Multiplying rationals is similar to multiplying fractions with rational terms. It works as such:
NOTE: When adding rationals, we can determine all restrictions simply by identifying those in the denominator!!
Simplify the expression \(\cfrac{k²+k}{k²-k} \times \cfrac{3k-21}{2k²-11k-21}\) and state its restrictions.
First, we can factor the numerator and denominators of each expression. We can take out a common factor from numerator and denominator of the first rational:
\(= \cfrac{\textcolor{red}{k}²+\textcolor{red}{k}}{\textcolor{red}{k}²-\textcolor{red}{k}}\)
\(= \cfrac{\textcolor{red}{k}(k+1)}{\textcolor{red}{k}(k-1)}\)
We can factor the numerator of the second rational by taking out a common factor. We can factor its denominator using Trinomial Factoring:
\(= \cfrac{\textcolor{red}{3}k-\textcolor{red}{21}}{2k²\textcolor{blue}{-11k}-21}\)
\(= \cfrac{\textcolor{red}{3}(k-7)}{2k²\textcolor{blue}{-14k}+\textcolor{blue}{3k}-21}\)
\(= \cfrac{3(k-7)}{(\textcolor{red}{2k}²-\textcolor{red}{14k})+(\textcolor{red}{3}k-\textcolor{red}{21})}\)
\(= \cfrac{3(k-7)}{\textcolor{red}{2k}(k-7)+ \textcolor{red}{3}(k-7)}\)
\(= \cfrac{3(k-7)}{\textcolor{red}{(2k+3)}(k-7)}\)
We can now combine the factored rationals into one expression:
We can now check for restrictions. To do so, we must identify any values that would cause the denominator to be set to 0. First, we can find the restriction for \(k\):
Then, we can find the restriction for \(k - 1\):
\(k - 1 \ne 0\)
\(k \ne 1\)
Next, we can find the restriction for \(2k + 3\):
\(2k + 3 \ne 0\)
\(k \ne \cfrac{-3}{2}\)
After, we can find the restriction for \(k - 7\):
\(k - 7 \ne 0\)
\(k \ne 7\)
We can now multiply the expressions:
Finally, we can cancel out like terms:
\(= \cfrac{3\textcolor{red}{k}(k+1)(\textcolor{green}{k-7})}{\textcolor{red}{k}(k-1)(2k+3)(\textcolor{green}{k-7})}\)
\(= \cfrac{3\cancel{\textcolor{red}{k}}(k+1){(\cancel{\textcolor{green}{k-7}})}}{\cancel{\textcolor{red}{k}}(k-1)(2k+3)(\cancel{\textcolor{green}{k-7}})}\)
\(= \cfrac{3(k+1)}{(k-1)(2k+3)}\)
Therefore, we can determine that the simplified expression is \(\boldsymbol{\cfrac{3(k+1)}{(k-1)(2k+3)}}\).
The restrictions are \(\boldsymbol{k \ne 0, 1, \cfrac{-3}{2}, 7}\).
\(\cfrac{3a²(b+2)}{b(3-a)} \times \cfrac{10b(a-3)}{a³b⁴}\)
The expression cannot be factored any further. As a result, we can determine its restrictions. To do so, we must identify any values that would cause the denominator to be set to \(0\):
First, we can determine the restriction for \(3-a\):
\(3 - a \ne 0\)
\(a ≠ 3\)
Next, we can determine the restriction for \(a\):
Then, we can determine the restriction for \(b\):
We can now multiply the expressions:
\(= \cfrac{\textcolor{red}{3a²(b+2)}}{\textcolor{green}{b(3-a)}} \times \cfrac{\textcolor{red}{10b(a-3)}}{\textcolor{green}{a³b⁴}}\)
\(= \cfrac{\textcolor{red}{3a²(b+2)(10b)(a-3)}}{\textcolor{green}{b(3-a)(a³b⁴)}}\)
Next, we can cancel like terms to get our simplified expression:
\(= \cfrac{3\textcolor{red}{a²}(b+2)(10\textcolor{green}{b})(\textcolor{blue}{a-3})}{\textcolor{green}{b}(\textcolor{blue}{3-a})(\textcolor{red}{a³}b⁴)}\)
\(= \cfrac{3\cancel{\textcolor{red}{a²}}(b+2)(10\cancel{\textcolor{green}{b}})(\cancel{\textcolor{blue}{a-3})}}{(-1)(\cancel{\textcolor{blue}{a-3}})(\cancel{\textcolor{red}{a^3}}.a\cancel{\textcolor{green}b}.b⁴)}\)
\(= \cfrac{-30(b+2)}{ab⁴}\)
Therefore, we can determine that the simplified expression is \(\boldsymbol{\cfrac{-30(b+2)}{ab⁴}}\).
