Question

1. simplify the rational expression \\(\frac{x - 5}{x - 7} - \frac{x^2 - 6x + 8}{x^2 + 8x + 7} \cdot \frac{x^2 + 3x + 2}{x^2 - 11x + 28}\\). identify any non - permissible values. show all work.
2. using an example, explain why non - permissible values must be noted before simplifying rational expressions and equations.

Explanation:

Question 5

Step 1: Factor all quadratic expressions

First, factor each quadratic numerator and denominator:

• \(x^2 - 6x + 8=(x - 2)(x - 4)\) (since \(-2\times -4 = 8\) and \(-2 + (-4)=-6\))
• \(x^2 + 8x + 7=(x + 1)(x + 7)\) (since \(1\times7 = 7\) and \(1 + 7 = 8\))
• \(x^2 + 3x + 2=(x + 1)(x + 2)\) (since \(1\times2 = 2\) and \(1 + 2 = 3\))
• \(x^2 - 11x + 28=(x - 4)(x - 7)\) (since \(-4\times -7 = 28\) and \(-4+(-7)=-11\))

So the expression becomes:
\(\frac{x - 5}{x - 7}-\frac{(x - 2)(x - 4)}{(x + 1)(x + 7)}\cdot\frac{(x + 1)(x + 2)}{(x - 4)(x - 7)}\)

Step 2: Simplify the multiplication part

Cancel out common factors in the multiplication:
The \((x - 4)\) in the numerator and denominator cancels, and the \((x + 1)\) in the numerator and denominator cancels. So:
\(\frac{(x - 2)(x - 4)}{(x + 1)(x + 7)}\cdot\frac{(x + 1)(x + 2)}{(x - 4)(x - 7)}=\frac{(x - 2)(x + 2)}{(x + 7)(x - 7)}\) (using the difference of squares: \((a - b)(a + b)=a^2 - b^2\), so \((x - 2)(x + 2)=x^2 - 4\))

Now the expression is:
\(\frac{x - 5}{x - 7}-\frac{x^2 - 4}{(x + 7)(x - 7)}\)

Step 3: Find a common denominator

The common denominator of \(\frac{x - 5}{x - 7}\) and \(\frac{x^2 - 4}{(x + 7)(x - 7)}\) is \((x + 7)(x - 7)\). Rewrite the first fraction:
\(\frac{x - 5}{x - 7}=\frac{(x - 5)(x + 7)}{(x + 7)(x - 7)}\) (multiply numerator and denominator by \((x + 7)\))

Step 4: Subtract the fractions

Now subtract the two fractions:

$$
LATEXBLOCK0
$$

Step 5: Identify non - permissible values

Non - permissible values are values that make the denominator of any original fraction equal to zero.

• For \(\frac{x - 5}{x - 7}\), \(x-7 = 0\Rightarrow x = 7\)
• For \(\frac{(x - 2)(x - 4)}{(x + 1)(x + 7)}\), \(x + 1=0\Rightarrow x=-1\) and \(x + 7 = 0\Rightarrow x=-7\)
• For \(\frac{(x + 1)(x + 2)}{(x - 4)(x - 7)}\), \(x - 4=0\Rightarrow x = 4\) and \(x - 7=0\Rightarrow x = 7\) (already considered)

So the non - permissible values are \(x=-7,-1,4,7\)

Consider the rational expression \(\frac{x - 1}{x^2 - 1}\cdot\frac{x + 1}{x - 2}\). First, factor \(x^2-1=(x - 1)(x + 1)\). If we simplify without noting non - permissible values first, we might cancel \((x - 1)\) and \((x + 1)\) and get \(\frac{1}{x - 2}\). But the original expression has denominators \(x^2 - 1=(x - 1)(x + 1)\) and \(x - 2\). So \(x
eq1,-1,2\) (since these values make the denominators zero). If we don't note the non - permissible values before simplifying, when we cancel \((x - 1)\) and \((x + 1)\), we might forget that \(x = 1\) and \(x=-1\) were originally not allowed, even though they don't make the simplified denominator \(x - 2\) zero. So non - permissible values must be noted before simplifying to ensure that we exclude all values that make the original expression undefined, because simplification can sometimes eliminate factors from the denominator that were originally zero - making, but the original expression is still undefined at those values.

Answer:

The simplified expression is \(\frac{2x - 31}{(x + 7)(x - 7)}\) and the non - permissible values are \(x=-7,-1,4,7\)

Question 6