Example Question - lcd
Here are examples of questions we've helped users solve.
Finding the Least Common Denominator (LCD) for Rational Equations
To find the least common denominator (LCD) for the given rational equation, we must first determine the denominators of each of the individual fractions in the equation and factor them if necessary. The denominators in the equation are: 1. \(x - 2\) 2. \(x^2 - x - 2\) 3. \(x + 1\) The second denominator \(x^2 - x - 2\) can be factored to identify its components. Factoring \(x^2 - x - 2\) we get: \(x^2 - x - 2 = (x - 2)(x + 1)\) Now we have the fully factored denominators: 1. \(x - 2\) 2. \((x - 2)(x + 1)\) 3. \(x + 1\) The least common denominator (LCD) must contain all factors present in these individual denominators in such a way that each denominator divides the LCD without a remainder. In this case, since \((x - 2)(x + 1)\) already contains the other factors \((x - 2)\) and \((x + 1)\), it serves as the LCD. So the LCD for the given rational equation is: \((x - 2)(x + 1)\)
Finding Least Common Denominator of Fractions
The image shows three fractions: \[ \frac{x}{x - 2}, \quad \frac{3x}{x - 2}, \quad \frac{4}{x + 1}. \] The question asks to find the least common denominator (LCD) of these fractions. The denominators are \( x - 2 \), \( x - 2 \), and \( x + 1 \). To find the LCD, we look for a common denominator that each of these can divide into without leaving a remainder. Since \( x - 2 \) and \( x + 1 \) are different linear factors and neither divides the other, the LCD is simply the product of the unique factors \( x - 2 \) and \( x + 1 \). Therefore, the LCD is: \[ (x - 2)(x + 1) = x^2 + x - 2x - 2 = x^2 - x - 2. \] This is the LCD for the given fractions.
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