The restrictions are \(\boldsymbol{a \ne 3, 0}\) and \(\boldsymbol{b \ne 0}\).
\(\cfrac{3y²-7y+2}{y²+3y+2} \times \cfrac{8y+8}{4y-8}\)
First, we can factor the numerator and denominators of each expression. We can factor the numerator and denominator of the first rational by using trinomial factoring:
\(= \cfrac{3y²\textcolor{red}{-7y}+2}{y²+\textcolor{blue}{3y}+2}\)
\(= \cfrac{3y²\textcolor{red}{-6y}\textcolor{red}{-1y}+2}{y²+\textcolor{blue}{2y}+\textcolor{blue}{1y}+2}\)
\(= \cfrac{(\textcolor{red}{3y}²-6y)+(\textcolor{red}{-1}y+2)}{(\textcolor{blue}{y}²+2y)+(\textcolor{blue}{1}y+2)}\)
\(= \cfrac{\textcolor{red}{3y}(y-2)\textcolor{red}{-1}(y-2)}{\textcolor{blue}{y}(y+2)+\textcolor{blue}{1}(y+2)}\)
\(= \cfrac{(3y-1)(y-2)}{(y+1)(y+2)}\)
Next, we can factor the numerator and denominator of the second rational by taking out common factors:
\(= \cfrac{\textcolor{red}{8}y+8}{\textcolor{blue}{4}y-8}\)
\(= \cfrac{\textcolor{red}{8}(y+1)}{\textcolor{blue}{4}(y-2)}\)
We can now combine the factored rationals into one expression:
Then, we can check for restrictions. To do so, we must identify any values that would cause the denominator to be set to \(0\).
First, we can find the restriction for \(y + 2\):
\(y + 2 \ne 0\)
\(y \ne -2\)
Next, we can find the restriction for \(y + 1\):
\(y + 1 \ne 0\)
\(y \ne -1\)
Then, we can find the restriction for \(y - 2\):
\(y - 2 \ne 0\)
\(y \ne 2\)
We can now multiply the expressions:
\(= \cfrac{\textcolor{red}{(3y-1)(y-2)}}{\textcolor{green}{(y+1)(y+2)}} \times \cfrac{\textcolor{red}{8(y+1)}}{\textcolor{green}{4(y-2)}}\)
\(= \cfrac{\textcolor{red}{8(3y-1)(y-2)(y+1)}}{\textcolor{green}{4(y+1)(y+2)(y-2)}}\)
Finally, we can cancel like terms to get our simplified expression:
\(= \cfrac{8(3y-1)(\textcolor{red}{y-2})(\textcolor{green}{y+1})}{4(\textcolor{green}{y+1})(y+2)(\textcolor{red}{y-2})}\)
\(= \cfrac{8(3y-1)(\cancel{\textcolor{red}{y-2}})(\cancel{\textcolor{green}{y+1}})}{4(\cancel{\textcolor{green}{y+1}})(y+2)(\cancel{\textcolor{red}{y-2}})}\)
\(= \cfrac{2(3y-1)}{y+2}\)
Therefore, we can determine that the simplified expression is \(\boldsymbol{\cfrac{2(3y-1)}{y+2}}\) and the restrictions are \(\boldsymbol{y \ne -2, -1, 2}\).
