4x+12)/(5x+15)

(4x+12)/(5x+15)=(4(x+3))/(5(x+3)) &emsp&emspFactor 4 from the numerator. Factor 5 from the denominator.

EXAMPLE 3 Simplify. (x^2+9x+14)/(x^2-4)

= ((x+7)(x+2))/((x-2)(x+2)) &emsp&emspFactor both the numerator and the denominator.

= (x+7)/(x-2) &emsp&emsp Apply the basic rule of fractions.

Some problems may involve more than one step of factoring. Always remember to remove any common factors as the first step, if it is possible to do so.

EXAMPLE 4 Simplify. (9x-x^3)/(x^3+x^2-6x)

Remove a common factor from the polynomial in the numerator and in the denominator.

Factor each polynomial and apply the basic rule of fractions. Note that (3 + x) is equivalent to (x + 3) since addition is commutative.

Be on the lookout for a special simplifying situation. Watch for a situation where each term in the denominator is opposite in sign from each term in the numerator. In such cases you should factor -1 or another negative number from one polynomial so that the expression in each set of parentheses is equivalent. Look carefully at the following two examples.

EXAMPLE 5 Simplify. (5x-15)/(6-2x)

Notice that the variable term in the numerator, 5x , and the variable term in the denominator, -2x , are opposite in sign. Likewise, the numerical terms -15 and 6 are opposite in sign. Factor out a negative number from the denominator.

Factor 5 from the numerator. Factor -2 from the denominator. Note that (x-3) and (-3 + x) are equivalent since +x -3 = -3 + x .

= &minus 5/2 &emsp&emsp Apply the basic rule of fractions.

Note that 5/-2 is not considered to be in simple form. We usually avoid leaving a negative number in the denominator. Therefore, to simplify, give the result as &minus 5/2 or -5/2 .

EXAMPLE 6 Simplify. (2x^2-11x+12)/(16-x^2)

Factor numerator and denominator. Observe that (x-4) and (4-x) are opposite in sign.

Factor -1 out of (+4-x) to obtain -1(-4 + x)

Apply the basic rule of fractions, since (x-4) and (-4 + x) are equivalent.

After doing Examples 5 and 6, you will notice a pattern. Whenever the factor in the numerator and the factor in the denominator are exactly opposite in sign, the value -1 results. We could actually make this a definition of that property.

You may use this definition in reducing fractions if it is helpful to you. Otherwise, you may use the factoring method discussed earlier.

&emsp&emspSome problems will involve two or more variables. In such cases, you will need to factor carefully and make sure that each set of parentheses contains the correct letters.

EXAMPLE 7 Simplify, (x^2-7xy+12y^2)/(2x^2-7xy-4y^2

Factor both the numerator and the denominator.

Apply the basic rule of fractions.

Multiplication and Division of Rational Expressions

After studying this section, you will be able to:

1. Multiply algebraic fractions and write the answer in simplest form.

2. Divide algebraic fractions and write the answer in simplest form.

Multiplying Algebraic Fractions

To multiply two fractional expressions, we multiply the numerators and we multiply the denominators. As before, the denominators cannot equal zero.

Simplifying or reducing fractions prior to multiplying them usually makes the problem easier to do. To do the problem the long way would certainly increase the chance for error! This long approach should be avoided.
&emsp&emspAs an example, let&rsquos do the same problem two ways to see which way seems easier. First let&rsquos multiply by the &lsquo&lsquolong way&rsquo&rsquo:

Multiply the numerators and multiply the denominators

Reduce the fraction. (Note: It takes a bit of trial and error to discover how to reduce it.)

Compare this with the method of reducing the fractions prior to multiplying them. Let&rsquos multiply by simplifying before multiplication:

Step 1. It is easier to factor first. We factor the numerator and denominator of the second fraction.

Step 2. We express the product as one fraction (by the definition of multiplication of fractions).

Step 3. Then we apply the basic rule of fractions to divide common factors of 5 and 7 that appear in the denominator and in the numerator.

A similar approach can be used with the multiplication of rational algebraic expressions. We first factor the numerator and denominator of each fraction wherever possible. Then we cancel any factor that is common to a numerator and a denominator. Finally, we multiply the remaining numerators and the remaining denominators.

