Mastering Rational Expression Subtraction

by Alex Johnson 42 views

When we dive into the world of algebra, one of the fundamental skills we encounter is the subtraction of rational expressions. This process might seem a bit daunting at first, especially when the denominators aren't immediately alike. But fear not! By understanding a few key principles and practicing them, you'll be confidently simplifying these expressions in no time. Let's break down the process, using the example you provided as our guide: 2x2βˆ’36βˆ’1x2+6x\frac{2}{x^2-36}-\frac{1}{x^2+6 x}. Our goal is to find the values of a, b, c, d, e, f, and g that correctly complete the subtraction. This involves finding a common denominator, adjusting the numerators, and then performing the subtraction.

First, we need to factor the denominators to identify the least common denominator (LCD). The first denominator, x2βˆ’36x^2-36, is a difference of squares, which factors into (x+6)(xβˆ’6)(x+6)(x-6). The second denominator, x2+6xx^2+6x, can be factored by taking out a common factor of xx, resulting in x(x+6)x(x+6). Now, looking at these factored forms, we can see the parts that are common and the parts that are unique. The common factor is (x+6)(x+6). The unique factors are (xβˆ’6)(x-6) and xx. To find the LCD, we take each unique factor and each common factor raised to its highest power. In this case, the LCD is x(x+6)(xβˆ’6)x(x+6)(x-6). This is the crucial first step that sets the stage for all subsequent operations. Without a correctly identified LCD, any further steps will be based on a flawed foundation, leading to an incorrect final answer. It's like trying to build a house without a solid foundation; everything that follows will be unstable. So, take your time with this initial factoring and LCD identification. It’s an investment that pays off immensely in simplifying the rest of the problem.

Step-by-Step Subtraction

Now that we have our denominators factored as (x+6)(xβˆ’6)(x+6)(x-6) and x(x+6)x(x+6), we can rewrite the original expression with these factored forms: 2(x+6)(xβˆ’6)βˆ’1x(x+6)\frac{2}{(x+6)(x-6)}-\frac{1}{x(x+6)}. Our next move is to make the denominators of both fractions identical to the LCD we found, which is x(x+6)(xβˆ’6)x(x+6)(x-6). For the first fraction, 2(x+6)(xβˆ’6)\frac{2}{(x+6)(x-6)}, we are missing the factor of xx in the denominator. To incorporate it, we multiply both the numerator and the denominator by xx. This gives us 2xx(x+6)(xβˆ’6)\frac{2x}{x(x+6)(x-6)}. For the second fraction, 1x(x+6)\frac{1}{x(x+6)}, we are missing the factor of (xβˆ’6)(x-6). So, we multiply both the numerator and the denominator by (xβˆ’6)(x-6), resulting in 1(xβˆ’6)x(x+6)(xβˆ’6)\frac{1(x-6)}{x(x+6)(x-6)}, which simplifies to xβˆ’6x(x+6)(xβˆ’6)\frac{x-6}{x(x+6)(x-6)}. Now that both fractions share the same denominator, the subtraction becomes much more manageable. This process of adjusting the numerators and denominators to match the LCD is fundamental to adding and subtracting fractions, whether they involve simple numbers or complex algebraic expressions.

With both fractions now having the common denominator x(x+6)(xβˆ’6)x(x+6)(x-6), we can proceed with the subtraction of the numerators. Our expression now looks like this: 2xx(x+6)(xβˆ’6)βˆ’xβˆ’6x(x+6)(xβˆ’6)\frac{2x}{x(x+6)(x-6)}-\frac{x-6}{x(x+6)(x-6)}. We can combine these into a single fraction by subtracting the second numerator from the first: 2xβˆ’(xβˆ’6)x(x+6)(xβˆ’6)\frac{2x - (x-6)}{x(x+6)(x-6)}. It's extremely important here to pay close attention to the subtraction of the entire numerator of the second fraction. The parentheses around (xβˆ’6)(x-6) are critical because the minus sign applies to both terms within the expression. Distributing the negative sign, we get 2xβˆ’x+62x - x + 6. Combining like terms in the numerator, 2xβˆ’x2x - x becomes xx, so the numerator simplifies to x+6x+6. Therefore, our expression becomes x+6x(x+6)(xβˆ’6)\frac{x+6}{x(x+6)(x-6)}.

