Problem 27 - Intermediate Algebra

Question

Solve equation. \(\frac{p-1}{2}+1=\frac{3}{p}\)

Step-by-Step Solution

Answer

The solutions are \(p = -3\) and \(p = 2\).

Step 1: Eliminate the Fraction by Finding a Common Denominator

To solve the equation \(\frac{p-1}{2}+1=\frac{3}{p}\), start by eliminating the fractions. Notice that the common denominator between the fractions is \(2p\). Multiply every term by \(2p\) to clear the denominators. This yields:\[2p \left(\frac{p-1}{2}\right) + 2p(1) = 2p \left(\frac{3}{p}\right)\]Simplifying this, we get:\[p(p-1) + 2p = 6\]

Step 2: Expand and Simplify the Equation

Expand and simplify the terms from the equation obtained:\[p^2 - p + 2p = 6\]Combine like terms:\[p^2 + p = 6\]

Step 3: Rearrange Into a Standard Quadratic Equation Form

Rearrange the terms so the equation takes the standard form of a quadratic equation:\[p^2 + p - 6 = 0\]

Step 4: Factor the Quadratic Equation

Now, you need to factor the quadratic equation \(p^2 + p - 6 = 0\). Look for two numbers whose product is \(-6\) and whose sum is \(1\). These numbers are \(3\) and \(-2\). So, the equation can be factored as:\[(p + 3)(p - 2) = 0\]

Step 5: Solve for p

Using the factored form \((p+3)(p-2)=0\), set each factor equal to zero and solve for \(p\):1. \(p + 3 = 0\) implies \(p = -3\)2. \(p - 2 = 0\) implies \(p = 2\)

Step 6: Verify the Solutions

Substitute \(p = -3\) and \(p = 2\) back into the original equation to verify:1. For \(p = -3\): \(\frac{-3 - 1}{2} + 1 = \frac{3}{-3}\), simplify to check if both sides equal.2. For \(p = 2\): \(\frac{2 - 1}{2} + 1 = \frac{3}{2}\), simplify to check if both sides equal.Both solutions satisfy the original equation, so \(p = -3\) and \(p = 2\) are correct.

Key Concepts

Factoring Quadratic Equations

Factoring quadratic equations is a method used to solve equations that are in the form of a polynomial, specifically a quadratic polynomial. A quadratic polynomial is typically given in the form \(ax^2 + bx + c = 0\). The goal is to rewrite the quadratic as a product of two binomials.To factor a quadratic equation:

• You need to find two numbers that multiply to the constant term \(c\) and at the same time add up to the coefficient of the middle term \(b\).
• For example, consider the equation \(p^2 + p - 6 = 0\). Here, we need two numbers whose product is \(-6\) and whose sum is \(1\).
• These numbers are \(3\) and \(-2\), hence the equation can be rewritten as \((p+3)(p-2) = 0\).

Factoring makes it easier to find solutions since each binomial can be set to zero. This leads us to possible solutions for \(p\). This is one of the most intuitive ways to solve quadratic equations, especially when the quadratic can be easily factored, as seen in our example.

Eliminating Fractions

Eliminating fractions from an equation can simplify the process of solving it. Fractions can sometimes make equations look more complex than they actually are. By removing fractions, you make equations easier to handle and solve.Here's how you do it:

• Identify the common denominator of all fractions involved in the equation. For example, in the equation \(\frac{p-1}{2} + 1 = \frac{3}{p}\), the common denominator would be \(2p\).
• Multiply each term in the equation by this common denominator to clear the fractions. This action simplifies the equation to something more familiar, like \(p(p-1) + 2p = 6\).

This method of elimination transforms an equation with fractions into an algebraic equation without fractions, making it much simpler to solve. This step is crucial as it paves the way to solving the equation more efficiently.

Verifying Solutions

Verifying solutions in any mathematical problem is a significant step to ensure accuracy. After solving equations, it is important to substitute the solutions back into the original equation to confirm that they work.To verify solutions:

• Take each potential solution and plug it back into the original equation.
• For example, with the solutions \(p = -3\) and \(p = 2\), substitute them back into the equation \(\frac{p-1}{2} + 1 = \frac{3}{p}\).
• For \(p = -3\), check whether \(\frac{-3 - 1}{2} + 1 = \frac{3}{-3}\) holds true.
• Similarly, verify \(p = 2\) by checking if \(\frac{2-1}{2} + 1 = \frac{3}{2}\) follows.

By ensuring that these substituted values satisfy the original equation, you confirm the correctness of each solution. Always remember, verification is key to accurate problem solving and reinforces your confidence in the solutions obtained.