Dividing rationals is similar to dividing fractions with rational terms. It works as such:
NOTE: Do NOT flip the second expression before factoring or you won't be able to identify the restrictions!!
Simplify the expression \(\cfrac{6x²-15x}{3x²+5x-12} \div \cfrac{4x²-25}{3x+9}\) and state its restrictions.
First, we can factor the numerators and denominators of each expression. We can factor the numerator of the first rational by taking out a common factor and the its denominator by using Trinomial Factoring:
\(= \cfrac{\textcolor{red}{6x}²-15x}{3x²+\textcolor{blue}{5x}-12}\)
\(= \cfrac{\textcolor{red}{3x}(2x-5x)}{3x²+\textcolor{blue}{9x}\textcolor{blue}{-4x}-12}\)
\(= \cfrac{3x(2x-5x)}{(\textcolor{blue}{3x}²+9x)(\textcolor{blue}{-4x}-12)}\)
\(= \cfrac{3x(2x-5x)}{\textcolor{blue}{3x}(x+3)\textcolor{blue}{-4}(x+3)}\)
\(= \cfrac{3x(2x-5x)}{(3x-4)(x+3)}\)
We can factor the numerator of the second rational by using Difference of Squares and its denominator by taking out a commmon factor:
\(= \cfrac{\textcolor{red}{4x²-25}}{\textcolor{blue}{3}x+9}\)
\(= \cfrac{\textcolor{red}{(2x-5)(2x+5)}}{\textcolor{blue}{3}(x+3)}\)
We can now combine both rationals into a common expression:
Next, we can multiply by reciprocol and cancel like terms:
\(= \cfrac{3x(\textcolor{red}{2x-5x})}{(3x-4)(\textcolor{green}{x+3})} \times \cfrac{3(\textcolor{green}{x+3})}{(\textcolor{red}{2x-5})(2x+5)}\)
\(= \cfrac{9x\cancel{(\textcolor{red}{2x-5x})}\cancel{(\textcolor{green}{x+3})}}{(3x-4)(\cancel{\textcolor{green}{x+3}})\cancel{(\textcolor{red}{2x-5})}(2x+5)}\)
\(= \cfrac{9x}{(3x-4)(2x+5)}\)
Finally, we can check for restrictions. To do so, we must identify any values that would cause the denominator to be set to \(0\). First, we can find the restriction for \(3x-4\):
\(3x-4 \ne 0\)
\(x \ne \cfrac{4}{3}\)
Next, we can find the restriction for \(x+3\):
\(x+3 \ne 0\)
\(x \ne -3\)
Then, we can find the restriction for \(2x-5\):
\(2x-5 ≠ 0\)
\(x \ne \cfrac{5}{2}\)
After, we can find the restriction for \(2x+5\):
\(2x+5 ≠ 0\)
\(x \ne \cfrac{-5}{2}\)
Therefore, we can determine that the simplified expression is \(\boldsymbol{\cfrac{9x}{(3x-4)(2x+5)}}\).
The restrictions are \(\boldsymbol{x \ne \cfrac{4}{3}, -3, \pm \cfrac{5}{2}}\).