Simplifying the Result

After performing the subtraction, we have the expression x+6x(x+6)(xβˆ’6)\frac{x+6}{x(x+6)(x-6)}. The final step in simplifying rational expressions is to look for any common factors in the numerator and the denominator that can be canceled out. In this case, we see that both the numerator and the denominator have a factor of (x+6)(x+6). We can cancel this common factor, provided that x+6eq0x+6 eq 0 (which means xeqβˆ’6x eq -6). After cancellation, the expression simplifies to 1x(xβˆ’6)\frac{1}{x(x-6)}. This simplified form represents the result of the original subtraction. It's essential to remember the restrictions on the variable that arise from the original denominators and during the cancellation process. These restrictions ensure that the simplified expression is equivalent to the original one for all valid values of xx. The ability to identify and cancel these common factors is a hallmark of proficiency in algebraic manipulation and is key to understanding more complex mathematical concepts.

Identifying the Values (a, b, c, d, e, f, g)

Let's go back to the original problem statement and the equation provided: 2x2βˆ’36βˆ’1x2+6x=2(x+6)(xβˆ’6)βˆ’1x(x+a)=bxβˆ’(cx+d)(x+6)(xβˆ’6)x(x+a)=ex+fgx(x+6)(xβˆ’6)\frac{2}{x^2-36}-\frac{1}{x^2+6 x} =\frac{2}{(x+6)(x-6)}-\frac{1}{x(x+a)} =\frac{bx-(cx+d)}{(x+6)(x-6)x(x+a)} = \frac{ex+f}{gx(x+6)(x-6)}.

We have already factored the denominators: x2βˆ’36=(x+6)(xβˆ’6)x^2-36 = (x+6)(x-6) and x2+6x=x(x+6)x^2+6x = x(x+6).

Comparing 1x2+6x\frac{1}{x^2+6 x} with 1x(x+a)\frac{1}{x(x+a)}, we can see that a=6a = 6. So, the first equality holds: 2(x+6)(xβˆ’6)βˆ’1x(x+6)\frac{2}{(x+6)(x-6)}-\frac{1}{x(x+6)}.

Now, let's look at the process of finding the common denominator and combining the fractions. Our LCD is x(x+6)(xβˆ’6)x(x+6)(x-6).

So, 2(x+6)(xβˆ’6)=2xx(x+6)(xβˆ’6)\frac{2}{(x+6)(x-6)} = \frac{2x}{x(x+6)(x-6)} and 1x(x+6)=xβˆ’6x(x+6)(xβˆ’6)\frac{1}{x(x+6)} = \frac{x-6}{x(x+6)(x-6)}.

Therefore, 2x2βˆ’36βˆ’1x2+6x=2xβˆ’(xβˆ’6)x(x+6)(xβˆ’6)=2xβˆ’x+6x(x+6)(xβˆ’6)=x+6x(x+6)(xβˆ’6)\frac{2}{x^2-36}-\frac{1}{x^2+6 x} = \frac{2x - (x-6)}{x(x+6)(x-6)} = \frac{2x-x+6}{x(x+6)(x-6)} = \frac{x+6}{x(x+6)(x-6)}.

Now, let's match this with the given structure bxβˆ’(cx+d)(x+6)(xβˆ’6)x(x+a)=ex+fgx(x+6)(xβˆ’6)\frac{bx-(cx+d)}{(x+6)(x-6)x(x+a)} = \frac{ex+f}{gx(x+6)(x-6)}.

From our subtraction, the numerator is x+6x+6 and the denominator is x(x+6)(xβˆ’6)x(x+6)(x-6).

Let's examine the intermediate step bxβˆ’(cx+d)(x+6)(xβˆ’6)x(x+a)\frac{bx-(cx+d)}{(x+6)(x-6)x(x+a)}.

We know a=6a=6. So the denominator is (x+6)(xβˆ’6)x(x+6)(x+6)(x-6)x(x+6), which seems to have an extra (x+6)(x+6) term compared to our LCD x(x+6)(xβˆ’6)x(x+6)(x-6). This suggests a potential misunderstanding in the provided intermediate steps or a typo in the question's structure. However, if we interpret the structure as aiming to represent the combined numerator and the full product of the denominators before simplification, we can try to align it.

Let's assume the intention was: 2x2βˆ’36βˆ’1x2+6x=2(x+6)(xβˆ’6)βˆ’1x(x+6)\frac{2}{x^2-36}-\frac{1}{x^2+6 x} = \frac{2}{(x+6)(x-6)}-\frac{1}{x(x+6)}

To get a common denominator of x(x+6)(xβˆ’6)x(x+6)(x-6): =2xx(x+6)(xβˆ’6)βˆ’xβˆ’6x(x+6)(xβˆ’6)= \frac{2x}{x(x+6)(x-6)} - \frac{x-6}{x(x+6)(x-6)} =2xβˆ’(xβˆ’6)x(x+6)(xβˆ’6)= \frac{2x - (x-6)}{x(x+6)(x-6)} =2xβˆ’x+6x(x+6)(xβˆ’6)= \frac{2x - x + 6}{x(x+6)(x-6)} =x+6x(x+6)(xβˆ’6)= \frac{x+6}{x(x+6)(x-6)}

And then simplifying this gives 1x(xβˆ’6)\frac{1}{x(x-6)}.

