This is a process similar to adding and subtracting fractions with rational terms. It works as such:
To help get things started, let's try simplifying the expression \(\cfrac{5}{x^2} + \cfrac{3}{xy}\).
First, we can combine all the terms by finding a common denominator for both rationals. The first term has two \(x\)'s and the second term has an \(x\) and \(y\). This means that the first term needs a \(y\) while the second term needs an \(x\). Therefore, we can multiply the first rational by \(y\) and the second rational by \(x\) to get the common denominator, \(x^2y\):
\(= \cfrac{5}{x^2} + \cfrac{3}{xy}\)
\(= \left(\cfrac{\textcolor{red}{y}}{\textcolor{red}{y}}\right)\left(\cfrac{5}{x^2}\right) + \left(\cfrac{\textcolor{blue}{x}}{\textcolor{blue}{x}}\right)\left(\cfrac{3}{xy}\right)\)
\(= \cfrac{(5)(\textcolor{red}{y})}{(x^2)(\textcolor{red}{y})} + \cfrac{(3)(\textcolor{blue}{x})}{(xy)(\textcolor{blue}{x})}\)
\(= \cfrac{5y}{x^2y} + \cfrac{3x}{x^2y}\)
Now that both rationals have the exact same denominator, we can combine all terms in the numerator under the same denominator:
\(= \cfrac{(5y + 3x)}{x^2y}\)
Therefore, we can determine that \(\cfrac{5}{x^2} + \cfrac{3}{xy}\) simplified is \(\boldsymbol{\cfrac{(5y + 3x)}{x^2y}}\).
Simplify the expression \(\cfrac{6-x}{x²y} - \cfrac{3+y}{xy²} + \cfrac{1}{3x³y²}\) and state its restrictions.
First, we need to manipulate the denominators of each expression to give them a common denominator. In this case, our LCD will be \(3x³y²\). This means that we will need to multiply the first rational by \(3xy\) and the second rational by \(3x^2\):
\(= \cfrac{6-x}{x²y} - \cfrac{3+y}{xy²} + \cfrac{1}{3x³y²}\)
\(= \left(\cfrac{\textcolor{red}{3xy}}{\textcolor{red}{3xy}}\right)\left(\cfrac{6-x}{x²y}\right) - \left(\cfrac{\textcolor{blue}{3x²}}{\textcolor{blue}{3x^2}}\right)\left(\cfrac{3+y}{xy²}\right) + \cfrac{1}{3x³y²}\)
\(= \cfrac{(\textcolor{red}{3xy})(6-x)}{3x³y²} - \cfrac{(\textcolor{blue}{3x²})(3+y)}{3x³y²} + \cfrac{1}{3x³y²}\)
Now that both rationals have the exact same denominator, we can combine all terms in the numerator under the same denominator:
\(= \cfrac{[(3xy)(6-x) - (3x²)(3+y) + 1]}{3x³y²}\)
We can now simplify the numerator by expanding terms:
\(= \cfrac{[(3xy)(\textcolor{red}{6-x}) - (3x²)(\textcolor{green}{3+y}) + 1]}{3x³y²}\)
\(= \cfrac{[(3xy)(\textcolor{red}{6}) + (3xy)(\textcolor{red}{-x}) - [(3x²)(\textcolor{green}{3}) + (3x²)(\textcolor{green}{y})] + \textcolor{black}{1}]}{3x³y²}\)
Now that the numerator is expanded, we can simplify further by collecting like terms:
\(= \cfrac{[18xy - 3x^2y - [9x² + 3x²y] + 1]}{3x³y²}\)
\(= \cfrac{(18xy - \textcolor{green}{3x²y} - \textcolor{green}{3x²y} - 9x² + 1)}{3x³y²}\)
\(= \cfrac{(18xy - 6x²y - 9x² + 1)}{3x³y²}\)
Since we can't factor the expression any further, we can now check for restrictions. To do so, we must identify any values that would cause the denominator to be set to \(0\):
\(x \ne 0\)
\(y \ne 0\)
Therefore, we can determine that the simplified expression is \(\boldsymbol{\cfrac{(18xy - 6x²y - 9x² + 1)}{3x³y²}}\).