EXAMPLE 1 Multiply. (x^2-x-12)/(x^2-16)*(2x^2+7x-4)/(x^2-4x-21)

Factor wherever possible is always the first step.

Apply the basic rule of fractions. (Three pairs of factors divide out.)

In some cases, a given numerator can be factored more than once. You should always be sure that you remove the common factor first wherever possible.

EXAMPLE 2 Multiply. (x^4-16)/(x^3+4x)*(2x^2-8x)/(4x^2+2x-12)

Factor each numerator and denominator. Factoring out the common factor first is very important.

Factor again where possible.

Remove factors that appear in both numerator and denominator. (There are four such pairs to be removed.)

Write the answer as one fraction. (Usually, if there is more than one set of parentheses in a numerator, the answer is left in factored form.)

Dividing Algebraic Fractions

For any two fractions a/b and c/d the operation of division can be performed by inverting the second fraction and multiplying it by the first fraction.

This property holds whether a , b , c , and d are polynomials or numerical values. (It is assumed, of course, that no denominator is zero.)

&emsp&emspIn the first step for dividing two algebraic fractions, you should invert the second fraction and write the problem as a multiplication. Then you follow the procedure for multiplying algebraic fractions.

EXAMPLE 3. Divide. (6x+12y)/(2x-6y) ÷ (9x^2-36y^2)/(4x^2-36y^2)

Invert the second fraction and write the problem as the product of two fractions.

Factor each numerator and denominator.

Factor again where possible

Remove factors that appear in both numerator and denominator.

Write the result as one fraction and simplify.

Usually, answers are left in this form.

Although it is correct to write this answer as (4x+12y)/(3x-6y) it is customary to leave the answer in factored form in order to ensure that the final answer is simplified.

A polynomial that is not in fraction form can be written as a fraction if you write a denominator of 1 .

EXAMPLE 4 Divide. (15-3x)/(x+6) ÷ (x^2-9x + 20)

Note that x^2-9x + 20 can be written as (x^2-9x + 20)/1

Factor where possible. Note we had to factor -3 from the first numerator so that we would have a common factor with the second denominator.

Divide out the common factor. (-5 + x) is equivalent to (x-5) and the final answer

Note that the answer can be written in several equivalent forms.

Take a minute to study each. Each is equivalent to the others.

A word of caution:

It is logical to assume that the types of problems encountered in Section 6.2 have at least one common factor that can be divided out. Therefore if after factoring you do not observe any common factors, you should be somewhat suspicious. In such cases, it would be wise to double check your factoring steps to see if an error has been made.

Addition and Subtraction of Rational Expressions

After studying this section, you will be able to:

1. Add and subtract algebraic fractions with the same denominator.

2. Determine the LCD for two or more algebraic fractions with different denominators.

3. Add or subtract algebraic fractions with different denominators.

Adding and Subtracting Algebraic Fractions with the Same Denominator

If fractional algebraic expressions have the same denominator, they can be combined in a fashion similar to that used in arithmetic. The numerators are added or subtracted and the denominator remains the same.

EXAMPLE 1 Add. (5a)/(a+2b)+(6a)/(a+2b)

Note the denominators are the same. Add the numerators.

EXAMPLE 2 Subtract, (3x)/((x+y)(x-2y))-(8x)/((x+y)(x-2y))

Write as one fraction and simplify.

Determining the LCD

How do we add or subtract fractional expressions when the denominators are not the same? First we must find the least common denominator (LCD). You need to be clear on how to find a least common denominator and how to add and subtract fractions from arithmetic before you attempt this lesson.

EXAMPLE 3 Find the LCD of the following algebraic fractions. 5/(2x-4) , 6/(3x-6)

The different factors are 2 , 3 , and (x-2) . Since no factor is repeated more than once in any one denominator, the LCD is the product of these three factors.

EXAMPLE 4 Find the LCD.

If a factor occurs more than once in any one denominator, the LCD will contain that factor repeated the greatest number of times that it occurs in any One denominator.

Adding and Subtracting Fractions with Different Denominators

If two fractions have different denominators, we first change them to equivalent fractions with the least common denominator. Then we add or subtract the numerators and keep the common denominator.

EXAMPLE 5 Add. 5/(xy)+2/y

The denominators are different. Find the LCD. The two factors are x and y . We observe that the LCD is xy .

