Problem 27 - Algebra A Combined Function

Question

Solve each equation. $$ \frac{a}{a-6}=\frac{-2}{a-1} $$

Step-by-Step Solution

Answer

The solutions are \( a = -4 \) and \( a = 3 \).

Step 1: Cross-Multiply

To solve the equation \( \frac{a}{a-6} = \frac{-2}{a-1} \), begin by cross-multiplying. This means you multiply the numerator of one fraction by the denominator of the other, giving us: \[ a(a-1) = -2(a-6) \]

Step 2: Expand Both Sides

Next, expand both sides of the equation: - Left side: \( a(a-1) = a^2 - a \)- Right side: \( -2(a-6) = -2a + 12 \)This transforms the equation into: \[ a^2 - a = -2a + 12 \]

Step 3: Move All Terms to One Side

Bring all terms to one side to set the equation to zero. Add \(2a\) and subtract \(12\) from both sides:\[ a^2 - a + 2a - 12 = 0 \]Simplifying gives:\[ a^2 + a - 12 = 0 \]

Step 4: Factor the Quadratic Equation

Factor the quadratic equation \( a^2 + a - 12 = 0 \). Look for two numbers that multiply to \(-12\) and add to \(1\). These numbers are \(4\) and \(-3\). So, the factorization is:\[ (a+4)(a-3) = 0 \]

Step 5: Solve for the Roots

Set each factor equal to zero and solve for \(a\):1. \( a+4 = 0 \) gives \( a = -4 \)2. \( a-3 = 0 \) gives \( a = 3 \)

Step 6: Verify the Solutions

Check each solution in the original equation to ensure they are valid. Substitute \( a = -4 \) and \( a = 3 \) into the original equation:1. For \( a = -4 \): The left side becomes \( \frac{-4}{-10} = \frac{2}{5} \) and the right side becomes \( \frac{-2}{-5} = \frac{2}{5} \), so \( a = -4 \) is valid.2. For \( a = 3 \): The left side becomes \( \frac{3}{-3} = -1 \) and the right side becomes \( \frac{-2}{2} = -1 \), so \( a = 3 \) is valid.

Key Concepts

Cross-Multiplication

Cross-multiplication is a fundamental method used to eliminate fractions when solving rational equations. This technique allows us to confidently manage and simplify complex fractional relationships.
To apply cross-multiplication, you take the numerator of one fraction and multiply it by the denominator of the other fraction. So, for our given equation \( \frac{a}{a-6} = \frac{-2}{a-1} \), cross-multiplying results in the equation \( a(a-1) = -2(a-6) \).

• This step involves rearranging the terms such that the equation no longer contains any fractions.
• Cross-multiplication helps set up the equation for further algebraic manipulation.

After cross-multiplying, you end up with a simpler equation where you can proceed to expand the expressions and continue solving.

Quadratic Equations

Quadratic equations are polynomial equations of degree two, usually in the form \( ax^2 + bx + c = 0 \). Solving these equations often involves transforming the equation to this standard form.
In the given exercise, following cross-multiplication and expansion, the equation \( a^2 - a = -2a + 12 \) is rearranged by moving all terms to one side, resulting in \( a^2 + a - 12 = 0 \). This equation is now in a standard quadratic form, ready to be factored or solved using other methods, such as completing the square or the quadratic formula.

• Recognizing the standard form helps identify the next steps for solving.
• Setting the equation to zero is crucial for applying various solving techniques.

Factoring Polynomials

Factoring polynomials is an essential algebraic process that involves expressing a polynomial as a product of its factors.
Once the equation \( a^2 + a - 12 = 0 \) is in standard form, the next step is to factor it. Here, we look for two numbers that multiply to \(-12\) (the constant term) and add to 1 (the coefficient of the linear term).

• For this quadratic, the numbers are \( 4 \) and \( -3 \).
• Thus, we can factor the equation into \( (a+4)(a-3) = 0 \).

This factorization allows us to set each factor equal to zero and solve for the variable \( a \), simplifying the process to finding the roots of the equation.

Verification of Solutions

Verification of solutions is the final and crucial step in solving equations to ensure the obtained answers are correct.
After solving for \( a \), which gives \( a = -4 \) or \( a = 3 \), it's important to substitute these values back into the original equation \( \frac{a}{a-6} = \frac{-2}{a-1} \) to verify correctness.

• For \( a = -4 \), both sides of the equation simplify to \( \frac{2}{5} \), confirming it as a valid solution.
• For \( a = 3 \), both sides become \( -1 \), confirming this solution as well.

Verification helps confirm accuracy and ensure no solutions make any original denominators zero, which would invalidate the solution.