Solve the rational equation by multiplying both sides by the LCD. Check your results for extraneous solutions.
$\frac{3}{x^2+5 x+6}+\frac{x-1}{x+2}=\frac{7}{x+3}$
x= $\square$ is a solution.
x= $\square$ is an extraneous solution.
Answer (2)
Factor the denominator and find the LCD: x 2 + 5 x + 6 = ( x + 2 ) ( x + 3 ) , LCD is ( x + 2 ) ( x + 3 ) .
Multiply both sides by the LCD: 3 + ( x − 1 ) ( x + 3 ) = 7 ( x + 2 ) .
Simplify and solve the quadratic equation: x 2 − 5 x − 14 = 0 , which factors to ( x − 7 ) ( x + 2 ) = 0 , giving potential solutions x = 7 and x = − 2 .
Check for extraneous solutions: x = − 2 makes the denominator zero, so it is extraneous. x = 7 is a valid solution. Thus, the solution is x = 7 and the extraneous solution is x = − 2 .
x = 7 is a solution. x = − 2 is an extraneous solution.
Explanation
Understanding the Problem We are given the rational equation x 2 + 5 x + 6 3 + x + 2 x − 1 = x + 3 7 Our goal is to solve for x and identify any extraneous solutions. Extraneous solutions are solutions that satisfy the transformed equation but not the original equation, usually because they make a denominator equal to zero.
Factoring and Finding the LCD First, we factor the quadratic expression in the denominator: x 2 + 5 x + 6 = ( x + 2 ) ( x + 3 ) So the equation becomes ( x + 2 ) ( x + 3 ) 3 + x + 2 x − 1 = x + 3 7 The least common denominator (LCD) is ( x + 2 ) ( x + 3 ) .
Multiplying by the LCD Next, we multiply both sides of the equation by the LCD ( x + 2 ) ( x + 3 ) :
( x + 2 ) ( x + 3 ) ( ( x + 2 ) ( x + 3 ) 3 + x + 2 x − 1 ) = ( x + 2 ) ( x + 3 ) ( x + 3 7 ) This simplifies to 3 + ( x − 1 ) ( x + 3 ) = 7 ( x + 2 ) Expanding the terms, we get 3 + ( x 2 + 3 x − x − 3 ) = 7 x + 14 3 + x 2 + 2 x − 3 = 7 x + 14 x 2 + 2 x = 7 x + 14 x 2 − 5 x − 14 = 0
Solving the Quadratic Equation Now, we solve the quadratic equation x 2 − 5 x − 14 = 0 . We can factor this equation as ( x − 7 ) ( x + 2 ) = 0 So the possible solutions are x = 7 and x = − 2 .
Checking for Extraneous Solutions We need to check for extraneous solutions. The original equation is ( x + 2 ) ( x + 3 ) 3 + x + 2 x − 1 = x + 3 7 If x = − 2 , the denominators x + 2 and ( x + 2 ) ( x + 3 ) become zero, so x = − 2 is an extraneous solution. If x = 7 , none of the denominators are zero, so x = 7 is a valid solution. Therefore, the solution is x = 7 and the extraneous solution is x = − 2 .
Final Answer The solution to the rational equation is x = 7 , and the extraneous solution is x = − 2 .
x = 7 is a solution. x = − 2 is an extraneous solution.
Examples
Rational equations are used in various real-world applications, such as calculating the time it takes to complete a task when multiple people are working together. For example, if one person can complete a job in x hours and another person can complete the same job in y hours, the rational equation x 1 + y 1 = t 1 can be used to find the time t it takes for them to complete the job together. Understanding how to solve rational equations is crucial for solving problems related to rates, work, and proportions in various fields, including engineering, physics, and economics.
The solution to the equation is x = 7 , while x = − 2 is an extraneous solution since it makes a denominator zero. We solved the equation by multiplying through by the least common denominator and simplifying the resulting quadratic equation. Finally, we checked for extraneous solutions by reviewing which values make the denominators zero.
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