5/(xy)+2/y=5/(xy)+2/y*x/x &emsp&emspMultiply the second fraction by x/x .

&emsp&emsp&emsp&emsp&emsp = 5/(xy)+(2x)/(xy) &emsp&emspNow each fraction has a common denominator of xy .

EXAMPLE 6 Add. (3x)/((x+y)(x-y))+5/(x+y)

The two factors are (x + y) and (x-y) . We observe that the LCD = (x + y)(x-y) . Multiply the second fraction by (x-y)/(x-y) .

Now each fraction has a common denominator of (x + y)(x-y) .

Write the sum of the numerators over one common denominator.

It is important to remember that the LCD is the smallest algebraic expression into which each denominator can be divided. For algebraic expressions the LCD must contain each factor that appears in any denominator. If the factor is repeated, the LCD must contain each factor the greatest number of times that it appears in any one denominator.

EXAMPLE 7 Add. 5/(xy^2)+3/(x^2y)+2/y^3

The x factor is squared in one fraction. The y factor is cubed in one fraction. Therefore, the LCD is x^2y^3 .
Multiply each fraction by the appropriate value to obtain x^2y^3 in each denominator. (Note: Here we always multiply the numerator and denominator of any one fraction by the same value to obtain an equivalent fraction.)

Now all fractions have a common denominator.

Write the sum as one fraction.

In many cases, the denominators in an addition or subtraction problem are not in factored form. You must factor each denominator in order to determine the LCD. Collect like terms in the numerator then look to see if that final numerator can be factored. If so, the fraction can sometimes be simplified.

EXAMPLE 8 Add. 5/(x^2-y^2)+(3x)/(x^3+x^2y)

Factor the two denominators. Observe that the LCD is x^2(x + y)(x-y) .

Multiply each fraction by the appropriate value to obtain a common denominator of x^2(x + y)(x-y) .

Write the sum of the numerators over one common denominator.

Remove the common factor x in the numerator and denominator and simplify.

It is very easy to make a mistake in sign when subtracting two fractions. You will find it helpful to place parentheses around the numerator of the second fraction so that you will not forget to subtract the entire numerator.

EXAMPLE 9 Subtract. (3x+4)/(x-2)-(x-3)/(2x-4)

Factor the second denominator and observe that the LCD is 2(x-2) .

Multiply the first fraction by 2/2 so that the resulting fraction will have the common denominator.

Write the indicated subtraction as one fraction. Note the parentheses around (x-3) .

Remove the parentheses in the numerator.

To avoid making errors when subtracting two fractions, some students find it helpful to change subtraction to addition of the opposite of the second fraction.

EXAMPLE 10 Subtract. (8x)/(x^2-16)-4/(x-4)

Factor the first denominator. Use the property that a/b-c/b=a/b+(-c)/b

Multiply the second fraction by (x+4)/(x+4)

Write the addition of the numerators over one common denominator.

Collect like terms. Note that the numerator can be factored.

Divide out. Since (x-4) is a factor of the numerator and the denominator, we may divide out the common factor.

Simplify Complex Rational Expressions

After studying this section, you will be able to:

1. Simplify complex fractions by adding or subtracting in the numerator and denominator.

2. Simplify complex fractions by multiplying by the LCD of all the denominators.

Simplifying Complex Fractions by Adding or Subtracting in the Numerator and the Denominator

A complex fraction has a fraction in the numerator or in the denominator, or both.

The bar in a complex fraction is both a grouping symbol and a symbol for division.

((a+b)/3)/((x-2y)/4) is equivalent to ((a+b)/3) ÷ ((x-2y)/4)

We need a procedure for simplifying complex fractions.

EXAMPLE 1 Simplify. (1/x)/(2/y^2+1/y)

Step 1 Add the two fractions in the denominator.

Step 2 Divide the fraction in the numerator by the fraction in the denominator.

In some problems there may be two or more fractions in the numerator and the denominator.

EXAMPLE 2 Simplify. (1/x+1/y)/(3/a-2/b)

We observe that the LCD of the fractions in the numerator is xy . The LCD of the fractions in the denominator is ab .

Multiply each fraction by the appropriate value to obtain common denominators.

Add the two fractions in the numerator and subtract the two fractions in the denominator.

Invert the fraction in the denominator and multiply it by the numerator.