The restrictions are \(\boldsymbol{x \neq 0}\) and \(\boldsymbol{y \neq 0}\).
First, we need to manipulate the denominators of each expression to give them a common denominator. In this case, our LCD will be \(12\). This means that we will need to multiply the second rational by \(4\) and the third rational by \(3\):
\(= \cfrac{2x-1}{12} + \cfrac{3x-2}{3} - \cfrac{x+1}{2}\)
\(= \cfrac{2x-1}{12} + \left(\cfrac{\textcolor{red}{4}}{\textcolor{red}{4}}\right)\left(\cfrac{3x-2}{3}\right) - \left(\cfrac{\textcolor{blue}{6}}{\textcolor{blue}{6}}\right)\left(\cfrac{x+1}{2}\right)\)
\(= \cfrac{2x-1}{12} + \cfrac{(\textcolor{red}{4})(3x-2)}{12} - \cfrac{(\textcolor{blue}{6})(x+1)}{12}\)
Now that both rationals have the exact same denominator, we can combine all terms in the numerator under the same denominator:
\(= \cfrac{[2x - 1 + (4)(3x-2) - (6)(x+1)]}{12}\)
We can now simplify the numerator by expanding terms:
\(= \cfrac{[2x - 1 + (4)(\textcolor{red}{3x-2}) - (6)(\textcolor{green}{x+1})]}{12}\)
\(= \cfrac{[2x - 1 + (4)(\textcolor{red}{3x}) + (4)(\textcolor{red}{-2}) - [(6)(\textcolor{green}{x}) + (6)(\textcolor{green}{1})]]}{12}\)
Now that the numerator is expanded, we can simplify further by collecting like terms:
\(= \cfrac{(\textcolor{magenta}{2x} - 1 + \textcolor{magenta}{12x} - 8 - \textcolor{magenta}{6x} - 6)}{12}\)
\(= \cfrac{(\textcolor{magenta}{8x} - 15)}{12}\)
Since we can't factor the expression any further, we can now check for restrictions. However, since the denominator is a whole number and not a term, there are no restrictions.
Therefore, we can determine that the simplified expression is \(\boldsymbol{\cfrac{(8x - 15)}{12}}\) and there are no restrictions.
Simplify the expression \(\cfrac{5x+2}{25x²-1} - \cfrac{3x-1}{25x²+10x+1}\) and state its restrictions.
First, we can factor the denominators to find the LCD:
\(= \cfrac{5x+2}{\textcolor{red}{25x²-1}} - \cfrac{3x-1}{\textcolor{blue}{25x²+10x+1}}\)
\(= \cfrac{5x+2}{\textcolor{red}{(5x-1)(5x+1)}} - \cfrac{3x-1}{\textcolor{blue}{(5x+1)(5x+1)}}\)
We can then manipulate the denominators of each expression to give them a common denominator. In this case, our LCD will be \((5x+1)(5x+1)(5x-1)\). This means that we will need to multiply the first rational by \(5x + 1\) and the second rational by \(5x - 1\):
