Question

cphs : advanced algebra: concepts and connections - block (27.0831030)
adding and subtracting rational expressions
subtracting rational expressions
enter the values for the variables a through g that show how to
subtract the rational expressions correctly:
$\frac{2}{x^2 - 36} - \frac{1}{x^2 + 6x} = \frac{2}{(x + 6)(x - 6)} - \frac{1}{x(x + a)}$
$= \frac{bx}{(x + 6)(x - 6)x} - \frac{(x - c)}{(x + 6)(x - 6)x}$
$= \frac{dx - x + e}{(x + 6)(x - 6)x}$
$= \frac{(x + f)}{(x + 6)(x - 6)x}$
$= \frac{g}{x(x - 6)}$
$a = \square$
$b = \square$
$c = \square$
$d = \square$
$e = \square$
$f = \square$
$g = \square$

Explanation:

Step1: Factor denominators

First, factor the denominators of the rational expressions. We know that \(x^{2}-36=(x + 6)(x - 6)\) (by the difference of squares formula \(a^{2}-b^{2}=(a + b)(a - b)\) where \(a=x\) and \(b = 6\)) and \(x^{2}+6x=x(x + 6)\).

Step2: Find common denominator

The common denominator for \(\frac{2}{x^{2}-36}\) and \(\frac{1}{x^{2}+6x}\) is \(x(x + 6)(x - 6)\) (since \(x^{2}-36=(x + 6)(x - 6)\) and \(x^{2}+6x=x(x + 6)\)).

Step3: Rewrite fractions with common denominator

Rewrite \(\frac{2}{x^{2}-36}\) as \(\frac{2x}{x(x + 6)(x - 6)}\) (multiply numerator and denominator by \(x\)) and \(\frac{1}{x^{2}+6x}\) as \(\frac{x - 6}{x(x + 6)(x - 6)}\) (multiply numerator and denominator by \((x - 6)\)).

Step4: Subtract the fractions

Now, subtract the two fractions: \(\frac{2x}{x(x + 6)(x - 6)}-\frac{x - 6}{x(x + 6)(x - 6)}=\frac{2x-(x - 6)}{x(x + 6)(x - 6)}\).

Step5: Simplify the numerator

Simplify the numerator: \(2x-(x - 6)=2x-x + 6=x + 6\).

Step6: Simplify the fraction

So the fraction becomes \(\frac{x + 6}{x(x + 6)(x - 6)}\), and we can cancel out the \((x + 6)\) terms (assuming \(x eq - 6\)) to get \(\frac{1}{x(x - 6)}\). Wait, no, wait, let's check again. Wait, original problem is \(\frac{2}{x^{2}-36}-\frac{1}{x^{2}+6x}\). Let's do it step by step with the given steps in the image.

Looking at the image steps:

First, \(\frac{2}{x^{2}-36}=\frac{2}{(x + 6)(x - 6)}=\frac{2x}{x(x + 6)(x - 6)}\) (so \(b = 2\), because in the step \(\frac{bx}{(x + 6)(x - 6)x}\), so \(b = 2\)).

Then, \(\frac{1}{x^{2}+6x}=\frac{1}{x(x + 6)}=\frac{x - 6}{x(x + 6)(x - 6)}\) (so the numerator here is \(x - 6\), so in the step \(\frac{(x - c)}{(x + 6)(x - 6)x}\), so \(c = 6\) because \(x - c=x - 6\) implies \(c = 6\)).

Then, subtracting: \(\frac{2x}{x(x + 6)(x - 6)}-\frac{x - 6}{x(x + 6)(x - 6)}=\frac{2x-(x - 6)}{x(x + 6)(x - 6)}=\frac{2x-x + 6}{x(x + 6)(x - 6)}=\frac{x + 6}{x(x + 6)(x - 6)}\) (so \(d = 1\) (coefficient of \(x\) in numerator? Wait, no, looking at the steps:

First step: \(\frac{2}{x^{2}-36}=\frac{2}{(x + 6)(x - 6)}=\frac{bx}{(x + 6)(x - 6)x}\) ⇒ \(b = 2\) (since numerator 2 times x is \(2x\), so \(bx = 2x\) ⇒ \(b = 2\)).