Write the answer as one fraction.

In some problems, factoring may be necessary to determine the LCD and to combine fractions.

EXAMPLE 3 Simplify. (1/(x^2-1)+2/(x+1))/x

We observe the need for factoring x^2-1 .

The LCD for the fractions in the numerator is (x + 1)(x-1) .

Add the two fractions in the numerator.

Simplify the numerator. Invert the fraction in the denominator and multiply.

Write the answer as one fraction.

In simplifying complex fractions, always be alert to see if the final fraction can be reduced or simplified.

EXAMPLE 4 Simplify. (3/(a+b)-3/(a-b))/(5/(a^2-b^2))

The LCD of the two fractions in the numerator is (a + b)(a-b) .

Study carefully how we combine the two fractions in the numerator. Do you see how we obtain -6b ?

Note that since (a + b)(a-b) are factors in both numerator and denominator they may be divided out.

Simplifying Complex Fractions by Multiplying by the LCD of All the Denominators

There is another way to simplify complex fractions: Multiply the numerator and denominator of the complex fraction by the least common denominator of all the denominators appearing in the complex fraction.

EXAMPLE 5 Simplify by multiplying by the LCD. (5/(ab^2)-2/(ab))/(3-5/(2a^2b))

The LCD of all the denominators in the complex fraction is 2a^2b^2 . Multiply each term by 2a^2b^2 .

So that you can compare the two methods, we will redo Example 4 by the alternative method.

EXAMPLE 6 Simplify. (3/(a+b)-3/(a-b))/(5/(a^2-b^2)) &emsp&emspUse alternative method.

The LCD of all individual fractions contained in the complex fraction is (a+b)(a-b) . Multiply each term by the LCD.

Equations Involving Rational Expressions

After studying this section, you will be able to:

1. Solve equations involving algebraic fractions that have solutions.

2. Determine if an equation involving algebraic fractions has no solution.

Solving Equations Involving Algebraic Fractions

In earlier lesson we developed procedures to solve linear equations containing fractions whose denominators were numerical values. In this section we use a similar approach to solve equations containing fractions whose denominators are polynomials.

EXAMPLE 1 Solve for x and check your solution. 5/x+2/3=2-2/x-1/6

We observe that the LCD is 6x . Multiply each term by 6x .

Simplify. Do you see how each term is obtained?

Subtract 4x from both sides.

Check 5/6+2/3 ≟ 2-2/6-1/6 &emsp&emspReplace each x by 6 .

It is sometimes necessary to factor denominators before the correct LCD can be determined.

EXAMPLE 2 Solve and check your solution. 3/(x+5)-1=(4-x)/(2x+10)

We observe a need to factor the 2x + 10 . We determine that the LCD is 2(x + 5) .

Multiply each term by the LCD.

Subtract 4 from both sides.

Check : Replace each x in the original equation by -8 .

Determining If an Equation Involving Algebraic Fractions Has No Solution

Linear equations containing fractions that have variables in denominators sometimes appear to have solutions when in fact they do not. By this we mean that the "solutions" we get by using completely correct methods are, in actuality, not solutions.

&emsp&emspIn the case where a value makes a denominator in the equation equal to zero, we say there is no solution to the equation. Such a value is called an extraneous solution. An extraneous solution is an apparent solution that does not satisfy the original equation. It is important that you check the number in the original equation.

EXAMPLE 3 Solve and check. y/(y-2)-4=2/(y-2)

We observe that the LCD is y-2 . Multiply each term by (y-2) .

Simplify. Do you see how this is done?

Subtract 8 from both sides.

Divide both sides by negative 3 .

y=2 is only an apparent solution.

There is no solution to this problem.

Why? We can see immediately that y = 2 is not a possible solution for the original equation. The use of the value y = 2 in a denominator would make the denominator equal to zero, and the expression is undefined.

Check. Suppose that you try to check the apparent solution by substituting y = 2 .

This does not check since you do not obtain a real number when you divide by zero.

There is no such number as 2 ÷ 0 . We see that y = 2 does not check. There is no solution to this problem.

Ratio, Proportion, and Other Applied Problems

After studying this section, you will be able to:

1. Solve problems involving ratio and proportion.

2. Solve problems involving similar triangles.

3. Solve certain distance problems.

Solving Problems Involving Ratio and Proportion

A ratio is a comparison of two quantities. You may be familiar with ratios that compare miles to hours or miles to gallons. A ratio is often written as a quotient in the form of a fraction. For example, the ratio of 7 to 9 can be written as 7/9 .