\(= \left(\cfrac{\textcolor{purple}{5x+1}}{\textcolor{purple}{5x+1}}\right)\left(\cfrac{5x+2}{(5x-1)(5x+1)}\right) - \left(\cfrac{\textcolor{magenta}{5x-1}}{\textcolor{magenta}{5x-1}}\right)\left(\cfrac{3x-1}{(5x+1)(5x+1)}\right)\)
\(= \cfrac{(\textcolor{purple}{5x+1})(5x+2)}{(\textcolor{purple}{5x+1})(5x+1)(5x-1)} - \cfrac{(\textcolor{magenta}{5x-1})(3x-1)}{(\textcolor{magenta}{5x-1})(5x+1)(5x+1)}\)
Now that both rationals have the exact same denominator, we can combine all terms in the numerator under the same denominator:
\(= \cfrac{[(5x+1)(5x+2) - (5x - 1)(3x - 1)]}{(5x+1)(5x+1)(5x-1)}\)
We can simplify the numerator by expanding terms:
\(= \cfrac{[(\textcolor{red}{5x}+\textcolor{green}{1})(\textcolor{red}{5x}+\textcolor{green}{2}) - (\textcolor{red}{5x} - \textcolor{green}{1})(\textcolor{red}{3x} \textcolor{green}{-1})]}{(5x+1)(5x+1)(5x-1)}\)
\(= \cfrac{[(\textcolor{red}{5x})(\textcolor{red}{5x}) + (\textcolor{red}{5x})(\textcolor{green}{2}) + (\textcolor{green}{1})(\textcolor{red}{5x}) + (\textcolor{green}{1})(\textcolor{green}{2}) - [(\textcolor{red}{5x})(\textcolor{red}{3x}) + (\textcolor{red}{5x})(\textcolor{green}{-1}) + (\textcolor{green}{-1})(\textcolor{red}{3x}) + (\textcolor{green}{-1})(\textcolor{green}{-1})]}{(5x+1)(5x+1)(5x-1)}\)
\(= \cfrac{[\textcolor{red}{25x²} + \textcolor{red}{10}\textcolor{green}{x} + \textcolor{red}{5}\textcolor{green}{x} + \textcolor{green}{2} - (\textcolor{red}{15x²} - \textcolor{red}{5}\textcolor{green}{x} - \textcolor{red}{3}\textcolor{green}{x} + \textcolor{green}{1})]}{(5x+1)(5x+1)(5x-1)}\)
Now that the numerator is expanded, we can simplify further by collecting like terms:
\(= \cfrac{(\textcolor{red}{25x²} - \textcolor{red}{15x²} + \textcolor{red}{15}\textcolor{green}{x} + \textcolor{red}{5}\textcolor{green}{x} + \textcolor{red}{3}\textcolor{green}{x} \textcolor{green}{- 1} + \textcolor{green}{2})}{(5x+1)(5x+1)(5x-1)}\)
\(= \cfrac{(10x² + 23x + 1)}{(5x+1)(5x+1)(5x-1)}\)
Since we can't factor the expression any further, 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 \(5x - 1\):
\(5x - 1 \ne 0\)
\(x \ne \cfrac{1}{5}\)
Next, we can find the restriction for \(5x + 1\):
\(5x + 1 \ne 0\)
\(x \ne -\cfrac{1}{5}\)
Therefore, we can determine that the simplified expression is \(\boldsymbol{\cfrac{(10x² + 23x + 1)}{(5x+1)(5x+1)(5x-1)}}\).
The restrictions are \(\boldsymbol{x \ne \pm \cfrac{-1}{5}}\).
