Problem 32 - Intermediate Algebra

Question

Solve equation. \(\frac{3}{2 x+4}=\frac{x-2}{2}+\frac{x-5}{2 x+4}\)

Step-by-Step Solution

Answer

The solution to the equation is \(x = 3\).

Step 1: Identify the Equation Components

The equation is \( \frac{3}{2x+4} = \frac{x-2}{2} + \frac{x-5}{2x+4} \). Notice that it includes fractions, and the denominators are \(2x+4\) and \(2\).

Step 2: Find a Common Denominator

The common denominator for the fractions is \(2(2x+4)\). We will multiply every term in the equation by \(2(2x+4)\) to eliminate the fractions.

Step 3: Eliminate Denominators

Multiply each term:\\[ \frac{3}{2x+4} \times 2(2x+4) = 3 \times 2 = 6 \] \\[ \frac{x-2}{2} \times 2(2x+4) = (x-2)(2x+4) \] \\[ \frac{x-5}{2x+4} \times 2(2x+4) = 2(x-5) \]

Step 4: Distribute and Simplify

Expand the terms: \- The left side is simply \(6\). \- The right side becomes: \ \[ (x-2)(2x+4) = 2x^2 + 4x - 4x - 8 = 2x^2 - 8 \] \ and \ \[ 2(x-5) = 2x - 10 \]. \Combine all terms: \\(6 = 2x^2 - 8 + 2x - 10\)

Step 5: Rearrange and Simplify the Equation

Combine all terms on one side to form a quadratic equation: \\[ 2x^2 + 2x - 8 - 10 - 6 = 0 \] \Simplify to get: \\[ 2x^2 + 2x - 24 = 0 \] \Divide every term by 2 to simplify: \\[ x^2 + x - 12 = 0 \]

Step 6: Solve the Quadratic Equation

Factor the quadratic equation \(x^2 + x - 12 = 0\). Think of two numbers that multiply to -12 and add to 1. These numbers are 4 and -3. \This gives us: \\[ (x+4)(x-3) = 0 \]

Step 7: Solve for x

Setting each factor equal to zero gives: \- \(x+4=0 \rightarrow x = -4\) \- \(x-3=0 \rightarrow x = 3\)

Step 8: Verify Solutions

Substitute \(x = -4\) and \(x = 3\) back into the original equation to ensure they both work. Only \(x = 3\) satisfies the original equation, as \(x = -4\) introduces division by zero terms. The solution is \(x = 3\).

Key Concepts

Fractions in Algebra

Fractions in algebra are similar to regular fractions, but they often involve variables in the numerator, denominator, or both. Understanding how to handle them is crucial for solving algebraic equations. The key idea when dealing with fractions in an algebra context is to eliminate them by finding a common denominator. This simplifies the problem and makes the equation easier to solve.

• Identify the denominators in the equation. In our exercise, the denominators are \(2x+4\) and \(2\).
• Find the least common denominator (LCD). This is a common multiple of all denominators, and it allows you to clear fractions easily. For the given equation, the LCD is \(2(2x+4)\).
• Multiply each term in the equation by the LCD. This process will clear the fractions and simplify the algebraic manipulation.

Simply multiplying through by the common denominator helps avoid separate fractional operations and provides a clear path to solving the equation further.

Quadratic Equations

Quadratic equations are polynomial equations of degree 2, typically represented as \(ax^2 + bx + c = 0\). These equations have distinct characteristics:

• The power of the variable \(x\) is 2.
• The graph of a quadratic equation forms a parabola.
• Quadratics can have two solutions, one solution, or no real solution based on the discriminant \(b^2 - 4ac\).

Solving a quadratic equation involves finding the values of \(x\) that make the equation true.
In our original exercise, once we eliminate the fractions, the resulting equation simplifies to \(x^2 + x - 12 = 0\). This is clearly a quadratic equation. To solve, we need to find the roots of this equation, which can be achieved through factoring, completing the square, or using the quadratic formula.

Factoring Quadratic Equations

Factoring is a method used to solve quadratic equations by expressing them as a product of two binomials. For an equation like \(x^2 + bx + c = 0\), the challenge is to find two numbers that multiply to \(c\) and add to \(b\).Here are steps to successfully factor the quadratic equation from the exercise:

• Identify the constant term \(c\) and the linear coefficient \(b\). For our equation \(x^2 + x - 12\), \(b = 1\) and \(c = -12\).
• Look for two numbers that multiply to \(-12\) and add to \(1\). These numbers are \(4\) and \(-3\).
• Write the equation as \((x + 4)(x - 3) = 0\).

Using factoring, the solutions \(x = -4\) and \(x = 3\) emerge. However, verifying these solutions in the context of the original equation is imperative, as seen where \(x = 3\) is the valid solution that does not result in division by zero.