&emsp&emspA proportion is an equation that states that two ratios are equal. For example,

Let&rsquos take a closer look at the last proportion. We can see that the LCD of the fractional equation is bd .

(bd)a/b =(bd)c/d &emsp&emspMultiply each side by the LCD.

ad = bc &emsp&emspSince multiplication is commutative, da = ad .

Thus we have proved the following:

This is sometimes called cross-multiplying. It can be applied only if you have one fraction and nothing else on each side of the equation.
&emsp&emspWe can use a proportion and the technique of cross-multiplying to solve a variety of
applied problems involving two ratios.

EXAMPLE 1 Michael took 5 hours to drive 245 miles on the turnpike. If he continues at the same rate, how many hours will it take him to drive a distance of 392 miles?

Let x = the number of hours it will take to drive 392 miles. If 5 hours are needed to drive 245 miles, then x hours is needed to drive 392 .

We can write this as a proportion. Compare time to distance in each ratio.

&emsp&emsp&emsp&emsp&emsp&emsp&emsp&emsp&emsp&emsp

3. Solve the equation and state the answer.

1960/245=x &emsp&emsp Divide both sides by 245

8=x
It would take Michael 8 hours to drive 392 miles.

EXAMPLE 2 If 3/4 inch on a map represents an actual distance of 20 miles, how long is the distance represented by 4 1/8 inches on the same map?

Let x = the distance represented by 4 1/8 inches.

Write 4 1/8 as 33/8 and simplify.

Multiplication of fractions.

4 1/8 inches on the map represents an actual distance of 110 miles.

Solving Problems Involving Similar Triangles

Similar triangles are triangles that have the same shape, but may be a different size. For example, if you draw a triangle on a sheet of paper and place the paper in a photocopy machine and make a copy that is reduced by 25%, you would create a triangle that was similar to the original triangle. The two triangles would have the same shape. The corresponding sides of the triangles would be proportional.

You can use the proportion equation to show that the corresponding sides of the above triangles are proportional. In fact, you can use the proportion equation to find the length of a side of a triangle if it is not known.

EXAMPLE 3 Find the length of side x in the two similar triangles pictured below.

x= 135/32 &emsp&emsp Divide both sides by 32 .

We can also use similar triangles for indirect measurement, for instance, to find the measure of an object that is too tall to measure using standard measuring devices. When the sun shines on two vertical objects at the same time, the shadows and the objects form similar triangles.

EXAMPLE 4 A woman who is 5 feet tall casts a shadow that is 8 feet long. At the same time of day, a building casts a shadow that is 72 feet long. How tall is the building?

We do not know the height of the building, so we call it x .

2. Write an equation and solve.

The height of the building is 45 feet.

In problems such as Example 4 we are assuming that the building and the person are standing exactly perpendicular to the ground. In other words, each triangle is assumed to be a right triangle. In some similar problems of this type, if the triangle is not a right triangle you must be careful to be sure that the angle between the object and the ground is the same in each case.

Solving Distance Problems Involving Algebraic Fractions

Some distance problems are solved using rational equations. We will need the formula

Distance = Rate × Time, D = RT , which we can write in the form T = D/R

EXAMPLE 5 Plane A flies at a speed that is 50 kilometers per hour faster than plane B. Plane A flies 500 kilometers in the amount of time that plane B flies 400 kilometers. Find the speed of each plane.

Let s = the speed of plane B in kilometers per hour.

Let s + 50 = the speed of plane A in kilometers per hour.

Make a simple table for D , R , and T .

&emsp&emsp&emsp&emsp&emsp&emsp&emsp&emsp&emsp&emsp&emsp&emsp

Since T = D/R we divide the expression for D by the expression for R and write it in the table in the column for time.

&emsp&emsp&emsp&emsp&emsp&emsp&emsp&emsp&emsp&emsp&emsp&emsp

2. Write an equation and solve.

Each plane flies for the same amount of time. That is, the time for plane A equals the time for plane B.

You can solve this equation using the methods you learnt in earlier lesson or you may cross- multiply. Here we will cross-multiply.