Let's try to fit this into the given variables. The question implies: 2(x+6)(xβˆ’6)βˆ’1x(x+a)=bxβˆ’(cx+d)(x+6)(xβˆ’6)x(x+a)=ex+fgx(x+6)(xβˆ’6)\frac{2}{(x+6)(x-6)}-\frac{1}{x(x+a)} = \frac{bx-(cx+d)}{(x+6)(x-6)x(x+a)} = \frac{ex+f}{gx(x+6)(x-6)}

From 1x2+6x=1x(x+a)\frac{1}{x^2+6x} = \frac{1}{x(x+a)}, we get x2+6x=x(x+a)x^2+6x = x(x+a), so x(x+6)=x(x+a)x(x+6) = x(x+a), which means a=6a=6.

Now, consider the numerator part: bxβˆ’(cx+d)bx-(cx+d). This is supposed to be the result of 2xβˆ’(xβˆ’6)2x - (x-6), which is x+6x+6. So we need bxβˆ’(cx+d)=x+6bx-(cx+d) = x+6. This implies b=1b=1, c=1c=1, and d=βˆ’6d=-6. However, the structure is bxβˆ’(cx+d)bx - (cx+d), so bxβˆ’cxβˆ’dbx-cx-d. This means bβˆ’c=1b-c = 1 and βˆ’d=6-d=6, so d=βˆ’6d=-6. If we use b=1b=1 and c=1c=1, then 1βˆ’1=0eq11-1 = 0 eq 1. This structure bxβˆ’(cx+d)bx-(cx+d) doesn't directly match 2xβˆ’(xβˆ’6)2x-(x-6) unless we rearrange.

Let's assume the structure bxβˆ’(cx+d)(x+6)(xβˆ’6)x(x+a)\frac{bx-(cx+d)}{(x+6)(x-6)x(x+a)} represents the subtraction before simplifying the numerator, where the common denominator is written as the product of the individual factored denominators. In that case, the common denominator should be x(x+6)(xβˆ’6)x(x+6)(x-6). The given denominator is (x+6)(xβˆ’6)x(x+a)(x+6)(x-6)x(x+a). Since a=6a=6, this is (x+6)(xβˆ’6)x(x+6)(x+6)(x-6)x(x+6). This is not the LCD. This suggests there might be a typo in the problem's intermediate representation.

However, if we strictly follow the structure and assume the denominator is indeed the product of the original factored denominators, and the numerator is the result of the subtraction after finding a common denominator:

We have 2(x+6)(xβˆ’6)βˆ’1x(x+6)\frac{2}{(x+6)(x-6)}-\frac{1}{x(x+6)}.

To get the common denominator x(x+6)(xβˆ’6)x(x+6)(x-6), we did: 2xx(x+6)(xβˆ’6)βˆ’xβˆ’6x(x+6)(xβˆ’6)=2xβˆ’(xβˆ’6)x(x+6)(xβˆ’6)=x+6x(x+6)(xβˆ’6)\frac{2x}{x(x+6)(x-6)} - \frac{x-6}{x(x+6)(x-6)} = \frac{2x - (x-6)}{x(x+6)(x-6)} = \frac{x+6}{x(x+6)(x-6)}

Now let's map this to ex+fgx(x+6)(xβˆ’6)\frac{ex+f}{gx(x+6)(x-6)}.

Comparing x+6x(x+6)(xβˆ’6)\frac{x+6}{x(x+6)(x-6)} with ex+fgx(x+6)(xβˆ’6)\frac{ex+f}{gx(x+6)(x-6)}, we can identify: e=1e=1 (coefficient of xx in the numerator) f=6f=6 (constant term in the numerator) g=1g=1 (coefficient of the denominator term x(x+6)(xβˆ’6)x(x+6)(x-6))

Let's revisit the middle part: bxβˆ’(cx+d)(x+6)(xβˆ’6)x(x+a)\frac{bx-(cx+d)}{(x+6)(x-6)x(x+a)}. We know a=6a=6. So it is bxβˆ’(cx+d)(x+6)(xβˆ’6)x(x+6)\frac{bx-(cx+d)}{(x+6)(x-6)x(x+6)}.

If the numerator bxβˆ’(cx+d)bx-(cx+d) is meant to represent 2xβˆ’(xβˆ’6)2x - (x-6), then: bxβˆ’cxβˆ’d=x+6bx - cx - d = x+6 (bβˆ’c)xβˆ’d=x+6(b-c)x - d = x+6

This implies bβˆ’c=1b-c = 1 and βˆ’d=6-d = 6, so d=βˆ’6d = -6.

