Problem 26
Question
Solve the rational equation. Check your solutions. $$-\frac{3 x}{x+2}+\frac{1}{x}=2$$
Step-by-Step Solution
Verified Answer
The solution to the equation \(-\frac{3 x}{x+2}+\frac{1}{x}=2\) is \(x = -1\).
1Step 1: Identify the Least Common Denominator (LCD)
The denominators of the fractions in the equation are \(x+2\) and \(x\). Therefore, the least common denominator (LCD) is \(x(x+2)\).
2Step 2: Multiply through by the LCD and simplify
Multiply each term of the equation by the LCD \(x(x+2)\) to eliminate the denominators. This gives: \(-3x^2 + x(x+2) = 2x(x+2)\). Simplify this to \(-3x^2 + x^2 + 2x = 2x^2 + 4x\).
3Step 3: Simplify the equation and solve
Combine like terms to obtain: \(-2x^2 - 2x = 0\). Factor out an \(x\) to obtain: \(-2x(x+1) = 0\). From this equation, the two possible solutions for \(x\) are \(0\) and \(-1\).
4Step 4: Check the solutions
Substitute each of the two solutions back into the original equation. Upon substituting \(x = 0\), the expression \(-\frac{3 x} {x +2} + \frac{1} {x} = 2\) turns into \(-\frac{3*0}{0 +2} + \frac{1}{0}\), which is not defined because division by zero is undefined, hence it is an extraneous solution and must be discarded. Upon substituting \(x = -1\), the expression turns into \(-\frac{3*(-1)}{-1 +2} + \frac{1}{-1} = 2\), which simplifies to \(2 = 2\). Hence, \(x = -1\) is a valid solution for the equation.
Key Concepts
Least Common DenominatorSimplify EquationCheck Extraneous Solutions
Least Common Denominator
When solving rational equations, finding the least common denominator (LCD) is a crucial step. The LCD is the smallest expression that each denominator in the equation can divide into without leaving a remainder. In our example, the denominators are
\( x + 2 \) and \( x \). To combine these two, we search for a single denominator that includes all factors from both terms, resulting in \( x(x + 2) \).
This approach allows us to multiply each term by the LCD to eliminate the fractions, which makes the equation much simpler to handle. By clearing the denominators, we prevent errors related to fraction operations and set up the equation for easier manipulation and solving. It's similar to finding common ground when trying to negotiate something complex - once you have a common base, everything becomes more straightforward.
\( x + 2 \) and \( x \). To combine these two, we search for a single denominator that includes all factors from both terms, resulting in \( x(x + 2) \).
This approach allows us to multiply each term by the LCD to eliminate the fractions, which makes the equation much simpler to handle. By clearing the denominators, we prevent errors related to fraction operations and set up the equation for easier manipulation and solving. It's similar to finding common ground when trying to negotiate something complex - once you have a common base, everything becomes more straightforward.
Simplify Equation
Simplifying an equation is like tidying up a messy room; it involves reducing clutter to make the problem more manageable. Once we've multiplied through by the LCD, the next step is to combine like terms and organize the equation.
In the given exercise, after clearing the denominators, we are left with a quadratic equation. We combine the like terms to simplify it to \( -2x^2 - 2x = 0 \). We factor this to further simplify the equation to \( -2x(x + 1) = 0 \), which helps us see the possible solutions more clearly: \( x = 0 \) and \( x = -1 \). Simplifying equations is crucial as it lays out the path to solving for the unknowns by making the patterns and solutions more visible.
In the given exercise, after clearing the denominators, we are left with a quadratic equation. We combine the like terms to simplify it to \( -2x^2 - 2x = 0 \). We factor this to further simplify the equation to \( -2x(x + 1) = 0 \), which helps us see the possible solutions more clearly: \( x = 0 \) and \( x = -1 \). Simplifying equations is crucial as it lays out the path to solving for the unknowns by making the patterns and solutions more visible.
Check Extraneous Solutions
After finding potential solutions for a rational equation, it is essential to check for extraneous solutions. These are 'false' solutions that emerge from the steps of solving the equation but don't actually satisfy the original equation.
Checking for extraneous solutions involves substituting the solutions back into the original equation to see if they hold true. For example, in our exercise, when we substitute \( x = 0 \) back into the equation, we find that it leads to a division by zero, which is undefined. Therefore, \( x = 0 \) is an extraneous solution and must be discarded. However, substituting \( x = -1 \) confirms that it is a valid solution because the original equation holds true. This step ensures the accuracy of our work, much like proofreading a story to make sure it makes sense from beginning to end.
Checking for extraneous solutions involves substituting the solutions back into the original equation to see if they hold true. For example, in our exercise, when we substitute \( x = 0 \) back into the equation, we find that it leads to a division by zero, which is undefined. Therefore, \( x = 0 \) is an extraneous solution and must be discarded. However, substituting \( x = -1 \) confirms that it is a valid solution because the original equation holds true. This step ensures the accuracy of our work, much like proofreading a story to make sure it makes sense from beginning to end.
Other exercises in this chapter
Problem 25
Factor to find the \(x\)-intercepts of the parabola described by the quadratic function. Also find the real zeros of the function. $$G(t)=2 t^{2}-t-3$$
View solution Problem 26
Solve the inequality by factoring. $$12 x^{2}+5 x-2 \geq 0$$
View solution Problem 26
In Exercises \(17-40,\) let \(f(x)=-x^{2}+x, g(x)=\frac{2}{x+1},\) and \(h(x)=-2 x+1 .\) Evaluate each of the following. $$(g-h)(3)$$
View solution Problem 26
Identify the underlying basic function, and use transformations of the basic function to sketch the graph of the given function. $$f(x)=\left|\frac{x}{2}\right|
View solution