Example 1

Determine the least common multiple of [latex]x^2-4,\;x^2-4x-5[/latex] and [latex]x^2-7x+10[/latex].

Solution

We factor each expression:

[latex]x^2-4=\color{blue}{(x-2)}(x+2)[/latex]

[latex]x^2-4x-5=\color{red}{(x-5)}(x+1)[/latex]

[latex]x^2-7x+10=\color{red}{(x-5)}\color{blue}{(x-2)}[/latex]

The LCM must contain the highest power of every factor:

LCM [latex]=\color{blue}{(x-2)}(x+2)\color{red}{(x-5)}(x+1)[/latex]

The colors indicate factors that are common to two expressions. Notice that they are written in the LCM only once.

Example 2

Add: [latex]\dfrac{5x-3}{x}+\dfrac{6x+4}{x}[/latex]. State any restrictions on the variable.

Solution

In this example we already have a common denominator, [latex]x[/latex].

[latex]\begin{aligned}\dfrac{5x-3}{x}+\dfrac{6x+4}{x}&=\dfrac{5x-3+6x+4}{x}&&\text{Combine numerators with the common denominator. }x\neq 0\\\\&=\dfrac{11x+1}{x}&&\text{Simplify numerator}\end{aligned}[/latex]

Example 3

Subtract: [latex]\dfrac{4x}{x-1}-\dfrac{x-5}{x-1}[/latex]

Solution

In this example we have a common denominator, [latex]x-1[/latex], with a restriction that [latex]x\neq 1[/latex].

[latex]\begin{aligned}\dfrac{4x}{x-1}-\dfrac{x-5}{x-1}&=\dfrac{4x-(x-5)}{x-1}&&\text{Combine the numerators}\\\\&=\dfrac{4x-x+5}{x-1}&&\text{Subtract using the distributive property}\\\\&=\dfrac{3x+5}{x-1}&&\text{Simplify}\end{aligned}[/latex]

Example 4

Add: [latex]\dfrac{2x^2}{x^2-4}+\dfrac{x}{x-2}[/latex]. State any restrictions on the variable.

Solution

[latex]\begin{aligned}\dfrac{2x^2}{x^2-4}+\dfrac{x}{x-2}&=\dfrac{2x^2}{(x-2)(x+2)}+\dfrac{x}{(x-2)}&&\text{Factor. LCM }=(x-2)(x+2)\;\;x\neq\pm2\\\\&=\dfrac{2x^2}{(x-2)(x+2)}+\dfrac{x\color{blue}{(x+2)}}{(x-2)\color{blue}{(x+2)}}&&\text{Build fractions with a common denominator}\\\\&=\dfrac{2x^2+x(x+2)}{(x-2)(x+2)}&&\text{Combine numerators; use common denominator}\\\\&=\dfrac{2x^2+x^2+2x}{(x-2)(x+2)}&&\text{Distribute }x\text{ into }(x+2)\\\\&=\dfrac{3x^2+2x}{(x-2)(x+2)}&&\text{Simplify}\\\\&=\dfrac{x(3x+2)}{(x-2)(x+2)}&&\text{Factor numerator to see if fraction will simplify}\end{aligned}[/latex]

Example 5

Subtract: [latex]\dfrac{2x^2}{x^2-4x-5}-\dfrac{3x}{x-5}[/latex]. State any restrictions on the variable.

Solution

[latex]\begin{aligned}\dfrac{2x^2}{x^2-4x-5}-\dfrac{3x}{x-5}&=\dfrac{2x^2}{(x-5)(x+1)}-\dfrac{3x}{(x-5)}&&\text{Factor. LCM }=(x-5)(x+1)\;\;x\neq5,\,-1\\\\&=\dfrac{2x^2}{(x-5)(x+1)}-\dfrac{3x\color{blue}{(x+1)}}{(x-5)\color{blue}{(x+1)}}&&\text{Build fractions using LCM as common denominator}\\\\&=\dfrac{2x^2-3x(x+1)}{(x-5)(x+1)}&&\text{Combine fractions}\\\\&=\dfrac{2x^2-3x^2-3x}{(x-5)(x+1)}&&\text{Distribute }-3x\text{ into }(x+1)\\\\&=\dfrac{-x^2-3x}{(x-5)(x+1)}&&\text{Simplify}\\\\&=\dfrac{-x(x+3)}{(x-5)(x+1)}&&\text{Factor to look for common factors}\end{aligned}[/latex]

Example 6

Add the two functions [latex]f(x)=\dfrac{x+2}{x^2-1}[/latex] and [latex]g(x)=\dfrac{x-3}{x^2+x-2}[/latex]. State the domain of the function. State the restrictions and the domain.

Solution

Before we perform the addition, we need to discuss the domain. The denominators of the two functions cannot be zero because it is not defined for a fraction with zero in the denominator. The denominator [latex]x^2-1[/latex] can be factored as [latex](x-1)(x+1)[/latex]. Setting [latex](x-1)(x+1)=0[/latex] results in [latex]x=1, -1[/latex]. Therefore, [latex]x \neq 1, -1[/latex]. The denominator [latex]x^2+x-2[/latex] can be factored as [latex](x+2)(x-1)[/latex]. Setting [latex](x+2)(x-1)=0[/latex] results in [latex]x=-2, 1[/latex]. Therefore, [latex]x \neq -2, 1[/latex]. In summary, the domain of [latex](f+g)(x)[/latex] is all real numbers except for the numbers [latex]1, -1, -2[/latex].

