If \(A + B = \rm \frac{x^2 - 8}{x + 2} \ \ and \ A - B = \frac{-x^2 + 2x + 4}{x + 2}\) then what is B equal to ?
Solving for B in a System of Algebraic Equations
The problem asks us to find the value of B given two equations involving A and B in terms of x. The given equations are:
- \(A + B = \frac{x^2 - 8}{x + 2}\)
- \(A - B = \frac{-x^2 + 2x + 4}{x + 2}\)
We have a system of two linear equations with variables A and B. To find the value of B, we can eliminate A from these equations. A common method for this is to subtract the second equation from the first equation.
Step-by-Step Solution for Finding B
Let's subtract Equation 2 from Equation 1:
\((A + B) - (A - B) = \left(\frac{x^2 - 8}{x + 2}\right) - \left(\frac{-x^2 + 2x + 4}{x + 2}\right)\)
Simplify the left side:
\(A + B - A + B = 2B\)
Simplify the right side. Since both fractions have the same denominator \((x+2)\), we can combine the numerators:
\(\frac{(x^2 - 8) - (-x^2 + 2x + 4)}{x + 2}\)
Carefully distribute the negative sign in the numerator:
\(\frac{x^2 - 8 + x^2 - 2x - 4}{x + 2}\)
Combine like terms in the numerator:
\(\frac{(x^2 + x^2) + (-2x) + (-8 - 4)}{x + 2}\)
\(\frac{2x^2 - 2x - 12}{x + 2}\)
So, we have \(2B = \frac{2x^2 - 2x - 12}{x + 2}\).
Now, divide both sides by 2 to find B:
\(B = \frac{1}{2} \left(\frac{2x^2 - 2x - 12}{x + 2}\right)\)
\(B = \frac{2(x^2 - x - 6)}{2(x + 2)}\)
Assuming \(x+2 \neq 0\), we can cancel out the factor of 2 in the numerator and denominator:
\(B = \frac{x^2 - x - 6}{x + 2}\)
Simplifying the Expression for B
Let's try to simplify the rational expression \(\frac{x^2 - x - 6}{x + 2}\) further by factoring the quadratic expression in the numerator, \(x^2 - x - 6\). We look for two numbers that multiply to -6 and add up to -1. These numbers are -3 and +2.
So, the numerator factors as \(x^2 - x - 6 = (x - 3)(x + 2)\).
Substitute this factored form back into the expression for B:
\(B = \frac{(x - 3)(x + 2)}{x + 2}\)
Assuming \(x + 2 \neq 0\), we can cancel the \((x + 2)\) term from the numerator and the denominator:
\(B = x - 3\)
Comparing with the Options
Now, let's check the given options to see which one is equivalent to \(x - 3\).
Option 1: \(\frac{x^2 - 4}{x^2 + 4x + 4} = \frac{(x-2)(x+2)}{(x+2)^2} = \frac{x-2}{x+2}\) (Not \(x-3\))
Option 2: \(\frac{x^2 - 4}{x^2 - 4x + 4} = \frac{(x-2)(x+2)}{(x-2)^2} = \frac{x+2}{x-2}\) (Not \(x-3\))
Option 3: \(\frac{2x^2 - 7x + 3}{2x - 1}\) Let's factor the numerator \(2x^2 - 7x + 3\). We need two numbers that multiply to \(2 \times 3 = 6\) and add up to -7. These numbers are -1 and -6.
\(2x^2 - 7x + 3 = 2x^2 - 6x - x + 3 = 2x(x - 3) - 1(x - 3) = (2x - 1)(x - 3)\)
So, Option 3 is \(\frac{(2x - 1)(x - 3)}{2x - 1}\). Assuming \(2x - 1 \neq 0\), this simplifies to \(x - 3\). This matches our result.
Option 4: \(\frac{2x^2 + 7x - 3}{2x - 1}\) Let's factor the numerator \(2x^2 + 7x - 3\). We need two numbers that multiply to \(2 \times (-3) = -6\) and add up to 7. These numbers are 6 and -1.
\(2x^2 + 7x - 3 = 2x^2 + 6x - x - 3 = 2x(x + 3) - 1(x + 3) = (2x - 1)(x + 3)\)
So, Option 4 is \(\frac{(2x - 1)(x + 3)}{2x - 1}\). Assuming \(2x - 1 \neq 0\), this simplifies to \(x + 3\). (Not \(x-3\))
Therefore, the expression for B is equivalent to Option 3.
| Step | Description | Mathematical Operation |
|---|---|---|
| 1 | Write down the given equations. | \(A + B = \dots, A - B = \dots\) |
| 2 | Subtract the second equation from the first. | \((A+B) - (A-B) = (\dots) - (\dots)\) |
| 3 | Simplify both sides to find 2B. | \(2B = \frac{\text{Simplified Numerator}}{x+2}\) |
| 4 | Divide by 2 to find B. | \(B = \frac{\text{Further Simplified Numerator}}{x+2}\) |
| 5 | Factor the numerator and simplify the rational expression. | \(B = \frac{(x-3)(x+2)}{x+2} = x-3\) |
| 6 | Check which option simplifies to the result. | Compare \(x-3\) with options. |
Revision Table: Key Concepts
| Concept | Explanation | Relevance to Problem |
|---|---|---|
| System of Equations | A set of two or more equations containing two or more variables. | The problem provides two equations with variables A and B. |
| Elimination Method | A method for solving systems of equations by adding or subtracting equations to eliminate one variable. | We subtracted the second equation from the first to eliminate A and solve for B. |
| Rational Expressions | A fraction where the numerator and/or the denominator are polynomials. | The given equations involve rational expressions of x. Simplifying these expressions is crucial. |
| Factoring Polynomials | Breaking down a polynomial into a product of simpler polynomials (factors). | Factoring the numerator of the expression for B and the numerator of Option 3 allowed for simplification. |
Additional Information on Algebraic Simplification
When working with algebraic fractions and rational expressions like those in this problem, simplifying them is a key skill. Simplification often involves factoring polynomials in the numerator and denominator and cancelling out common factors.
Remember that cancelling factors is only valid when the factor is not equal to zero. For instance, when cancelling \((x+2)\) in \(\frac{(x - 3)(x + 2)}{x + 2}\), we assume \(x+2 \neq 0\), meaning \(x \neq -2\). Similarly, when cancelling \((2x - 1)\) in Option 3, we assume \(2x - 1 \neq 0\), meaning \(x \neq \frac{1}{2}\). These values would be excluded from the domain where the simplification holds true.
Polynomial long division is another method that could be used to simplify rational expressions, especially when factoring is not immediately obvious or possible. However, factoring is often quicker when applicable, as shown in the solution for Option 3.
Understanding how to manipulate and simplify algebraic expressions is fundamental for solving more complex problems in algebra and calculus.
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