Second step: \(\frac{1}{x^{2}+6x}=\frac{1}{x(x + 6)}=\frac{(x - c)}{(x + 6)(x - 6)x}\) ⇒ \(x - c\) is the numerator when we multiply 1 by \((x - 6)\), so \(x - c=x - 6\) ⇒ \(c = 6\).

Third step: \(\frac{2x}{x(x + 6)(x - 6)}-\frac{x - 6}{x(x + 6)(x - 6)}=\frac{2x-(x - 6)}{x(x + 6)(x - 6)}=\frac{2x-x + 6}{x(x + 6)(x - 6)}=\frac{x + 6}{x(x + 6)(x - 6)}\) (so the numerator after subtraction is \(x + 6\), which is \(dx - x+e\)? Wait, the step shows \(\frac{dx - x+e}{(x + 6)(x - 6)x}\). So \(dx - x+e=(d - 1)x+e\). And we have \(2x-(x - 6)=x + 6\), so \((d - 1)x+e=x + 6\). Therefore, \(d - 1 = 1\) ⇒ \(d = 2\)? No, wait, no, \(2x-(x - 6)=x + 6\), so \(dx - x+e=x + 6\) ⇒ \(d = 1\)? Wait, maybe I misread the step. The step is \(\frac{dx - x+e}{(x + 6)(x - 6)x}\). So \(dx - x+e=(d - 1)x+e\). And \(2x-(x - 6)=x + 6\), so \((d - 1)x+e=x + 6\). Therefore, \(d - 1 = 1\) ⇒ \(d = 2\)? No, \(2x - x= x\), so \(d = 2\)? Wait, no, original numerator of first fraction is 2, multiplied by x is 2x. Second fraction numerator is 1, multiplied by (x - 6) is (x - 6). So subtraction: 2x - (x - 6)=2x - x + 6=x + 6. So \(dx - x+e=x + 6\) ⇒ \(d = 1\) (coefficient of x is 1), so \(d - 1 = 1\) ⇒ \(d = 2\)? No, \(dx - x=(d - 1)x\). So (d - 1)x+e=x + 6. Therefore, \(d - 1 = 1\) ⇒ \(d = 2\)? No, 1x + 6, so \(d - 1 = 1\) ⇒ \(d = 2\), and \(e = 6\).

Then next step: \(\frac{(x + f)}{(x + 6)(x - 6)x}\). Since we have \(\frac{x + 6}{(x + 6)(x - 6)x}\), so \(f = 6\) (because numerator is \(x + 6\), so \(x + f=x + 6\) ⇒ \(f = 6\)).

Then next step: \(\frac{g}{x(x - 6)}\). After canceling (x + 6) from numerator and denominator, we get \(\frac{1}{x(x - 6)}\)? Wait, no, \(\frac{x + 6}{(x + 6)(x - 6)x}=\frac{1}{x(x - 6)}\) when we cancel (x + 6). Wait, but numerator is \(x + 6\), denominator is (x + 6)(x - 6)x, so cancel (x + 6) gives \(\frac{1}{x(x - 6)}\), so \(g = 1\)? But wait, let's match with the variables:

- \(a\): In the step \(\frac{1}{x(x + a)}\), the second fraction is \(\frac{1}{x^{2}+6x}=\frac{1}{x(x + 6)}\), so \(a = 6\).

- \(b\): In \(\frac{bx}{(x + 6)(x - 6)x}\), the first fraction \(\frac{2}{x^{2}-36}=\frac{2}{(x + 6)(x - 6)}=\frac{2x}{x(x + 6)(x - 6)}\), so \(b = 2\).

