Problem 70 - Intermediate Algebra

Question

Solve equation. If a solution is extraneous, so indicate. \(\frac{-5}{c+2}=\frac{3}{2-c}+\frac{2 c}{c^{2}-4}\)

Step-by-Step Solution

Answer

The solution is \(c = \frac{2}{3}\), and it is not extraneous.

Step 1: Identify common denominators

The denominators in the equation are \(c+2\), \(2-c\), and \(c^2-4\). Notice that \(c^2-4\) is a difference of squares and can be factored as \((c+2)(2-c)\). Thus, the common denominator is \((c+2)(2-c)=(c+2)(-(c-2))=-(c^2-4)\).

Step 2: Multiply through by the common denominator

Multiply every term in the equation by the common denominator \(-(c^2-4)\) to eliminate the fractions. The equation becomes: \[-5\cdot(-(c-2))=3\cdot(- (c+2))+ 2c.\] This simplifies to: \[5(c-2) = -3(c+2) + 2c.\]

Step 3: Distribute and simplify

Simplify the equation by distributing the terms:\[5c - 10 = -3c - 6 + 2c.\] Then combine like terms on the right:\[5c - 10 = -c - 6.\]

Step 4: Solve for c

Add \(c\) to both sides to combine \(c\) terms on one side:\[5c + c - 10 = -6.\] This simplifies to:\[6c - 10 = -6.\] Add 10 to both sides to isolate the \(c\) term:\[6c = 4.\] Finally, divide by 6:\[c = \frac{4}{6} = \frac{2}{3}.\]

Step 5: Check for extraneous solutions

Substitute \(c = \frac{2}{3}\) back into the original equation to check if it creates a zero in the denominator or an untrue statement:The denominators are \(c+2=\frac{2}{3}+2=\frac{8}{3}\), \(2-c=2-\frac{2}{3}=\frac{4}{3}\), and \((c+2)(2-c)=\frac{8}{3}\times\frac{4}{3}eq0.\)Since no denominator becomes zero, the solution is valid.

Key Concepts

Equation Solving

When we talk about solving equations, it simply means finding the unknown, or the variable, that makes the equation true. In the exercise we looked at, the task was to find the value of the variable \(c\) that satisfies the equality: \[ \frac{-5}{c+2} = \frac{3}{2-c} + \frac{2c}{c^2-4} \]Basic steps in equation solving usually include identifying like terms, combining them, and performing basic operations such as addition, subtraction, multiplication, or division to isolate the variable. The core idea is to perform the same operations on both sides of the equation until you can solve for the unknown.In our particular exercise, the tricky part is managing the complex fractions and finding a common denominator to simplify the process. This allows each term in the equation to be resolved in sequence, finally revealing the solution for \(c\). The operations must be carefully carried out to maintain equality on both sides.

Extraneous Solutions

Extraneous solutions are solutions that emerge from the process of solving the equation but do not satisfy the original equation itself.Essentially, when you solve an equation, sometimes the steps involved in manipulating the equation can introduce 'false' solutions. These solutions might appear mathematically valid at first but don't hold up when substituted back into the initial equation.To identify these, it's crucial to substitute your potential solutions back into the original equation. As discussed in the exercise, once \(c = \frac{2}{3}\) is derived as a potential solution, plugging it back in helps confirm whether it is indeed a true solution or an extraneous one. If none of the denominators become zero and the equation holds true, then the solution is valid and not extraneous. Be alert for situations where solving the equation might result in denominators equaling zero, as this often leads to extraneous solutions.

Rational Equations

Rational equations are equations that involve fractions, where the numerator and the denominator are polynomials.These equations require special attention due to their inherent potential for providing extraneous solutions. Managing the fractions carefully — particularly by finding a common denominator as shown in our exercise — is key to successfully solving these equations.The original equation given:\[ \frac{-5}{c+2} = \frac{3}{2-c} + \frac{2c}{c^2-4} \]needs factoring and careful manipulation to manage the terms effectively. Rational equations often require you to factor the denominators accurately, as demonstrated here, where recognizing \(c^2-4\) as a difference of squares plays a crucial role.By multiplying through by the common denominator, the equation is transformed into a simpler form without fractions, which can more easily be solved. Always remember to check for extraneous solutions, as manipulating rational equations can sometimes yield solutions that don't actually work when reinserted into the original equation.