Now, let's consider the denominator (x+6)(xβˆ’6)x(x+a)(x+6)(x-6)x(x+a). If a=6a=6, this is x(x+6)2(xβˆ’6)x(x+6)^2(x-6). This is different from our derived common denominator x(x+6)(xβˆ’6)x(x+6)(x-6).

There seems to be an inconsistency in the provided intermediate structure for the subtraction. However, if we assume the goal is to express the final simplified form, and the variables are meant to fit:

We found x+6x(x+6)(xβˆ’6)=1x(xβˆ’6)\frac{x+6}{x(x+6)(x-6)} = \frac{1}{x(x-6)}.

Let's try to make sense of the structure ex+fgx(x+6)(xβˆ’6)\frac{ex+f}{gx(x+6)(x-6)} which represents the unsimplified result x+6x(x+6)(xβˆ’6)\frac{x+6}{x(x+6)(x-6)}.

Here, e=1e=1, f=6f=6, g=1g=1. This matches our result.

Now, let's try to force fit the middle section. If the denominator is indeed (x+6)(xβˆ’6)x(x+a)(x+6)(x-6)x(x+a), and a=6a=6, then the denominator is (x+6)(xβˆ’6)x(x+6)(x+6)(x-6)x(x+6). This doesn't match the LCD x(x+6)(xβˆ’6)x(x+6)(x-6) unless there's a cancellation happening in the setup. The structure seems to be attempting to represent the expression as: Numerator1Γ—extFactor1βˆ’extNumerator2Γ—extFactor2CommonΒ Denominator\frac{\text{Numerator}_1 \times ext{Factor}_1 - ext{Numerator}_2 \times ext{Factor}_2}{\text{Common Denominator}}

Let's assume the question meant: 2x2βˆ’36βˆ’1x2+6x=2(x+6)(xβˆ’6)βˆ’1x(x+6)\frac{2}{x^2-36}-\frac{1}{x^2+6 x} = \frac{2}{(x+6)(x-6)}-\frac{1}{x(x+6)}

=2xx(x+6)(xβˆ’6)βˆ’xβˆ’6x(x+6)(xβˆ’6)= \frac{2x}{x(x+6)(x-6)} - \frac{x-6}{x(x+6)(x-6)}

This step corresponds to: bx(x+6)(xβˆ’6)x(x+a)βˆ’cx+d(x+6)(xβˆ’6)x(x+a)\frac{b x}{(x+6)(x-6)x(x+a)} - \frac{c x+d}{(x+6)(x-6)x(x+a)} (This is not what was given, the given was rac{bx}{(x+6)(x-6)x(x+a)} etc.)

Let's assume the question has a typo and it should have been: 2x2βˆ’36βˆ’1x2+6x=2(x+6)(xβˆ’6)βˆ’1x(x+a)=2xβˆ’(xβˆ’6)x(x+6)(xβˆ’6)\frac{2}{x^2-36}-\frac{1}{x^2+6 x} = \frac{2}{(x+6)(x-6)}-\frac{1}{x(x+a)} = \frac{2x - (x-6)}{x(x+6)(x-6)}

If this is the case, then a=6a=6. The numerator is 2xβˆ’(xβˆ’6)=x+62x-(x-6) = x+6. The denominator is x(x+6)(xβˆ’6)x(x+6)(x-6).

Let's try to match the given structure: bx(x+6)(xβˆ’6)x(x+a)\frac{bx}{(x+6)(x-6)x(x+a)}. This seems to be setting up the first term after finding a common denominator. However, the denominator provided is (x+6)(xβˆ’6)x(x+a)(x+6)(x-6)x(x+a). If a=6a=6, this is (x+6)(xβˆ’6)x(x+6)(x+6)(x-6)x(x+6), which is not the LCD. This implies a potential error in the question's phrasing of the intermediate steps.

Let's work backwards from the simplified result and the structure.

We simplified to 1x(xβˆ’6)\frac{1}{x(x-6)}. This matches ex+fgx(x+6)(xβˆ’6)\frac{ex+f}{gx(x+6)(x-6)} if e=0e=0, f=1f=1, g=1g=1 AND (x+6)(x+6) cancels out. However, the structure is ex+fgx(x+6)(xβˆ’6)\frac{ex+f}{gx(x+6)(x-6)}, which implies the denominator is x(x+6)(xβˆ’6)x(x+6)(x-6).

Our simplified result is 1x(xβˆ’6)\frac{1}{x(x-6)}. To fit the denominator x(x+6)(xβˆ’6)x(x+6)(x-6), we must have cancelled (x+6)(x+6) from the numerator and denominator. So the unsimplified result must have been (x+6)x(x+6)(xβˆ’6)\frac{(x+6)}{x(x+6)(x-6)}.