- \(c\): In \(\frac{(x - c)}{(x + 6)(x - 6)x}\), the second fraction \(\frac{1}{x(x + 6)}=\frac{x - 6}{x(x + 6)(x - 6)}\), so \(x - c=x - 6\) ⇒ \(c = 6\).

- \(d\): In \(\frac{dx - x+e}{(x + 6)(x - 6)x}\), the numerator after subtraction is \(2x-(x - 6)=x + 6\), which is \(dx - x+e\). So \(dx - x+e=x + 6\) ⇒ \(d = 2\) (because \(2x - x=x\)), wait, no, \(dx - x=(d - 1)x\). So \((d - 1)x+e=x + 6\) ⇒ \(d - 1 = 1\) ⇒ \(d = 2\), and \(e = 6\).

- \(e\): From above, \(e = 6\).

- \(f\): In \(\frac{(x + f)}{(x + 6)(x - 6)x}\), the numerator is \(x + 6\), so \(f = 6\).

- \(g\): In \(\frac{g}{x(x - 6)}\), after canceling (x + 6), we have \(\frac{1}{x(x - 6)}\)? Wait, no, \(\frac{x + 6}{(x + 6)(x - 6)x}=\frac{1}{x(x - 6)}\), so \(g = 1\)? Wait, no, wait, original numerator after subtraction is \(x + 6\), so when we cancel (x + 6), we get \(\frac{1}{x(x - 6)}\), so \(g = 1\). But let's check again:

Wait, the problem is \(\frac{2}{x^{2}-36}-\frac{1}{x^{2}+6x}\).

1. Factor denominators:

\(x^{2}-36=(x + 6)(x - 6)\), \(x^{2}+6x=x(x + 6)\).

2. Rewrite with common denominator \(x(x + 6)(x - 6)\):

\(\frac{2}{(x + 6)(x - 6)}=\frac{2x}{x(x + 6)(x - 6)}\) (so \(b = 2\))

\(\frac{1}{x(x + 6)}=\frac{x - 6}{x(x + 6)(x - 6)}\) (so \(c = 6\) because the numerator is \(x - 6\))

3. Subtract:

\(\frac{2x}{x(x + 6)(x - 6)}-\frac{x - 6}{x(x + 6)(x - 6)}=\frac{2x-(x - 6)}{x(x + 6)(x - 6)}=\frac{2x - x+6}{x(x + 6)(x - 6)}=\frac{x + 6}{x(x + 6)(x - 6)}\) (so \(d = 2\) (coefficient of x in first term), \(e = 6\) (constant term), then numerator is \(x + 6\) so \(f = 6\) (since numerator is \(x + f\))

4. Cancel \((x + 6)\):

\(\frac{x + 6}{x(x + 6)(x - 6)}=\frac{1}{x(x - 6)}\) (so \(g = 1\))

And for \(a\), the second fraction is \(\frac{1}{x^{2}+6x}=\frac{1}{x(x + a)}\), so \(a = 6\) (since \(x^{2}+6x=x(x + 6)\))

So:

- \(a = 6\)

- \(b = 2\)

- \(c = 6\)

- \(d = 2\)

- \(e = 6\)

- \(f = 6\)

- \(g = 1\)

Wait, but let's check the step where it says \(\frac{dx - x+e}{(x + 6)(x - 6)x}\). The numerator is \(2x-(x - 6)=x + 6\), which can be written as \(2x - x+6\), so \(d = 2\), \(e = 6\). Then \(\frac{2x - x+6}{(x + 6)(x - 6)x}=\frac{x + 6}{(x + 6)(x - 6)x}\), so \(f = 6\) (since numerator is \(x + f\)). Then canceling \((x + 6)\) gives \(\frac{1}{x(x - 6)}\), so \(g = 1\). And \(a = 6\) because the second fraction is \(\frac{1}{x(x + a)}\) and \(x^{2}+6x=x(x + 6)\), so \(a = 6\).

Answer:

\(a = 6\)

\(b = 2\)

\(c = 6\)

\(d = 2\)

\(e = 6\)

\(f = 6\)

\(g = 1\)