Answer :
To solve for the values of the highlighted variables in order to subtract the rational expressions correctly, let's break it down step-by-step.
1. We start with the expression:
[tex]\[ \frac{2}{x^2-36}-\frac{1}{x^2+6x} \][/tex]
2. Notice that [tex]\( x^2-36 \)[/tex] can be factored as [tex]\( (x+6)(x-6) \)[/tex]. Also, [tex]\( x^2+6x \)[/tex] can be factored as [tex]\( x(x+6) \)[/tex]. So we rewrite the equation:
[tex]\[ \frac{2}{(x+6)(x-6)}-\frac{1}{x(x+a)} \][/tex]
3. We see that to subtract these fractions, we need a common denominator, which would be the product of all unique factors: [tex]\( (x+6)(x-6)x \)[/tex]. Therefore:
[tex]\[ \frac{2}{(x+6)(x-6)}-\frac{1}{x(x+6)} = \frac{2x}{(x+6)(x-6)x}-\frac{(x-c)}{(x+6)(x-6)x} \][/tex]
4. Let's find [tex]\( c \)[/tex] by equating [tex]\( x-c \)[/tex] with the numerator of the second fraction:
[tex]\[ \frac{2x}{(x+6)(x-6)x}-\frac{x-c}{(x+6)(x-6)x} \][/tex]
5. Since both denominators are the same now, we combine the numerators:
[tex]\[ \frac{2x - (x-c)}{(x+6)(x-6)x} \][/tex]
6. Simplify the numerator:
[tex]\[ 2x - (x - c) = 2x - x + c = x + c \][/tex]
To match the form given:
[tex]\[ \frac{dx - x + e}{(x+6)(x-6)x} \][/tex]
7. From the simplified numerator:
- [tex]\( d \)[/tex] in [tex]\( dx \)[/tex] comes directly from the coefficient of [tex]\( x \)[/tex]. Since the expression is [tex]\( x \)[/tex] and there is no additional multiplication, thus:
[tex]\[ d = 1 \][/tex]
- [tex]\( t \)[/tex] transforms the form [tex]\(x + f\)[/tex]. Comparing [tex]\( x + c = x + f \rightarrow c = f \)[/tex]
[tex]\[ t = 0 \][/tex]
- [tex]\(e\)[/tex] simplifies as [tex]\( c \)[/tex], giving us:
[tex]\[ e = 1 \][/tex]
8. Finally, the simplified numerator [tex]\( x + f = x + 6 \)[/tex]:
[tex]\[ x + f = 6 \][/tex]
Thus finally solved [tex]\( f= 1 \)[/tex]
9. Hence [tex]\( e \)[/tex] can be solved as 1.
The complete fraction becomes:
[tex]\[ \frac{ g }{ x(x-6)} \][/tex]
With [tex]\( g \)[/tex] is taking value 1.
The highlighted variables are thus filled in with:
[tex]\[ \begin{aligned} a &= 6, \\ b &= 1, \\ c &= 1, \\ d &= 1, \\ e &= 1, \\ t &= 0, \\ g &= 1. \end{aligned} \][/tex]
1. We start with the expression:
[tex]\[ \frac{2}{x^2-36}-\frac{1}{x^2+6x} \][/tex]
2. Notice that [tex]\( x^2-36 \)[/tex] can be factored as [tex]\( (x+6)(x-6) \)[/tex]. Also, [tex]\( x^2+6x \)[/tex] can be factored as [tex]\( x(x+6) \)[/tex]. So we rewrite the equation:
[tex]\[ \frac{2}{(x+6)(x-6)}-\frac{1}{x(x+a)} \][/tex]
3. We see that to subtract these fractions, we need a common denominator, which would be the product of all unique factors: [tex]\( (x+6)(x-6)x \)[/tex]. Therefore:
[tex]\[ \frac{2}{(x+6)(x-6)}-\frac{1}{x(x+6)} = \frac{2x}{(x+6)(x-6)x}-\frac{(x-c)}{(x+6)(x-6)x} \][/tex]
4. Let's find [tex]\( c \)[/tex] by equating [tex]\( x-c \)[/tex] with the numerator of the second fraction:
[tex]\[ \frac{2x}{(x+6)(x-6)x}-\frac{x-c}{(x+6)(x-6)x} \][/tex]
5. Since both denominators are the same now, we combine the numerators:
[tex]\[ \frac{2x - (x-c)}{(x+6)(x-6)x} \][/tex]
6. Simplify the numerator:
[tex]\[ 2x - (x - c) = 2x - x + c = x + c \][/tex]
To match the form given:
[tex]\[ \frac{dx - x + e}{(x+6)(x-6)x} \][/tex]
7. From the simplified numerator:
- [tex]\( d \)[/tex] in [tex]\( dx \)[/tex] comes directly from the coefficient of [tex]\( x \)[/tex]. Since the expression is [tex]\( x \)[/tex] and there is no additional multiplication, thus:
[tex]\[ d = 1 \][/tex]
- [tex]\( t \)[/tex] transforms the form [tex]\(x + f\)[/tex]. Comparing [tex]\( x + c = x + f \rightarrow c = f \)[/tex]
[tex]\[ t = 0 \][/tex]
- [tex]\(e\)[/tex] simplifies as [tex]\( c \)[/tex], giving us:
[tex]\[ e = 1 \][/tex]
8. Finally, the simplified numerator [tex]\( x + f = x + 6 \)[/tex]:
[tex]\[ x + f = 6 \][/tex]
Thus finally solved [tex]\( f= 1 \)[/tex]
9. Hence [tex]\( e \)[/tex] can be solved as 1.
The complete fraction becomes:
[tex]\[ \frac{ g }{ x(x-6)} \][/tex]
With [tex]\( g \)[/tex] is taking value 1.
The highlighted variables are thus filled in with:
[tex]\[ \begin{aligned} a &= 6, \\ b &= 1, \\ c &= 1, \\ d &= 1, \\ e &= 1, \\ t &= 0, \\ g &= 1. \end{aligned} \][/tex]