Domain = [latex]\{x\;|\;x\in\mathbb{R},\;x\neq\pm1,\,-2\}[/latex]

[latex]\begin{aligned}(f+g)(x)&=\dfrac{x+2}{x^2-1}+\dfrac{x-3}{x^2+x-2}&&\text{Factor}\\\\&=\dfrac{x+2}{(x-1)(x+1)}+\dfrac{x-3}{(x+2)(x-1)}&&\text{LCD }=(x-1)(x+1)(x+2)\\\\&=\dfrac{x+2}{(x-1)(x+1)}\color{blue}{\times\frac{(x+2)}{(x+2)}}+\dfrac{x-3}{(x+2)(x-1)}\color{blue}{\times \frac{(x+1)}{(x+1)}}&&\text{Build equivalent fractions}\\\\&=\dfrac{x^2+4x+4}{(x-1)(x+1)(x+2)}+\dfrac{x^2-2x-3}{(x-1)(x+1)(x+2)}&&\text{Combine numerators}\\\\&=\dfrac{(x^2+4x+4)+(x^2-2x-3)}{(x+1)(x-1)(x+2)}&&\text{Add like terms}\\\\&=\dfrac{2x^2+2x+1}{(x+1)(x-1)(x+2)}\end{aligned}[/latex]

Example 7

Subtract the function [latex]g(x)=\dfrac{x+3}{3x^2+9x+6}[/latex] from the function [latex]f(x)=\dfrac{x+2}{x^2-2x-8}[/latex]. State the restrictions and the domain.

Solution

Before we perform the addition, we need to discuss the domain first. The denominators of the two functions (or rational expressions) cannot be zero because it is not defined for a fraction with zero in the denominator. The denominator [latex]x^2-2x-8[/latex] can be factored as [latex](x-4)(x+2)[/latex]. Let [latex](x-4)(x+2)=0[/latex]. According to the zero product rule, [latex]x=4, -2[/latex]. Therefore, [latex]x \neq 4, -2[/latex]. The denominator [latex]3x^2+9x+6[/latex] can be factored as [latex]3(x+1)(x+2)[/latex]. Let [latex]3(x+1)(x+2)=0[/latex]. According to the zero product rule, [latex]x=-1, -2[/latex]. Therefore, [latex]x \neq -1, -2[/latex]. In summary, the domain of [latex](f-g)(x)[/latex] is all real numbers except for the numbers [latex]4, -1, -2[/latex].

[latex]\begin{aligned}(f-g)(x)&=\dfrac{x+2}{x^2-2x-8}-\dfrac{x+3}{3x^2+9x+6}\\\\&=\dfrac{x+2}{(x-4)(x+2)}-\dfrac{x+3}{3(x^2+3x+2)}&&\text{Factor}\\\\&=\dfrac{x+2}{(x-4)(x+2)}-\dfrac{x+3}{3(x+1)(x+2)}&&\text{LCD }=3(x+1)(x+2)(x-4)\\\\&=\dfrac{x+2}{(x-4)(x+2)} \times \color{blue}{\frac{3(x+1)}{3(x+1)}}-\dfrac{x+3}{3(x+1)(x+2)} \times \color{blue}{\frac{(x-4)}{(x-4)}}&&\text{Build equivalent fractions}\\\\&=\dfrac{3x^2+9x+6}{3(x-4)(x+2)(x+1)}-\dfrac{x^2-x-12}{3(x-4)(x+2)(x+1)}&&\text{Combine numerators}\\\\&=\dfrac{3x^2+9x+6-(x^2-x-12)}{3(x-4)(x+2)(x+1)}&&\text{Distribute the }-\text{ sign}\\\\&=\dfrac{3x^2+9x+6-x^2+x+12}{3(x-4)(x+2)(x+1)}&&\text{Combine like terms}\\&=\dfrac{2x^2+10x+18}{3(x-4)(x+2)(x+1)}&&\text{Factor}\\\\&=\dfrac{2(x^2+5x+9)}{3(x-4)(x+2)(x+1)}\end{aligned}[/latex]

Example 8

Combine [latex]f(x)+g(x)-h(x)[/latex], when [latex]f(x)=\dfrac{2x}{x-1},\;g(x)=\dfrac{1}{x},\;h(x)=\dfrac{2x-1}{x^2-x}[/latex]. State the domain of [latex]\left(f+g-h\right)(x)[/latex].

Solution

[latex]\begin{aligned}f(x)+g(x)-h(x)&=\dfrac{2x}{x-1}+\dfrac{1}{x}-\dfrac{2x-1}{x^2-x}\\\\&=\dfrac{2x}{(x-1)}+\dfrac{1}{x}-\dfrac{(2x-1)}{x(x-1)}&&\text{Factor. LCM}=x(x-1),\;x\neq0,\,1\\\\&=\color{blue}{\dfrac{x}{x}}\cdot\dfrac{2x}{(x-1)}+\color{blue}{\dfrac{(x-1)}{(x-1)}}\cdot\dfrac{1}{x}-\dfrac{(2x-1)}{x(x-1)}&&\text{Build fractions with common denominator}\\\\&=\dfrac{2x^2+(x-1)-(2x-1)}{x(x-1)}&&\text{Combine numerators}\\\\&=\dfrac{2x^2+x-1-2x+1}{x(x-1)}&&\text{distribute + and – signs}\\\\&=\dfrac{2x^2-x}{x(x-1)}&&\text{Combine like terms}\\\\&=\dfrac{x(2x-1)}{x(x-1)}&&\text{Factor numerator}\\\\&=\dfrac{2x-1}{x-1}&&\text{Cancel: }\dfrac{x}{x}=1\end{aligned}[/latex]

Domain of [latex]\left(f+g-h\right)(x)=\{x\;|\;x\in\mathbb{R},\;x\neq0,\,1\}[/latex]