500s = (s + 50)(400) &emsp&emsp Cross-multiply.

500s = 400s + 20,000 &emsp&emsp Remove parentheses.

100s = 20,000 &emsp&emsp Subtract 400s from each side.

s = 200 &emsp&emsp Divide each side by 100 .

Plane B travels 200 kilometers per hour. Since

plane A travels 250 kilometers per hour.

Solving Work Problems

Some applied problems involve the length of time needed to do a job. These problems are often referred to as work problems.

EXAMPLE 6 Reynaldo can sort a huge stack of mail on an old sorting machine in 9 hours. His brother Carlos can sort the same amount of mail using a newer sorting machine in 8 hours. How long would it take them to do the job working together? Express your answer in hours and minutes. Round to the nearest minute.

Let&rsquos do a little reasoning.
&emsp&emspIf Reynaldo can do the job in 9 hours, then in 1 hour he could do 1/9 of the job.
&emsp&emspIf Carlos can do the job in 8 hours, then in 1 hour he could do 1/8 of the job.
&emsp&emspLet x = the number of hours it takes Reynaldo and Carlos to do the job together.
&emsp&emspIn 1 hour together they could do 1/x of the job.

2. Write an equation and solve.

The amount of work Reynaldo can do in 1 hour plus the amount of work Carlos can do in 1 hour must be equal to the amount of work they could do together in 1 hour.

We observe the LCD is 72x . Multiply each term by the LCD.

To change 4/17 of an hour to minutes we multiply.

To the nearest minute this is 14 minutes. Thus doing the job together will take 4 hours and 14 minutes.


SOLUTION: Find the LCD of the rational expressions in the list. >>,>> thank you for the help.

You can put this solution on YOUR website!
To find the LCD we need to know what the factors of the denominators are. So we start by factoring the denominators. (From this point on I am going to ignore the numerators since they have no impact of what the LCD is or on how we find the LCD.)

The first denominator is a difference of squares so we can use the pattern to factor it:


The second denominator is a perfect trinomial square so we can use the pattern tp factor it:


So our two denoinators, in factored form are:
(h+k)(h-k) and (h-k)(h-k)

The LCD will be the smallest product that includes all the factors of both denominators. In this case the LCD will be:
(h+k)(h-k)(h-k)
Looking at the LCD you can see the first denominator in the first two factors and the second denominator in the last two factors.

(h+k)(h-k)(h-k)
is the LCD and this may be an acceptable form. If not, then multiply this out. Multiplying the first two factors is easy because they cam from the first denominator:


1) Make the denominators of the rational expressions the same by finding the Least Common Denominator (LCD).

Note: The Least Common Denominator is the same as the Least Common Multiple (LCM) of the given denominators.

2) Next, combine the numerators by the indicated operations (add and/or subtract) then copy the common denominator.

Note: Don’t forget to simplify further the rational expression by canceling common factors, if possible.

As they say, practice makes perfect. So we will go over six (6) worked examples in this lesson to illustrate how it is being done. Let’s get started!

Examples of Adding and Subtracting Rational Expressions

Example 1: Add and subtract the rational expressions below.

In this case, we are adding and subtracting rational expressions with unlike denominators. Our goal is to make them all the same.

Since I have monomials in the denominators, the LCD can be obtained by simply taking the Least Common Multiple of the coefficients, where LCM ( 3 , 6 ) = 6 , and multiply that to the variable x with the highest exponent.

The LCD should be (LCM of coefficients) times (LCM of variable x ) which gives us left( 6 ight)left( <> ight) = 6 .

The “blue fractions” are the appropriate multipliers to do the job!

Now that we have the same denominators, it is easy to simplify.

Combine similar terms (see the x variables?).

  • When you reach the point of having a single rational expression, your next critical step is to factor the top and the bottom completely.

The reason is that you may have common factors, which can be canceled out.

To make this a better answer, I will exclude the value of x that can make the original rational expression undefined.

I can add the condition that x e 0 .

Example 2: Add the rational expressions below.

This problem contains like denominators. We want this because it is the LCD itself – the given denominator of the rational expression.

So then the LCD that we are going to use is 2x + 1 .

Tip: Don’t rush by immediately doing all the calculations in your head. I suggest that you place each term inside the parenthesis before performing the required operation. This extra step may be your lifesaver to avoid careless mistakes.