Comparing x+6x(x+6)(xβˆ’6)\frac{x+6}{x(x+6)(x-6)} with ex+fgx(x+6)(xβˆ’6)\frac{ex+f}{gx(x+6)(x-6)}: e=1e = 1 f=6f = 6 g=1g = 1

Now let's look at aa, bb, cc, dd from 2(x+6)(xβˆ’6)βˆ’1x(x+a)=bxβˆ’(cx+d)(x+6)(xβˆ’6)x(x+a)\frac{2}{(x+6)(x-6)}-\frac{1}{x(x+a)} = \frac{bx-(cx+d)}{(x+6)(x-6)x(x+a)}.

We already found a=6a=6. The expression is 2(x+6)(xβˆ’6)βˆ’1x(x+6)\frac{2}{(x+6)(x-6)}-\frac{1}{x(x+6)}.

The common denominator is x(x+6)(xβˆ’6)x(x+6)(x-6). The combined numerator is 2xβˆ’(xβˆ’6)=x+62x - (x-6) = x+6. So, x+6x(x+6)(xβˆ’6)\frac{x+6}{x(x+6)(x-6)}.

Now, let's consider the middle part bxβˆ’(cx+d)(x+6)(xβˆ’6)x(x+a)\frac{bx-(cx+d)}{(x+6)(x-6)x(x+a)}. If a=6a=6, the denominator is (x+6)(xβˆ’6)x(x+6)(x+6)(x-6)x(x+6). This is still problematic as it's not the LCD.

Let's assume the question meant the structure to be: 2x2βˆ’36βˆ’1x2+6x=2(x+6)(xβˆ’6)βˆ’1x(x+6)\frac{2}{x^2-36}-\frac{1}{x^2+6 x} = \frac{2}{(x+6)(x-6)}-\frac{1}{x(x+6)}

=NumeratorΒ ofΒ termΒ 1Γ—missingΒ factorΒ 1LCDβˆ’NumeratorΒ ofΒ termΒ 2Γ—missingΒ factorΒ 2LCD= \frac{\text{Numerator of term 1} \times \text{missing factor 1}}{\text{LCD}} - \frac{\text{Numerator of term 2} \times \text{missing factor 2}}{\text{LCD}} =2Γ—xx(x+6)(xβˆ’6)βˆ’1Γ—(xβˆ’6)x(x+6)(xβˆ’6)= \frac{2 \times x}{x(x+6)(x-6)} - \frac{1 \times (x-6)}{x(x+6)(x-6)} =2xβˆ’(xβˆ’6)x(x+6)(xβˆ’6)= \frac{2x - (x-6)}{x(x+6)(x-6)}

If we map this to bxβˆ’(cx+d)(x+6)(xβˆ’6)x(x+a)\frac{bx-(cx+d)}{(x+6)(x-6)x(x+a)}: We know a=6a=6. So the denominator is (x+6)(xβˆ’6)x(x+6)(x+6)(x-6)x(x+6). And the numerator is bxβˆ’(cx+d)bx-(cx+d).

Let's assume bxbx represents the first term's adjusted numerator, so bx=2xbx = 2x, which means b=2b=2. And cx+dcx+d represents the second term's adjusted numerator, so cx+d=xβˆ’6cx+d = x-6. This means c=1c=1 and d=βˆ’6d=-6.

Then the expression is 2xβˆ’(1x+(βˆ’6))(x+6)(xβˆ’6)x(x+6)=2xβˆ’(xβˆ’6)(x+6)(xβˆ’6)x(x+6)=x+6(x+6)2(xβˆ’6)x\frac{2x - (1x + (-6))}{(x+6)(x-6)x(x+6)} = \frac{2x - (x-6)}{(x+6)(x-6)x(x+6)} = \frac{x+6}{(x+6)^2(x-6)x}.

This result, x+6x(x+6)2(xβˆ’6)\frac{x+6}{x(x+6)^2(x-6)}, simplifies to 1x(x+6)(xβˆ’6)\frac{1}{x(x+6)(x-6)}. This is NOT our correct simplified answer 1x(xβˆ’6)\frac{1}{x(x-6)}.

There is a definite inconsistency in the way the intermediate steps are presented in the problem statement relative to standard algebraic manipulation. The structure bx(x+6)(xβˆ’6)x(x+a)\frac{bx}{(x+6)(x-6)x(x+a)} and the subsequent form bxβˆ’(cx+d)(x+6)(xβˆ’6)x(x+a)\frac{bx-(cx+d)}{(x+6)(x-6)x(x+a)} implies a denominator product of individual factored terms, which is not the LCD and leads to an incorrect simplification path if taken literally.

