Again, since REs behave like fractions, to add or subtract fractions there MUST BE common denominators. Building a common denominator is the biggest challenge on these problems. Use factors from each RE to build the common denominator, then multiply the numerator by the factor that is missing. Watch this idea using simple fractions: $$\frac{5}{12}+\frac{13}{20}$$
| Step 1: Factor denominators: | \(3\cdot 4\) and \(4\cdot 5\) |
| Step 2: Build common denominator: | \(\color{red}{4}\) is a common factor of both \(\color{blue}{12}\) and \(\color{blue}{20}\), so CD is \(4\cdot 3\cdot 5=60\) |
| Step 3: Multiply by what is missing: | \(\large{\frac{5\cdot\color{green}{5}}{60}}\normalsize{+}\large{\frac{13\cdot\color{green}{3}}{60}}\) |
| Step 4: Now add numerators together: | \(\large{\frac{25+39}{60}}\normalsize{\implies}\large{\frac{64}{60}}\normalsize{\implies}\large{\frac{16}{15}}\) |
The above technique also applies to rational expressions:
Add the following rational expressions: \(\large{\frac{3}{x-2}}\normalsize{+} \large{\frac{5}{x}}\normalsize{-}\large{\frac{6}{x^2-2x}}\).
| Step 1: Factor denominators: | \(x-2,\;x\) and \(x(x-2)\) |
| Step 2: Build a common denominator: | CD = \(x(x-2)\) |
| Step 3: Multiply by the missing parts of the CD: | \(\large{\frac{3\cdot\color{blue}{x}}{\color{blue}{x}(x-2)}}\normalsize{+}\large{\frac{5\color{blue}{(x-2)}}{x\color{blue}{(x-2)}}}\normalsize{-}\large{\frac{6}{x^2-2x}}\) |
| Step 4: Combine numerators and simplify: | \(\begin{array}{l} \large{\frac{3x+5(x-2)-6}{x(x-2)}} \normalsize{\implies}\large{\frac{3x+5x-10-6}{x(x-2)}} \\ \normalsize{\implies}\large{\frac{8x-16}{x(x-2)}} \normalsize{\implies}\large{\frac{8(x-2)}{x(x-2)}} \normalsize{\implies}\large{\frac{8}{x}\;\color{green}{\checkmark}}\end{array}\) |
WARNING: When subtracting, the distributive property must be applied to the second RE!!
Subtract the following rational expressions: \(\large{\frac{m+4}{m^2-2m-3}}\normalsize{-}\large{\frac{2m-3}{m^2-5m+6}}\).
| Step 1: Factor denominators: | \((m-3)(m+1)\) and \((m-3)(m-2)\) |
| Step 2: Build a common denominator: | CD = \((m-3)(m+1)(m-2)\) |
| Step 3: Multiply by the missing parts of the CD: | \(\large{\frac{(m+4)\color{blue}{(m-2)}}{(m-3)(m+1)\color{blue}{(m-2)}}}\normalsize{-}\large{\frac{(2m-3)\color{blue}{(m+1)}}{(m-3)\color{blue}{(m+1)}(m-2)}}\) |
| Step 4: Distribute the subtraction, combine numerators and simplify: | \(\begin{array}{l} \large{\frac{(m+4)(m-2)-(2m-3)(m+1)}{(m-3)(m+1)(m-2)}} \\ \normalsize{\implies}\large{\frac{m^2-2m+4m-8-2m^2-2m+3m+3}{(m-3)(m+1)(m-2)}} \\ \normalsize{\implies}\large{\frac{-m^2+3m-5}{(m-3)(m+1)(m-2)}}\;\color{green}{\checkmark} \end{array}\) |
| The numerator cannot be factored, so the answer is in simplest terms. | |
Subtract the following REs: \(\large{\frac{5x}{x^2+xy-2y^2}}\normalsize{-}\large{\frac{3x}{x^2+5xy-6y^2}}\).
| Step 1: Factor denominators: | \((x+2y)(x-y)\) and \((x+6y)(x-y)\) |
| Step 2: Build the CD: | CD = \((x+2y)(x-y)(x+6y)\) |
| Step 3: Multiply by the missing parts of the CD: | \(\large{\frac{5x\color{blue}{(x+6y)}}{(x+2y)(x-y)\color{blue}{(x+6y)}}}\normalsize{-}\large{\frac{3x\color{blue}{(x+2y)}}{\color{blue}{(x+2y)}(x-y)(x+6y)}}\) |
| Step 4: Distribute the subtraction, combine numerators and simplify: | \(\begin{array}{l} \large{\frac{5x^2+30xy-3x^2-6xy}{(x+2y)(x-y)(x+6y)}} \\ \normalsize{\implies}\large{\frac{2x^2+24xy}{(x+2y)(x-y)(x+6y)}} \\ \normalsize{\implies}\large{\frac{2x(x+12y)}{(x+2y)(x-y)(x+6y)}}\;\color{green}{\checkmark}\end{array}\) |
| No factors cancel out, so the answer is in simplest terms. | |