Example Question - combining fractions
Here are examples of questions we've helped users solve.
Finding h(x) given f(x) and g(x)
The problem is asking to find h(x) = g(x) + f(x) - g(x) · f(x), given that f(x) = x / (x + 3) and g(x) = -3 / (2x + 3). First, let's find g(x) + f(x): g(x) + f(x) = (-3 / (2x + 3)) + (x / (x + 3)). We need to find a common denominator to combine these fractions. The least common denominator (LCD) will be the product of both denominators since they have no factors in common: (2x + 3)(x + 3). Now let's put each fraction over this common denominator: g(x) + f(x) = (-3)(x + 3) / ((2x + 3)(x + 3)) + (x)(2x + 3) / ((2x + 3)(x + 3)). Combine the terms: g(x) + f(x) = [(-3x - 9) + (2x² + 3x)] / ((2x + 3)(x + 3)). Simplify: g(x) + f(x) = (2x² - 3x - 9) / ((2x + 3)(x + 3)). Next, let's find g(x) · f(x): g(x) · f(x) = [(-3 / (2x + 3))] · [(x / (x + 3))]. Multiply the numerators and the denominators: g(x) · f(x) = (-3x) / ((2x + 3)(x + 3)). Now, let's find h(x) by subtracting g(x) · f(x) from g(x) + f(x): h(x) = g(x) + f(x) - g(x) · f(x) = (2x² - 3x - 9) / ((2x + 3)(x + 3)) - (-3x) / ((2x + 3)(x + 3)). Since the denominators are the same, we can combine the numerators: h(x) = (2x² - 3x - 9 + 3x) / ((2x + 3)(x + 3)). Simplify the numerator: h(x) = (2x² - 9) / ((2x + 3)(x + 3)). This is the simplified form of h(x). As for restrictions on the domain, since we cannot divide by zero, both denominators (2x + 3) and (x + 3) must be non-zero. Hence, x cannot be -3/2 or -3. Thus, the domain of h(x) is all real numbers except x ≠ -3/2 and x ≠ -3.
Solving a Fractional Mathematical Equation
This image shows a mathematical equation in fractions that will result in a whole number or a fraction: \[ \frac{13}{100} + \frac{5}{10} - \frac{1}{100} \] To solve this equation, we can simplify the fractions where possible and then combine them: - The fraction \(\frac{5}{10}\) simplifies to \(\frac{1}{2}\) because 5 is half of 10. - The fractions \(\frac{13}{100}\) and \(\frac{1}{100}\) are both over 100, so they can be combined easily. Now let's combine the simplified fractions: \[ \frac{13}{100} - \frac{1}{100} = \frac{13 - 1}{100} = \frac{12}{100} \] \[ \frac{12}{100}\] simplifies to \[\frac{3}{25}\] because both 12 and 100 are divisible by 4. Now we have \(\frac{3}{25} + \frac{1}{2}\). To combine these fractions, we need a common denominator. The smallest common denominator for 25 and 2 is 50. \[ \frac{3}{25} = \frac{3 \times 2}{25 \times 2} = \frac{6}{50} \] \[ \frac{1}{2} = \frac{1 \times 25}{2 \times 25} = \frac{25}{50} \] Now we can add these fractions: \[ \frac{6}{50} + \frac{25}{50} = \frac{6 + 25}{50} = \frac{31}{50} \] So, the result of the equation is \(\frac{31}{50}\), which is a fraction, not a whole number.
Combining Fractions with Common Denominator
To solve the equation given in the image, you need to combine the fractions on the left hand side of the equation. The first step is to find a common denominator for both fractions. The two denominators are \(x - 2\) and \(x^2 - 4\). Notice that \(x^2 - 4\) is a difference of two squares and can be factored as \((x + 2)(x - 2)\). Thus, the least common denominator (LCD) for both fractions is \(x^2 - 4\) or \((x + 2)(x - 2)\). Now write both fractions with the common denominator: \[\frac{x + 1}{x - 2} = \frac{(x + 1)(x + 2)}{x^2 - 4}\] The second fraction is already with the denominator \(x^2 - 4\), so leave it as is: \[\frac{x}{x^2 - 4}\] Now that you have a common denominator, you can combine the numerators: \[\frac{(x + 1)(x + 2) + x}{x^2 - 4}\] Expand the numerator: \[\frac{x^2 + 2x + x + 2 + x}{x^2 - 4}\] Combine like terms: \[\frac{x^2 + 4x + 2}{x^2 - 4}\] Set this equal to the right-hand side of the original equation: \[\frac{x^2 + 4x + 2}{x^2 - 4} = \frac{-1}{x^2 - 4}\] Since the denominators are the same, we can equate the numerators: \[x^2 + 4x + 2 = -1\] Bring the -1 to the left side of the equation: \[x^2 + 4x + 2 + 1 = 0\] Combine like terms: \[x^2 + 4x + 3 = 0\] Now, factor the quadratic equation: \[(x + 3)(x + 1) = 0\] Set each factor equal to zero and solve for x: \[x + 3 = 0 \quad \text{or} \quad x + 1 = 0\] \[x = -3 \quad \text{or} \quad x = -1\] So the solutions for x are -3 and -1. However, we must check for possible extraneous solutions because we're working with rational equations. Since \(x = -1\) makes the original denominators \((x - 2)\) and \((x^2 - 4)\) equal to zero, \(x = -1\) is an extraneous solution and must be discarded. Therefore, the solution to the equation is \(x = -3\).
Combining Fractions to Find Total Pretzels
To solve the question in the image, you need to combine the fractions that represent the amount of pretzels in Bag A and Bag B. Bag A is 1/3 full of pretzels. Bag B is 1/4 full of pretzels. To find the total amount of pretzels when combining both bags, you simply add the two fractions together: 1/3 + 1/4 To add these fractions, you need a common denominator. The smallest common denominator for 3 and 4 is 12. Convert each fraction to an equivalent fraction with the denominator of 12: (1/3) * (4/4) = 4/12 (1/4) * (3/3) = 3/12 Now with like denominators, you can add the numerators: 4/12 + 3/12 = 7/12 The combined amount of pretzels from both bags is 7/12 of a bag. Since 7/12 is more than 1/2 (which would be 6/12), the answer to the question is that you will have more than half a bag of pretzels when you combine the two bags.
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