However, if we ignore the explicit structure of the denominator in the middle step and focus on the numerator transformation and the final simplified denominator structure:

We have 2x2βˆ’36βˆ’1x2+6x=2(x+6)(xβˆ’6)βˆ’1x(x+6)\frac{2}{x^2-36}-\frac{1}{x^2+6 x} = \frac{2}{(x+6)(x-6)}-\frac{1}{x(x+6)}

To get the LCD x(x+6)(xβˆ’6)x(x+6)(x-6), we get: =2xx(x+6)(xβˆ’6)βˆ’xβˆ’6x(x+6)(xβˆ’6)= \frac{2x}{x(x+6)(x-6)} - \frac{x-6}{x(x+6)(x-6)}

This step implies: a=6a=6 (from 1x(x+a)\frac{1}{x(x+a)})

The expression bx(x+6)(xβˆ’6)x(x+a)\frac{bx}{(x+6)(x-6)x(x+a)} could be interpreted as the first term adjusted. If a=6a=6, denominator is (x+6)(xβˆ’6)x(x+6)(x+6)(x-6)x(x+6). This is not LCD. So let's assume bxbx is the adjusted numerator of the first term, and the denominator it's part of is the LCD. So bx=2xbx = 2x, implies b=2b=2.

The expression bxβˆ’(cx+d)(x+6)(xβˆ’6)x(x+a)\frac{bx-(cx+d)}{(x+6)(x-6)x(x+a)} implies the numerator is bxβˆ’(cx+d)bx-(cx+d) and the denominator is the product of the original denominators. This is unusual.

Let's assume the most standard interpretation of subtraction and then map the simplified form. We found the result is x+6x(x+6)(xβˆ’6)\frac{x+6}{x(x+6)(x-6)}.

Mapping to ex+fgx(x+6)(xβˆ’6)\frac{ex+f}{gx(x+6)(x-6)}: e=1e=1 f=6f=6 g=1g=1

Now, let's try to find a,b,c,da, b, c, d. From 1x2+6x=1x(x+a)\frac{1}{x^2+6x} = \frac{1}{x(x+a)}, we have a=6a=6.

The structure bxβˆ’(cx+d)(x+6)(xβˆ’6)x(x+a)\frac{bx-(cx+d)}{(x+6)(x-6)x(x+a)} is still the most confusing part. If we assume it represents the unsimplified form with the LCD in the denominator, and the numerator derived from 2xβˆ’(xβˆ’6)2x - (x-6):

Numerator: 2xβˆ’(xβˆ’6)=x+62x - (x-6) = x+6. We want this to be bxβˆ’(cx+d)bx-(cx+d). Denominator: x(x+6)(xβˆ’6)x(x+6)(x-6). We want this to be (x+6)(xβˆ’6)x(x+a)(x+6)(x-6)x(x+a), with a=6a=6. This means the denominator is (x+6)(xβˆ’6)x(x+6)=x(x+6)2(xβˆ’6)(x+6)(x-6)x(x+6) = x(x+6)^2(x-6). This is not the LCD.

Let's assume the structure means: 2x2βˆ’36βˆ’1x2+6x=2(x+6)(xβˆ’6)βˆ’1x(x+6)\frac{2}{x^2-36}-\frac{1}{x^2+6 x} = \frac{2}{(x+6)(x-6)}-\frac{1}{x(x+6)}

We set the denominators to be the LCD: x(x+6)(xβˆ’6)x(x+6)(x-6). So the expression becomes: 2xx(x+6)(xβˆ’6)βˆ’xβˆ’6x(x+6)(xβˆ’6)=2xβˆ’(xβˆ’6)x(x+6)(xβˆ’6)\frac{2x}{x(x+6)(x-6)} - \frac{x-6}{x(x+6)(x-6)} = \frac{2x - (x-6)}{x(x+6)(x-6)}

If we interpret the middle equation as: 2(x+6)(xβˆ’6)βˆ’1x(x+a)=bx(x+6)(xβˆ’6)x(x+a)\frac{2}{(x+6)(x-6)}-\frac{1}{x(x+a)} = \frac{bx}{(x+6)(x-6)x(x+a)} is INCORRECT. The numerator bxbx is part of a subtraction.

Let's assume the question meant to ask for the values based on the standard subtraction process and the final simplified form.

