Problem 35
Question
Contain rational equations with variables in denominators. For each equation, a. write the value or values of the variable that make a denominator zero. These are the restrictions on the variable. b. Keeping the restrictions in mind, solve the equation. $$\frac{2}{3 x}+\frac{1}{4}=\frac{11}{6 x}-\frac{1}{3}$$
Step-by-Step Solution
Verified Answer
The solution to the equation, considering restrictions for \(x\), is \(x = \frac{7}{3}\).
1Step 1: Determine the Restrictions
Set each denominator to zero and solve for \(x\). For this problem, the denominators are \(3x\) and \(6x\). Setting \(3x=0\) and \(6x=0\) respectively, the result yields \(x=0\) in both cases. Thus, \(x\) cannot be zero.
2Step 2: Clear the fractions
Multiply each term in the equation by \((3x)*(4)*(6x)\) to eliminate the denominators. This yields: \((2)(4)(6x) + (3x)(1)(6x) = (11)(4)(3x) - (3x)(6x)(1)\). After simplifying, the equation becomes \(48x + 18x^2 = 132x - 18x^2\).
3Step 3: Simplify the Equation
Bring all terms to one side of the equation to create a quadratic equation: \(48x + 18x^2 - 132x + 18x^2 = 0\). This simplifies to \(36x^2 - 84x = 0\).
4Step 4: Solve the Quadratic Equation
Now factor the quadratic equation: \(6x(6x - 14) = 0\). Set each factor equal to zero to solve for \(x\): if \(6x = 0\), then \(x = 0\) and if \(6x - 14 = 0\), then \(x = \frac{7}{3}\). However, recall the restriction aforementioned that \(x\) cannot be zero; hence, the only solution to the equation is \(x = \frac{7}{3}\).
Key Concepts
Denominator RestrictionsClearing FractionsQuadratic Equations
Denominator Restrictions
When dealing with rational equations—equations that involve fractions with variables in their denominators—it's crucial to identify the denominator restrictions before solving the problem. These restrictions are values that the variable cannot take as they would make the denominator zero, resulting in an undefined expression.
For example, consider the rational equation \[\begin{equation} \frac{2}{3x} + \frac{1}{4} = \frac{11}{6x} - \frac{1}{3} \end{equation}\]\. The denominators, in this case, are \(3x\) and \(6x\), both implying that \(x\) cannot be zero. That's because dividing by zero is undefined in mathematics. Identifying these restrictions is the first and a very important step when solving rational equations, as overlooking them can lead to incorrect solutions that do not hold true within the domain of the equation.
For example, consider the rational equation \[\begin{equation} \frac{2}{3x} + \frac{1}{4} = \frac{11}{6x} - \frac{1}{3} \end{equation}\]\. The denominators, in this case, are \(3x\) and \(6x\), both implying that \(x\) cannot be zero. That's because dividing by zero is undefined in mathematics. Identifying these restrictions is the first and a very important step when solving rational equations, as overlooking them can lead to incorrect solutions that do not hold true within the domain of the equation.
Clearing Fractions
Clearing fractions is a technique used to simplify the process of solving rational equations. This method involves multiplying each term in the equation by a common denominator—effectively the least common multiple of all denominators in the equation. This step eliminates the fractions, making the equation easier to solve.
Continuing with the given example, to clear the fractions from the equation \[\begin{equation} \frac{2}{3x} + \frac{1}{4} = \frac{11}{6x} - \frac{1}{3} \end{equation}\]\, we multiply each term by the least common denominator, which in this case would be \((3x)(4)(6x)\). The result is a simplified equation without fractions. Careful multiplication and combining like terms will lead to a quadratic equation, a polynominal equation where the highest degree of any term is 2, which is the next step in finding the solution.
Continuing with the given example, to clear the fractions from the equation \[\begin{equation} \frac{2}{3x} + \frac{1}{4} = \frac{11}{6x} - \frac{1}{3} \end{equation}\]\, we multiply each term by the least common denominator, which in this case would be \((3x)(4)(6x)\). The result is a simplified equation without fractions. Careful multiplication and combining like terms will lead to a quadratic equation, a polynominal equation where the highest degree of any term is 2, which is the next step in finding the solution.
Quadratic Equations
Once fractions are cleared from the rational equation, you often end up with a quadratic equation. A quadratic equation is a second-degree polynomial of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. Solving quadratic equations can be done through various methods such as factoring, completing the square, or using the quadratic formula.
When we transform our original rational equation to the standard quadratic form, we get \(36x^2 - 84x = 0\). The next step is to factor the quadratic equation to find the potential solutions for \(x\). By factoring out the greatest common factor, we may set each factor equal to zero and solve for \(x\). However, it is vital to recall the denominator restrictions from the first step and check if the solutions satisfy those conditions. If a solution causes a zero denominator, it must be discarded to ensure we have valid solutions within the domain defined by the original equation.
When we transform our original rational equation to the standard quadratic form, we get \(36x^2 - 84x = 0\). The next step is to factor the quadratic equation to find the potential solutions for \(x\). By factoring out the greatest common factor, we may set each factor equal to zero and solve for \(x\). However, it is vital to recall the denominator restrictions from the first step and check if the solutions satisfy those conditions. If a solution causes a zero denominator, it must be discarded to ensure we have valid solutions within the domain defined by the original equation.
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