  • Unless you have a good grasp on how to effectively combine like terms, I suggest you take another “baby step” as an additional precaution.

Do you see how I decided to place the like terms side-by-side on the numerator?

To prevent the original rational expression to have a denominator of zero, we say that x e - <1 over 2>.

Example 3: Add the rational expressions below.

This time I have the same trinomial in both denominators. This is similar to problem #2 but the quadratic trinomial adds a layer of fun. Later, I can factor out the denominator to see if there are common factors to cancel against the numerator.

  • Copy the common denominator and set it up just like this – placing each numerator in the parenthesis before adding them.
  • Rearrange the terms in such a way that similar terms are next to each other for ease of computation later.

You may say that x e - ,4 and x e + ,5 from the original denominator.

Example 4: Subtract the rational expressions below.

This is a good example because the denominators are different. I need to find the LCD by doing the following steps.

Factor each denominator completely, and line up the common factors. Identify each unique factor with the highest power.

Multiply together the ones with the highest exponents for each unique factor.

  • In this step, I haven’t done anything but factor out the denominator of the first rational expression.

The first denominator is okay but the second one is lacking left( ight) .

This is why I multiply it by the blue fraction .

  • Put them all together in one fraction with a common denominator of left( ight)left( ight) . However, keep each numerator inside a parenthesis.

Group similar terms together before simplifying them.

  • We got it! You may include the restrictions that x e 5 and x e - ,5 based on the original denominator of the given rational expression. This is to prevent the division of zero, which is not good.

Example 5: Subtract and add the rational expressions below.

This problem is definitely interesting. To solve this, hold on to the things that you already know. Find the LCD by doing the steps below.

Factor each denominator completely and neatly line up the common factors. Identify each unique factor with the highest power.

Multiply together the ones with the highest exponents for each unique factor.

  • Combine them in one fraction while keeping each numerator within a parenthesis. Make sure to copy the indicated operations correctly.
  • Now, we’ll factor out the numerator and hope to see common factors between the numerator and denominator that can be canceled.
  • We now have our final answer. Add the restrictions x e 4 and x e - ,3 to avoid dividing by zero.

Example 6: Subtract and add the rational expressions below.

This is our last example in this lesson. I must say this is very similar to example 5. By now, you should already have a solid understanding of how to add and subtract rational expressions.

Let’s start finding the LCD again.

Factor each denominator completely and neatly line up the common factors. Identify each unique factor with the highest power.

Multiply together the ones with the highest exponents for each unique factor.


INTERMEDIATE ALGEBRA

A course designed for students with strong arithmetic skills (without requiring a calculator) and an algebra
background, including solving linear equations in one variable and factoring polynomials. This course will
extend students’ algebra skills to include solving radical, rational, quadratic, & absolute-value equations,
and recognizing relationships between radical expressions and rational exponents . Complex numbers are
introduced in this course as well. Problem solving involving real-life scenarios is an integral part of this
course. In this course, students will enhance their problem-solving abilities and their ability to communicate
concepts of algebra in the language of mathematics , both orally and written.

UNIT TITLES

1. Algebra Review (OPTIONAL)

2. Compound Linear Inequalities in One Variable

3. Linear Equations in Two Variables : Algebra, Geometry, and Modeling

4. Systems of Linear Equations

5. Linear Absolute-Value Equations in One Variable

7. Rational Expressions & Equations

8. Radical Expressions & Equations

11. Introduction to Functions and Function Notation (OPTIONAL)

ASSESSMENT: Please provide a brief description (250 characters maximum) that details how students will be assessed on the course outcomes.

Written Quizzes/Examinations.
Cumulative Final Examination.

*** Complete the following only if course is seeking general education status ***

GENERAL EDUCATION Competencies and Skills *:

Please highlight in green font all Competencies / Skills from the list below that apply to this course. In the box to the right of the
Competency/Skill, enter all specific learning outcome numbers (i.e. 1.1, 2.7, 5.12) that apply.