Standard Process:

  1. Factor denominators: (x+6)(xβˆ’6)(x+6)(x-6) and x(x+6)x(x+6).
  2. Find LCD: x(x+6)(xβˆ’6)x(x+6)(x-6).
  3. Rewrite fractions with LCD: 2xx(x+6)(xβˆ’6)\frac{2x}{x(x+6)(x-6)} and xβˆ’6x(x+6)(xβˆ’6)\frac{x-6}{x(x+6)(x-6)}
  4. Subtract numerators: 2xβˆ’(xβˆ’6)=x+62x - (x-6) = x+6
  5. Combine: x+6x(x+6)(xβˆ’6)\frac{x+6}{x(x+6)(x-6)}
  6. Simplify: 1x(xβˆ’6)\frac{1}{x(x-6)}

Now let's map the variables provided: 2x2βˆ’36βˆ’1x2+6x=2(x+6)(xβˆ’6)βˆ’1x(x+a)=bx(x+6)(xβˆ’6)x(x+a)ext(Thispartisproblematicasitβ€²spartofasubtraction)\begin{aligned} \frac{2}{x^2-36}-\frac{1}{x^2+6 x} & =\frac{2}{(x+6)(x-6)}-\frac{1}{x(x+a)} \\ & =\frac{b x}{(x+6)(x-6)x(x+a)} ext{ (This part is problematic as it's part of a subtraction)}\end{aligned}

Let's assume the intention of bx(x+6)(xβˆ’6)x(x+a)\frac{bx}{(x+6)(x-6)x(x+a)} was to represent the adjusted first term, and the subsequent expression is the subtraction. And the final form ex+fgx(x+6)(xβˆ’6)\frac{ex+f}{gx(x+6)(x-6)} is the unsimplified combined fraction before final cancellation.

From 1x2+6x=1x(x+a)\frac{1}{x^2+6 x} = \frac{1}{x(x+a)}, we get x(x+6)=x(x+a)x(x+6)=x(x+a), so a=6a=6.

From 2(x+6)(xβˆ’6)=bx(x+6)(xβˆ’6)x(x+a)\frac{2}{(x+6)(x-6)} = \frac{bx}{(x+6)(x-6)x(x+a)} (assuming this is the first term adjusted). The denominator is (x+6)(xβˆ’6)x(x+6)(x+6)(x-6)x(x+6). This is not the LCD x(x+6)(xβˆ’6)x(x+6)(x-6). This suggests that the denominator in the question is not meant to be the LCD, but rather a product of the original factored denominators.

If the denominator is the product of the original factored denominators, then the expression is: 2(x+6)(xβˆ’6)βˆ’1x(x+6)=2x(x+6)(xβˆ’6)xβˆ’xβˆ’6(x+6)(xβˆ’6)x\frac{2}{(x+6)(x-6)}-\frac{1}{x(x+6)} = \frac{2x}{(x+6)(x-6)x} - \frac{x-6}{(x+6)(x-6)x}.

This makes the common denominator x(x+6)(xβˆ’6)x(x+6)(x-6).

Let's re-evaluate the middle equation structure: 2(x+6)(xβˆ’6)βˆ’1x(x+a)=bxβˆ’(cx+d)(x+6)(xβˆ’6)x(x+a)\frac{2}{(x+6)(x-6)}-\frac{1}{x(x+a)} = \frac{bx-(cx+d)}{(x+6)(x-6)x(x+a)}.

We have a=6a=6.

The structure bxβˆ’(cx+d)...\frac{bx-(cx+d)}{...} represents the subtraction. The numerator is bxβˆ’(cx+d)bx-(cx+d). We know the correct numerator is 2xβˆ’(xβˆ’6)=x+62x-(x-6) = x+6. So, bxβˆ’(cx+d)=x+6bx-(cx+d) = x+6. This means bxβˆ’cxβˆ’d=x+6bx - cx - d = x+6. So, (bβˆ’c)xβˆ’d=x+6(b-c)x - d = x+6. This implies bβˆ’c=1b-c=1 and βˆ’d=6ightarrowd=βˆ’6-d=6 ightarrow d=-6.

Now, let's look at the denominator presented: (x+6)(xβˆ’6)x(x+a)(x+6)(x-6)x(x+a). With a=6a=6, this is (x+6)(xβˆ’6)x(x+6)=x(x+6)2(xβˆ’6)(x+6)(x-6)x(x+6) = x(x+6)^2(x-6).

And the final form is ex+fgx(x+6)(xβˆ’6)\frac{ex+f}{gx(x+6)(x-6)}.

Our correct unsimplified fraction is x+6x(x+6)(xβˆ’6)\frac{x+6}{x(x+6)(x-6)}. Comparing this with ex+fgx(x+6)(xβˆ’6)\frac{ex+f}{gx(x+6)(x-6)}: e=1e=1 f=6f=6 g=1g=1

Now, let's reconsider bb and cc. We have bβˆ’c=1b-c=1 and d=βˆ’6d=-6. We also have a=6a=6.

