Problem 62 - Algebra A Combined Function

Question

Solve each equation. $$ \frac{-4}{a^{2}+2 a-8}+\frac{1}{a^{2}+9 a+20}=\frac{-4}{a^{2}+3 a-10} $$

Step-by-Step Solution

Answer

The solution is \(a = 6\).

Step 1: Factor the Denominators

Factor each of the denominators in the equation: For \(a^2 + 2a - 8\), factor it to \((a + 4)(a - 2)\).For \(a^2 + 9a + 20\), factor it to \((a + 5)(a + 4)\).For \(a^2 + 3a - 10\), factor it to \((a + 5)(a - 2)\).

Step 2: Rewrite the Equation with Factored Denominators

Substitute the factored forms into the equation:\[\frac{-4}{(a + 4)(a - 2)} + \frac{1}{(a + 5)(a + 4)} = \frac{-4}{(a + 5)(a - 2)}\]

Step 3: Find the Common Denominator

Recognize that the common denominator for all fractions is \((a + 4)(a - 2)(a + 5)\).

Step 4: Multiply Each Term by the Common Denominator

Multiply every term by the common denominator to eliminate fractions:\[-4(a + 5) + 1(a - 2) = -4(a + 4)\]

Step 5: Distribute and Simplify

Apply the distributive property:- Expanding gives: \(-4a - 20 + a - 2 = -4a - 16\)- Simplify the left side: \(-3a - 22 = -4a - 16\)

Step 6: Solve for 'a'

Combine like terms and solve for \(a\):- Add \(3a\) to both sides: \(-22 = -a - 16\)- Add 16 to both sides: \(-6 = -a\)- Multiply both sides by \(-1\): \(a = 6\)

Step 7: Verify the Solution

Substitute \(a = 6\) into the original denominators to check for validity:- \(a^2 + 2a - 8 = 36 + 12 - 8 = 40\) (not zero)- \(a^2 + 9a + 20 = 36 + 54 + 20 = 110\) (not zero)- \(a^2 + 3a - 10 = 36 + 18 - 10 = 44\) (not zero)The solution \(a = 6\) does not make any denominator zero, hence it is valid.

Key Concepts

Factoring Quadratics

Factoring quadratics is a method used to express a quadratic equation as the product of its linear factors. A quadratic equation is typically expressed as \(ax^2 + bx + c\). When factoring quadratics:

• We look for two binomials \((px + q)\) and \((rx + s)\) such that the product of these binomials equals the quadratic equation.
• To find these binomials, determine two numbers whose product is \(a \times c\) (the coefficient of \(x^2\) and the constant term) and whose sum is \(b\) (the coefficient of \(x\)).
• If the quadratic is factorable, it can typically be broken down into simple factors, making solving the equation more manageable.

In solving our equation, each quadratic denominator was factored into two binomials. Understanding how to factor is essential for simplifying complex algebraic equations.

Common Denominator

In an equation involving fractions, the common denominator provides a unified base to operate over.

• Think of it like a shared size in which all fractions can be compared and combined.
• To find a common denominator, factor each denominator and then determine the least common multiple (LCM) of these factors.
• This step is crucial for eliminating fractions, as multiplying all terms by the common denominator simplifies the equation to one without division.

In our exercise, the common denominator for the three fractions was found to be \((a + 4)(a - 2)(a + 5)\). Identifying this allows us to clear fractions from the equation by multiplying through by this expression, making the equation simpler to solve.

Distributive Property

The distributive property is a fundamental concept in algebra where multiplication is distributed over addition or subtraction.

• In its simplest form, the property is expressed as \(a(b + c) = ab + ac\). This means you multiply \(a\) by each term inside the parentheses.
• Using the distributive property allows us to expand expressions and remove parentheses, crucial for simplifying and solving equations.
• Applying this property ensures that all components of a term are accounted for, especially when working with polynomial expressions.

In our exercise, once the terms in the equations were multiplied by the common denominator, the distributive property was used to expand terms such as \(-4(a + 5)\), eventually helping to isolate and solve for the variable \(a\).

Like Terms

Combining like terms is a process where we simplify expressions by summing coefficients of terms with the same variable part.

• Terms are considered 'like' if they have the same variable raised to the same power, such as \(x^2\) and \(3x^2\).
• This process reduces the complexity of expressions, making it easier to handle and solve them.
• The goal is to consolidate terms to form a simpler expression, focusing on numerical coefficients.

In our solution steps, combining like terms came into play when simplifying expressions such as \(-4a - 20 + a - 2\). By identifying and merging like terms, it became possible to reach a more straightforward algebraic equation that was easily solvable for \(a\). Emphasizing this technique is vital for efficient algebraic manipulation.