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Magazine Chapter 7: Problem 27
Solve each equation. $$ \frac{a}{a-6}=\frac{-2}{a-1} $$
Short Answer
Expert verified
The solutions are \( a = -4 \) and \( a = 3 \).
Step by step solution
01
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: \[ a(a-1) = -2(a-6) \]
02
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: \[ a^2 - a = -2a + 12 \]
03
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:\[ a^2 + a - 12 = 0 \]
04
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:\[ (a+4)(a-3) = 0 \]
05
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 \( a = 3 \)
06
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \( a(a-1) = -2(a-6) \).
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.
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.
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).
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 \( (a+4)(a-3) = 0 \).
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.
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.