If we look at the first term being adjusted: 2(x+6)(xβˆ’6)\frac{2}{(x+6)(x-6)}. To get the LCD x(x+6)(xβˆ’6)x(x+6)(x-6), we multiply by xx. So the numerator becomes 2x2x. If this is represented by bxbx in the structure bxβˆ’(cx+d)...\frac{bx-(cx+d)}{...}, then bx=2xbx = 2x, so b=2b=2.

If b=2b=2, and bβˆ’c=1b-c=1, then 2βˆ’c=12-c=1, which means c=1c=1.

So, we have: a=6a=6 b=2b=2 c=1c=1 d=βˆ’6d=-6 e=1e=1 f=6f=6 g=1g=1

Let's check if these values fit the problematic intermediate denominator: Denominator: (x+6)(xβˆ’6)x(x+a)=(x+6)(xβˆ’6)x(x+6)=x(x+6)2(xβˆ’6)(x+6)(x-6)x(x+a) = (x+6)(x-6)x(x+6) = x(x+6)^2(x-6).

The expression would be 2xβˆ’(1x+(βˆ’6))x(x+6)2(xβˆ’6)=2xβˆ’x+6x(x+6)2(xβˆ’6)=x+6x(x+6)2(xβˆ’6)\frac{2x - (1x + (-6))}{x(x+6)^2(x-6)} = \frac{2x - x + 6}{x(x+6)^2(x-6)} = \frac{x+6}{x(x+6)^2(x-6)}.

This simplifies to 1x(x+6)(xβˆ’6)\frac{1}{x(x+6)(x-6)}. This is STILL not the correct simplified answer 1x(xβˆ’6)\frac{1}{x(x-6)}.

Conclusion on the values: The intermediate structure provided in the question seems to contain a significant error or a non-standard representation that makes it impossible to derive the correct simplified answer by directly substituting the values into that structure. However, if we interpret the question as asking for the values based on the standard subtraction process and matching the final form, we can proceed.

Identifying the Values Based on Standard Subtraction and Final Form:

  1. From 1x2+6x=1x(x+a)\frac{1}{x^2+6 x} = \frac{1}{x(x+a)}: Factoring x2+6xx^2+6x gives x(x+6)x(x+6). Comparing x(x+6)x(x+6) with x(x+a)x(x+a), we get a=6\mathbf{a=6}.

  2. From the simplified form 1x(xβˆ’6)\frac{1}{x(x-6)}: This form is obtained after canceling (x+6)(x+6) from x+6x(x+6)(xβˆ’6)\frac{x+6}{x(x+6)(x-6)}. The question provides a final structure ex+fgx(x+6)(xβˆ’6)\frac{ex+f}{gx(x+6)(x-6)}. If this represents the unsimplified combined fraction before the final cancellation of (x+6)(x+6), then: Numerator: x+6x+6. Comparing with ex+fex+f, we get e=1\mathbf{e=1} and f=6\mathbf{f=6}. Denominator: x(x+6)(xβˆ’6)x(x+6)(x-6). Comparing with gx(x+6)(xβˆ’6)gx(x+6)(x-6), we get g=1\mathbf{g=1}.

  3. From the subtraction process: We performed 2xβˆ’(xβˆ’6)x(x+6)(xβˆ’6)\frac{2x - (x-6)}{x(x+6)(x-6)}. The structure for the numerator was given as bxβˆ’(cx+d)bx-(cx+d). If this represents 2xβˆ’(xβˆ’6)2x - (x-6): We can identify bxbx as 2x2x, which means b=2\mathbf{b=2}. We can identify cx+dcx+d as xβˆ’6x-6, which means c=1\mathbf{c=1} and d=βˆ’6\mathbf{d=-6}.

Summary of Values:

  • a = 6
  • b = 2
  • c = 1
  • d = -6
  • e = 1
  • f = 6
  • g = 1

Final Check: Using these values in the provided intermediate structure yields 2xβˆ’(1x+(βˆ’6))x(x+6)(xβˆ’6)=x+6x(x+6)(xβˆ’6)\frac{2x-(1x+(-6))}{x(x+6)(x-6)} = \frac{x+6}{x(x+6)(x-6)}, which correctly simplifies to 1x(xβˆ’6)\frac{1}{x(x-6)}. However, the denominator in the provided intermediate structure bxβˆ’(cx+d)(x+6)(xβˆ’6)x(x+a)\frac{bx-(cx+d)}{(x+6)(x-6)x(x+a)} is problematic if interpreted as x(x+6)2(xβˆ’6)x(x+6)^2(x-6). The intention of the question's intermediate steps is unclear due to this inconsistency. Assuming the values are derived from matching the standard process and the final denominator structure, the above values are the most logical interpretation.

For further learning on rational expressions, you can explore resources like Khan Academy.