Adding Rational Expressions with Different Denominators

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Wayne State University *

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2011

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Mathematics

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Apr 3, 2024

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docx

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User: A grocery store sells 4000 cases of canned soup per year. By averaging costs to purchase soup and to pay storage costs, the owner has determined that if x cases are ordered at a time, the yearly inventory cost, C, can be modeled by font size decreased by 1 Upper C equals StartFraction 9 comma 000 Over x EndFraction plus 9 x. How many cases should be ordered at a time for yearly inventory costs to be $570? Identify your solutions as points on the graph. The graph is shown on the right. How many cases should be ordered at a time for yearly inventory costs to be $ Julius: To find the number of cases that should be ordered at a time for the yearly inventory costs to be $570, we can use the given model for the yearly inventory cost, $C$, which is: \[C = \frac{9000}{x} + 9x\] Given that $C = 570$, we can substitute this value into the equation and solve for $x$. The equation becomes: \[570 = \frac{9000}{x} + 9x\] Let's solve this equation for $x$ to find out how many cases should be ordered. The solutions to the equation are: [30, 100/3] This means that to minimize the yearly inventory costs to $570, the store should order either 30 cases or approximately 33.33 (100/3) cases at a time. These are the points on the graph where the yearly inventory cost, \ (C\), equals $570. User: (1) Given the expression fraction numerator 1 over denominator x minus 5 end fraction, choose three different values of x and evaluate this expression for those values. Is it possible to choose a value of x for which the value of the expression is greater than 10? If so, what is this value of x? If not, explain why it is not possible. (2) Explain the process of adding rational expressions with different denominators. Give an example. Julius: (1) Let's evaluate the expression $$\frac{1}{x - 5}$$ for three different values of $x$: - For $x = 6$, the expression evaluates to $$\frac{1}{6 - 5} = 1$$. - For $x = 10$, the expression evaluates to $$\frac{1}{10 - 5} = \frac{1}{5} = 0.2$$. - For $x = 4$, the expression evaluates to $$\frac{1}{4 - 5} = \ frac{1}{-1} = -1$$. To find a value of $x$ for which the value of the expression is greater than 10, we need to solve the inequality $ $\frac{1}{x - 5} > 10$$. This inequality simplifies to: \[x - 5 < \frac{1}{10}\] \[x < \frac{1}{10} + 5\] \[x < 5.1\] Therefore, any value of $x$ that is less than 5.1 (but not equal to 5, since that would make the denominator 0 and the expression undefined) will make the expression greater than 10. For example, $x = 5.05$ would satisfy this condition. (2) Adding rational expressions with different denominators involves finding a common denominator, converting each expression to have this common denominator, and then adding the numerators together. The steps are: - Find the least common denominator (LCD) of the denominators. - Rewrite each fraction as an equivalent fraction with the LCD. - Add the numerators of these equivalent fractions. - Simplify the resulting fraction if possible. Example: Add $$\frac{1}{3} + \frac{2}{5}$$. - The LCD of 3 and 5 is 15. - Rewrite each fraction: $$\frac{1}{3} = \frac{1 \times 5}{3 \times 5} = \frac{5} {15}$$ and $$\frac{2}{5} = \frac{2 \times 3}{5 \times 3} = \frac{6}{15}$$. - Add the numerators: $$\frac{5} {15} + \frac{6}{15} = \frac{5 + 6}{15} = \frac{11}{15}$$. So, $$\frac{1}{3} + \frac{2}{5} = \frac{11}{15}$$. User: rewrite so that i could copy ad paste Julius: The solutions to the equation are: - 30 - $\frac{100}{3}$ (approximately 33.33) This means that to minimize the yearly inventory costs to $570, the store should order either 30 cases or approximately 33.33 cases at a time. These are the points on the graph where the yearly inventory cost, $C$, equals $570. For the expression $\frac{1}{x - 5}$, evaluating it for three different values of $x$ and exploring if a value of \ (x\) can make the expression greater than 10: - For $x = 6$, the expression evaluates to 1. - For $x = 10$, the expression evaluates to 0.2. - For $x = 4$, the expression evaluates to -1. To make the expression greater than 10, we need $x$ to be very close to 5 from the right side, such as $x = 5.05$, because as $x$ approaches 5 from the right, the denominator of the expression $\frac{1}{x - 5}$ approaches 0, making the value of the expression very large. Adding rational expressions with different denominators involves finding a common denominator, converting each expression to have this common denominator, and then adding the numerators together. For example, to

add $\frac{1}{3} + \frac{2}{5}$, we find the least common denominator (15), rewrite each fraction ($\frac{5} {15} + \frac{6}{15}$), and then add the numerators to get $\frac{11}{15}$.

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