Chapter 11: Problem 20
Solve the equation by multiplying each side by the least common denominator. $$\frac{56}{x}=\frac{9-x}{2}$$
Short Answer
Expert verified
The solutions to the equation are x = 7 and x = 16.
Step by step solution
01
Identify the LCD
The LCD of the two fractions is the product of the two denominators, which are 'x' and '2'. So, the least common denominator (LCD) is \(2x\).
02
Multiply each side by the LCD
Now, multiply each side of the equation by the LCD, \(2x\). This gives:\(2x \cdot \frac{56}{x} = 2x \cdot \frac{9-x}{2}\).Simplifying both sides of the equation by cancelling out 'x' on the left side and '2' on the right side,yields to:\(2 \cdot 56 = x \cdot (9 - x)\).
03
Expand and Simplify the Equation
Expand the right side of the equation by multiplying x with each term within the bracket: 112 = 9x - \(x^2\). Rearrange the equation to get a quadratic equation, which looks like this:\(x^2\) - 9x + 112 = 0.
04
Factor the Quadratic Equation
The quadratic can be factored into: (x - 7)(x - 16) = 0.
05
Solve for x
Set each factor equal to zero and solve for x which gives:x - 7 = 0 or x - 16 = 0.From x-7=0, x = 7. From x - 16 = 0, x = 16.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Least Common Denominator
When we encounter an equation with fractions that have different denominators, we can make the process of solving the equation simpler by finding the least common denominator (LCD). The LCD is the smallest non-zero common multiple of the denominators. In simpler terms, it's the smallest number that each denominator can divide into without leaving a remainder.
To solve equations like our given example \(\frac{56}{x}=\frac{9-x}{2}\), we first determine the LCD which in this case, is the product of our two denominators, \(x\) and \(2\), giving us an LCD of \(2x\). We can now multiply each term of the equation by this LCD. This process clears the fractions, transforming our equation into one that's easier to work with and leading us to the next steps: simplifying, factoring, and finding the solution for \(x\).
Remember that when you multiply fractions by their LCD, you're not changing the value of the expression, just its form, which makes solving possible.
To solve equations like our given example \(\frac{56}{x}=\frac{9-x}{2}\), we first determine the LCD which in this case, is the product of our two denominators, \(x\) and \(2\), giving us an LCD of \(2x\). We can now multiply each term of the equation by this LCD. This process clears the fractions, transforming our equation into one that's easier to work with and leading us to the next steps: simplifying, factoring, and finding the solution for \(x\).
Remember that when you multiply fractions by their LCD, you're not changing the value of the expression, just its form, which makes solving possible.
Factoring Quadratics
One efficient method to solve quadratic equations like \(x^2 - 9x + 112 = 0\) is through factoring quadratics. Factoring involves breaking down the quadratic equation into a product of simpler binomials. This only works if the quadratic can be factored, which is often possible when the equation has integer solutions.
To factor a quadratic, we look for two numbers that not only multiply to give us the constant term (in this case, \(112\)) but also add up to the coefficient of the linear term \(x\), which is \(9\). These two numbers for our quadratic equation are \(7\) and \(16\) because \(7 \times 16 = 112\) and \(7 + 16 = 23\), noting a common mistake here—students often look for factors that sum to the linear term instead of the negative of it which in this instance would be \(9\), not \(23\).
The given quadratic factors into \(x - 7\) and \(x - 16\), allowing us to then set each factor equal to zero and solve for \(x\). This brings us to the solutions for the equation.
To factor a quadratic, we look for two numbers that not only multiply to give us the constant term (in this case, \(112\)) but also add up to the coefficient of the linear term \(x\), which is \(9\). These two numbers for our quadratic equation are \(7\) and \(16\) because \(7 \times 16 = 112\) and \(7 + 16 = 23\), noting a common mistake here—students often look for factors that sum to the linear term instead of the negative of it which in this instance would be \(9\), not \(23\).
The given quadratic factors into \(x - 7\) and \(x - 16\), allowing us to then set each factor equal to zero and solve for \(x\). This brings us to the solutions for the equation.
Multiplying Fractions
At the heart of solving equations with fractions often lies the operation of multiplying fractions. Whether combining like terms or clearing fractions from an equation, it's a fundamental skill to master.
To multiply fractions, we simply multiply the numerators (the top numbers) together and the denominators (the bottom numbers) together. For example, multiplying \(\frac{1}{2}\) by \(\frac{3}{4}\) would yield \(\frac{1 \times 3}{2 \times 4} = \frac{3}{8}\). It's quite straightforward.
However, when one of the terms is a variable, you multiply the coefficient (numerical part) by the numerator and the variable by the denominator. If the variable you're multiplying by appears in both the numerator and the denominator, they will cancel out. This is precisely what we did when we multiplied each side of our initial equation by the LCD, \(2x\), facilitating the simplification process. Understanding how to multiply fractions is crucial, as missteps here could derail the entire solution process.
To multiply fractions, we simply multiply the numerators (the top numbers) together and the denominators (the bottom numbers) together. For example, multiplying \(\frac{1}{2}\) by \(\frac{3}{4}\) would yield \(\frac{1 \times 3}{2 \times 4} = \frac{3}{8}\). It's quite straightforward.
However, when one of the terms is a variable, you multiply the coefficient (numerical part) by the numerator and the variable by the denominator. If the variable you're multiplying by appears in both the numerator and the denominator, they will cancel out. This is precisely what we did when we multiplied each side of our initial equation by the LCD, \(2x\), facilitating the simplification process. Understanding how to multiply fractions is crucial, as missteps here could derail the entire solution process.