\(\cfrac{6(x-1)}{x²} \div \cfrac{3(x-1)}{x(x+2)}\)
The expression cannot be factored any further. As a result, we can multiply by reciprocol and cancel like terms:
\(= \cfrac{6(x-1)}{x²} \div \cfrac{3(x-1)}{x(x+2)}\)
\(= \cfrac{6(\textcolor{red}{x-1})}{\textcolor{green}{x}²} \times \cfrac{\textcolor{green}{x}(x+2)}{3(\textcolor{red}{x-1})}\)
\(= \cfrac{6\cancel{\textcolor{green}{x}}(\cancel{\textcolor{red}{x-1}})(x+2)}{3x\cdot\cancel{\textcolor{green}{x}}(\cancel{\textcolor{red}{x-1}})}\)
\(= \cfrac{2(x+2)}{x}\)
Finally, we can check for restrictions. To do so, we must identify any values that would cause the denominator to be set to \(0\). First, we can find the restriction for \(x\):
Next, we can find the restriction for \(x+2\):
\(x+2 \ne 0\)
\(x \ne -2\)
Then, we can find the restriction for \(x-1\):
\(x-1 \ne 0\)
\(x \ne 1\)
Therefore, we can determine that the simplified expression is \(\boldsymbol{\cfrac{2(x+2)}{x}}\).
The restrictions are \(\boldsymbol{x \ne 0, -2, 1}\).
\(\cfrac{\cfrac{m+n}{5m⁴n}}{\cfrac{m²-n²}{15n²m}}\)
First, we can factor the numerators and denominators of each expression. In this instance, we must first put this expression into proper form:
We can factor the numerator of the second rational using Difference of Squares:
\(= \cfrac{\textcolor{red}{m²-n²}}{15n²m}\)
\(= \cfrac{\textcolor{red}{(m+n)(m-n)}}{15n²m}\)
Next, we can multiply by reciprocol and cancel like terms:
\(= \cfrac{\textcolor{red}{m+n}}{5m⁴n} \times \cfrac{15n²m}{(\textcolor{red}{m+n})(m-n)}\)
\(= \cfrac{15n\cdot \cancel{\textcolor{green}{n}}\cdot \cancel{\textcolor{blue}{m}}(\cancel{\textcolor{red}{m+n}})}{5m³\cdot \cancel{\textcolor{blue}{m}}\cdot \cancel{\textcolor{green}{n}}\cancel{(\textcolor{red}{m+n})}(m-n)}\)
\(= \cfrac{3n}{m³(m-n)}\)
Finally, we can check for restrictions. To do so, we must identify any values that would cause the denominator to be set to \(0\). First, we can find the restriction for \(n\):
Next, we can find the restriction for \(m\):
Then, we can find the restriction for \(m + n\):
\(m+n \ne 0\)
\(m \ne -n \; \text{OR} \; n \ne -m\)
After, we can find the restriction for \(m-n\):
\(m-n \ne 0\)
\(m \ne n \; \text{OR} \; n \ne m\)
Therefore, we can determine that the simplified expression is \(\boldsymbol{\cfrac{3n}{m³(m-n)}}\).
Additionally, the restrictions can be divided into two sets:
For the cylinder:
i. We can identify the cylinder's respective Surface Area and Volume formulas through these respective links:
\(SA = 2πr² + 2πrh\)
\(V = πr²h\)
We can now divide the cylinder's Surface Area formula to its Volume formula to determine the ratio:
\(\cfrac{SA}{V} = \cfrac{2πr² + 2πrh}{πr²h}\)
\(\cfrac{SA}{V} = \cfrac{2\cancel{πr}(r+h)}{\cancel{πr}\cdot rh}\)
\(\cfrac{SA}{V} = \cfrac{2(r+h)}{rh}\)
Therefore, we can determine that the cylinder's Surface Area to Volume ratio is \(\boldsymbol{\cfrac{2(r+h)}{rh}}\).
ii. We can use the same equation we determined in part one, just substituting \(h\) with \(3r\):
\(\cfrac{SA}{V} = \cfrac{2(r+3r)}{r(3r)}\)
\(\cfrac{SA}{V} = \cfrac{2(4\cancel{r})}{3r\cdot \cancel{r}}\)
\(\cfrac{SA}{V} = \cfrac{8}{3r}\)
Therefore, we can determine that the simplified ratio with the adjusted height is \(\boldsymbol{\cfrac{8}{3r}}\).