First, we can factor the denominators to find the LCD. We can remove a common factor from the first rational:
\(= \cfrac{x²-x}{2\textcolor{red}{x}²-3\textcolor{red}{x}}\)
\(= \cfrac{x²-x}{\textcolor{red}{x}(2x-3)}\)
We can factor the denominator for the second rational through Trinomial Factoring:
\(= \cfrac{3x²+1-4x}{3\textcolor{red}{-5x}+2x²}\)
\(=-\cfrac{3x²+1-4x}{2x²\textcolor{red}{-2x}\textcolor{red}{-3x}+3}\)
\(=-\cfrac{3x²+1-4x}{(2x²\textcolor{red}{-2x})(\textcolor{red}{-3x}+3)}\)
\(= -\cfrac{3x²+1-4x}{\textcolor{red}{2x}\textcolor{blue}{(x-1)}\textcolor{red}{-3}\textcolor{blue}{(x-1)}}\)
\(=-\cfrac{3x²+1-4x}{\textcolor{red}{(2x-3)}\textcolor{blue}{(x-1)}}\)
We can now put the factored rationals together:
Next, we can manipulate the denominators of each expression to give them a common denominator. In this case, our LCD will be \(x(x-1)(2x-3)\). This means that we will need to multiply the first rational by \(x-1\) and the second rational by \(x\):
\(= \left(\cfrac{\textcolor{red}{x-1}}{\textcolor{red}{x-1}}\right)\left(\cfrac{x²-x}{x(2x-3)}\right) - \left(\cfrac{ \textcolor{blue}{x}}{\textcolor{blue}{x}}\right)\left(\cfrac{3x²+1-4x}{(x-1)(2x-3)}\right)\)
\(= \cfrac{(\textcolor{red}{x-1})(x²-x)}{(\textcolor{red}{x-1})x(2x-3)} - \cfrac{(\textcolor{blue}{x})(3x²+1-4x)}{(\textcolor{blue}{x})(x-1)(2x-3)}\)
Now that both rationals have the exact same denominator, we can combine all terms in the numerator under the same denominator:
We can simplify the numerator by expanding terms:
\(= \cfrac{[(\textcolor{red}{x}-\textcolor{green}{1})(\textcolor{red}{x²}-\textcolor{green}{x}) - (\textcolor{red}{x})(\textcolor{red}{3x²}+\textcolor{green}{1}\textcolor{blue}{-4x})}{x(x-1)(2x-3)}\)
\(= \cfrac{[(\textcolor{red}{x})(\textcolor{red}{x²}) + (\textcolor{red}{x})(\textcolor{green}{-x}) + (\textcolor{green}{-1})(\textcolor{red}{x²}) + (\textcolor{green}{-1})(\textcolor{green}{-x}) - [(\textcolor{red}{x})(\textcolor{red}{3x²}) + (\textcolor{red}{x})(\textcolor{green}1) + (\textcolor{red}{x})(\textcolor{blue}{-4x})]]}{x(x-1)(2x-3)}\)
\(= \cfrac{[x³ - x² - x² + x - (3x³ + x - 4x²)]}{x(x-1)(2x-3)}\)
Now that the numerator is expanded, we can simplify further by collecting like terms:
\(= \cfrac{[\textcolor{red}{x³} - \textcolor{green}{x²} - \textcolor{green}{x²} + \textcolor{blue}{x} - (\textcolor{red}{3x³} + \textcolor{blue}{x} - \textcolor{green}{4x²})]}{x(x-1)(2x-3)}\)
\(= \cfrac{(\textcolor{red}{x³} - \textcolor{red}{3x³} - \textcolor{green}{2x²} + \textcolor{green}{4x^2} + \textcolor{blue}{x} - \textcolor{blue}{x})}{x(x-1)(2x-3)}\)
\(= \cfrac{(\textcolor{red}{-2x³} + \textcolor{green}{2x²})}{x(x-1)(2x-3)}\)
We can factor the numerator further which will allow us to cancel terms:
\(= \cfrac{(-2\textcolor{red}{x}³ + 2\textcolor{red}{x}²)}{\textcolor{red}{x}(x-1)(2x-3)}\)
\(= \cfrac{\textcolor{red}{x}(-2x^2 + 2x)}{\textcolor{red}{x}(x-1)(2x-3)}\)
\(= \cfrac{\cancel{\textcolor{red}{x}}[-2x + \cancel{(\textcolor{green}{x - 1})}]}{\cancel{\textcolor{red}{x}}(\cancel{\textcolor{green}{x-1}})(2x-3)}\)
\(= \cfrac{-2x}{2x-3}\)
Finally, we can check for restrictions. To do so, we must identify any values that would cause the denominator to be set to \(0\). In this case we can find the restriction for \(2x - 3\) by setting it equal to \(0\):
\(2x - 3 \ne 0\)
\(x \ne \cfrac{3}{2}\)
Therefore, we can determine that the simplified expression is \(\boldsymbol{\cfrac{{-2x}}{2x-3}}\) and the restriction is \(\boldsymbol{x \ne \cfrac{3}{2}}\).