1. Read with critical comprehension
2. Speak and listen effectively
3. Write clearly and coherently
4. Think creatively, logically, critically, and reflectively
(analyze, synthesize, apply, and evaluate)
2.1, 3.2, 3.5, 3.7, 3.9, 3.10, 3.14, 4.2, 4.4, 4.5, 4.6, 4.7, 5.2, 5.3, 5.4, 6.1, 6.2, 6.3, 6.4, 6.5, 6.6, 6.7. 6.9, 7.2, 7.3, 7.4, 7.5, 7.6, 7.7, 7.8, 7.9, 7.10, 7.11, 7.12, 8.3, 8.5, 8.6, 8.7, 8.9 8.10, 8.11, 8.12, 8.13, 8.14, 9.2, 9.3, 9.5, 9.6, 10.3, 10.4, 10.5, 10.,7, 10.8
5. Demonstrate and apply literacy in its various forms:
(highlight in green ALL that apply)
( 1. technological, 2. informational, 3. mathematical,
4. scientific, 5. cultural, 6. historical, 7. aesthetic and/or
8. environmental )
6. Apply problem solving techniques to real-world
experiences
2.3, 3.8, 3.15, 4.8, 10.9, 10.11
7. Apply methods of scientific inquiry
8. Demonstrate an understanding of the physical and
biological environment and how it is impacted by
human beings
9. Demonstrate an understanding of and appreciation
for human diversities and commonalities
10. Collaborate with others to achieve common goals.
11. Research, synthesize and produce original work
12. Practice ethical behavior
13. Demonstrate self-direction and self motivation
14. Assume responsibility for and understand the impact
of personal behaviors on self and society
15. Contribute to the welfare of the community

* General Education Competencies and Skills endorsed by 󈧉-󈧊 General Education Task Force


The given below is the online LCD calculator for you to do least common denominator calculation with ease. Just enter the data separated by a comma and click on calculate to get the result.

Least Common Denominator: The LCD is the least common multiple of the denominators of a set of fractions. It is the lowest positive integer that is divisible by all the denominators of a set of fractions. It can also be said as least common multiple of denominators of fractions.

LCD Calculation: To calculate the lowest common denominator you can refer the below-given example problem to understand the calculation concept. To make your calculations simple and easy you can try this least common denominator calculator.

Example

Least Common Denominator Calculation

For a set of fractions , 1/2, 5/6 and 3/7.
LCM of denominators 2,6 and 7 is 42 .
Hence LCD is 42 .
The rearranged fraction for the given series is 21/42, 35/42 and 18/42


How to calculate LCD?

The easiest way is, of course, to use our least common denominator calculator above, as it can handle LCD calculations for many denominators at once and you can enter them any way you like - separated by commas, spaces, tabs, new lines, etc. However, if you are not allowed to or don't want to use a calculator and need to do the math by yourself, you can use an algorithm or prime factorization, through which the calculations are simplified.

An algorithm that is simple to follow, but may take a large number of steps(!) is thus: given a finite sequence of positive integers X = (x1, x2, . xn) where n is larger than 1, start the algorithm and on each step m increase the least element of the sequence by adding to it the corresponding element from the initial sequence, resulting in a new sequence in which all elements remain the same, but the lowest value has been increased. Stop when all elements in the sequence are equal, as their common value is the LCD of sequence X (LCD(X)).


8.5: Building Rational Expressions and the LCD

Introduction to Rational Expressions

· Find values of a variable that make a rational expression undefined.

· Simplify rational expressions.

Rational expressions are fractions that have a polynomial in the numerator, denominator, or both. Although rational expressions can seem complicated because they contain variables, they can be simplified in the same way that numeric fractions, also called numerical fractions, are simplified.

Finding the Domain of an Expression

The first step in simplifying a rational expression is to determine the domain, the set of all possible values of the variables. The denominator in a fraction cannot be zero because division by zero is undefined. The reason is that when you multiply the answer 2, times the divisor 3, you get back 6. To be able to divide any number c by zero you would have to find a number that when you multiply it by 0 you would get back c . There are no numbers that can do this, so we say “division by zero is undefined”. In simplifying rational expressions you need to pay attention to what values of the variable(s) in the expression would make the denominator equal zero. These values cannot be included in the domain, so they're called excluded values. Discard them right at the start, before you go any further.

(Note that although the denominator cannot be equivalent to 0, the numerator can—this is why you only look for excluded values in the denominator of a rational expression.)

For rational expressions, the domain will exclude values for which the value of the denominator is 0. Two examples to illustrate finding the domain of an expression are shown below.