Problem 27 - Algebra 1: Concepts and Skills

Question

Solve the equation by multiplying each side by the least common denominator. Check your solutions. \(\frac{1}{s}+\frac{s}{s+2}=1\)

Step-by-Step Solution

Answer

The equation has no solutions.

Step 1: Multiply by the Least Common Denominator (LCD)

The LCD for \( s \) and \( s+2 \) is \( s(s+2) \). By multiplying each term on both sides of the equation by the LCD, the equation becomes \( s(s+2) \cdot \frac{1}{s} + s(s+2) \cdot \frac{s}{s+2} = s(s+2) \cdot 1. Simplifying this yields \( s+2 + s^2 = s^2 +2s \).

Step 2: Simplify and Solve

Now you need to combine all the like terms and rearrange the equation: \( s^2 +2s - s - 2 = 0 \). Therefore the equation becomes \( s^2 +s - 2 = 0 \). This equation can be factored to find the values of \( s \), resulting in \( (s-1)(s+2) = 0 \). Setting each factor equal to zero, we find the possible solutions are \( s=1\) or \( s=-2\).

Step 3: Check Your Solution

To verify the solutions, substitute each solution into the original equation: For \( s=1 \), \(\frac{1}{1} + \frac{1}{1+2} = 1. This simplifies to 1 + 1/3 = 4/3\). Since this does not equal 1, s=1 is not a solution. For \( s=-2 \), attempting to substitute \( s=-2 \) into the equation would result in division by zero, which is undefined, thus \( s=-2 \) is also not a solution. Therefore, the equation has no solutions.

Key Concepts

Least Common Denominator

When faced with rational equations, finding a least common denominator (LCD) is crucial. The LCD helps to eliminate the fractions and simplifies the solving process.
First, identify all the different denominators within the equation. In the given problem, the denominators are \( s \) and \( s+2 \).

• To find the LCD, take the product of these distinct denominators: \( s(s+2) \).
• Multiply every term in the equation by this LCD. This clears out the denominators, transforming the equation into a simpler, polynomial form.

This method not only simplifies calculations but also avoids errors commonly made with fractional arithmetic. Once the denominators are cleared, you can focus on solving a regular polynomial equation.

Factoring Quadratics

After obtaining a quadratic equation from a rational one, you'll often need to factor it. Factoring quadratics is about finding two binomials that multiply to give the original quadratic.
In the equation \( s^2 + s - 2 = 0 \), look for two numbers that multiply to \(-2\) (the constant term) and add to \(1\) (the coefficient of \(s\)).

• The numbers \(2\) and \(-1\) satisfy these conditions, so the quadratic factors to \((s-1)(s+2)\).

Setting each factor equal to zero gives potential solutions \(s = 1\) or \(s = -2\). Factorization is an efficient way to solve quadratics without using the quadratic formula. Remember, not every quadratic is factorable with integers, but many are.

Checking Solutions

Checking solutions is an essential step in solving equations, especially rational ones like our example. It ensures the solutions fit the original equation and are valid within its constraints.
For each potential solution, substitute it back into the initial equation:

• First, check \(s = 1\): Substituting \(s=1\) gives \(\frac{1}{1} + \frac{1}{1+2} = \frac{4}{3} \), which is not equal to 1. So \(s=1\) is not a valid solution.
• Next, check \(s = -2\): Substituting \(s=-2\) leads to division by zero, which is undefined. Therefore, \(s=-2\) is not valid too.

This process highlights the importance of substitution in finding valid solutions, as not every outcome from algebraic manipulation will satisfy the original problem's constraints.