Answer :
Let's determine the domains for each rational equation and match them accordingly.
1. For the equation [tex]\( y = \frac{x^2 - 5x - 14}{x^2 + 14x - 275} \)[/tex]:
- The denominator is [tex]\( x^2 + 14x - 275 \)[/tex].
- To find the values that make the denominator zero, solve [tex]\( x^2 + 14x - 275 = 0 \)[/tex].
- The solutions are [tex]\( x = 11 \)[/tex] and [tex]\( x = -25 \)[/tex].
- Therefore, the domain is all real numbers except [tex]\( x \neq 11 \)[/tex] and [tex]\( x \neq -25 \)[/tex].
2. For the equation [tex]\( y = \frac{4x}{x + 25} \)[/tex]:
- The denominator is [tex]\( x + 25 \)[/tex].
- To find the values that make the denominator zero, solve [tex]\( x + 25 = 0 \)[/tex].
- The solution is [tex]\( x = -25 \)[/tex].
- Therefore, the domain is all real numbers except [tex]\( x \neq -25 \)[/tex].
3. For the equation [tex]\( y = \frac{x^2 - 8x}{x^2 + 3} \)[/tex]:
- The denominator is [tex]\( x^2 + 3 \)[/tex].
- This expression is never zero for any real number [tex]\( x \)[/tex] since [tex]\( x^2 \geq 0 \)[/tex].
- Therefore, [tex]\( x^2 + 3 \)[/tex] is always positive.
- Hence, the domain is all real numbers.
4. For the equation [tex]\( y = \frac{x + 1}{x^2 - 10x - 11} \)[/tex]:
- The denominator is [tex]\( x^2 - 10x - 11 \)[/tex].
- To find the values that make the denominator zero, solve [tex]\( x^2 - 10x - 11 = 0 \)[/tex].
- The solutions are [tex]\( x = 11 \)[/tex] and [tex]\( x = -1 \)[/tex].
- Therefore, the domain is all real numbers except [tex]\( x \neq 11 \)[/tex] and [tex]\( x \neq -1 \)[/tex].
Now, let's match the domains to the rational equations.
- [tex]\( x \neq 11, x \neq -25 \)[/tex] matches with [tex]\( y = \frac{x^2 - 5x - 14}{x^2 + 14x - 275} \)[/tex].
- [tex]\( x \neq -25 \)[/tex] matches with [tex]\( y = \frac{4x}{x + 25} \)[/tex].
- [tex]\( \text{all real numbers} \)[/tex] matches with [tex]\( y = \frac{x^2 - 8x}{x^2 + 3} \)[/tex].
- [tex]\( x \neq 11, x \neq -1 \)[/tex] matches with [tex]\( y = \frac{x + 1}{x^2 - 10x - 11} \)[/tex].
So, the complete matching is:
- [tex]\( x \neq 11, x \neq -25 \)[/tex] → [tex]\( y = \frac{x^2 - 5x - 14}{x^2 + 14x - 275} \)[/tex]
- [tex]\( x \neq -25 \)[/tex] → [tex]\( y = \frac{4x}{x + 25} \)[/tex]
- [tex]\( \text{all real numbers} \)[/tex] → [tex]\( y = \frac{x^2 - 8x}{x^2 + 3} \)[/tex]
- [tex]\( x \neq 11, x \neq -1 \)[/tex] → [tex]\( y = \frac{x + 1}{x^2 - 10x - 11} \)[/tex]
1. For the equation [tex]\( y = \frac{x^2 - 5x - 14}{x^2 + 14x - 275} \)[/tex]:
- The denominator is [tex]\( x^2 + 14x - 275 \)[/tex].
- To find the values that make the denominator zero, solve [tex]\( x^2 + 14x - 275 = 0 \)[/tex].
- The solutions are [tex]\( x = 11 \)[/tex] and [tex]\( x = -25 \)[/tex].
- Therefore, the domain is all real numbers except [tex]\( x \neq 11 \)[/tex] and [tex]\( x \neq -25 \)[/tex].
2. For the equation [tex]\( y = \frac{4x}{x + 25} \)[/tex]:
- The denominator is [tex]\( x + 25 \)[/tex].
- To find the values that make the denominator zero, solve [tex]\( x + 25 = 0 \)[/tex].
- The solution is [tex]\( x = -25 \)[/tex].
- Therefore, the domain is all real numbers except [tex]\( x \neq -25 \)[/tex].
3. For the equation [tex]\( y = \frac{x^2 - 8x}{x^2 + 3} \)[/tex]:
- The denominator is [tex]\( x^2 + 3 \)[/tex].
- This expression is never zero for any real number [tex]\( x \)[/tex] since [tex]\( x^2 \geq 0 \)[/tex].
- Therefore, [tex]\( x^2 + 3 \)[/tex] is always positive.
- Hence, the domain is all real numbers.
4. For the equation [tex]\( y = \frac{x + 1}{x^2 - 10x - 11} \)[/tex]:
- The denominator is [tex]\( x^2 - 10x - 11 \)[/tex].
- To find the values that make the denominator zero, solve [tex]\( x^2 - 10x - 11 = 0 \)[/tex].
- The solutions are [tex]\( x = 11 \)[/tex] and [tex]\( x = -1 \)[/tex].
- Therefore, the domain is all real numbers except [tex]\( x \neq 11 \)[/tex] and [tex]\( x \neq -1 \)[/tex].
Now, let's match the domains to the rational equations.
- [tex]\( x \neq 11, x \neq -25 \)[/tex] matches with [tex]\( y = \frac{x^2 - 5x - 14}{x^2 + 14x - 275} \)[/tex].
- [tex]\( x \neq -25 \)[/tex] matches with [tex]\( y = \frac{4x}{x + 25} \)[/tex].
- [tex]\( \text{all real numbers} \)[/tex] matches with [tex]\( y = \frac{x^2 - 8x}{x^2 + 3} \)[/tex].
- [tex]\( x \neq 11, x \neq -1 \)[/tex] matches with [tex]\( y = \frac{x + 1}{x^2 - 10x - 11} \)[/tex].
So, the complete matching is:
- [tex]\( x \neq 11, x \neq -25 \)[/tex] → [tex]\( y = \frac{x^2 - 5x - 14}{x^2 + 14x - 275} \)[/tex]
- [tex]\( x \neq -25 \)[/tex] → [tex]\( y = \frac{4x}{x + 25} \)[/tex]
- [tex]\( \text{all real numbers} \)[/tex] → [tex]\( y = \frac{x^2 - 8x}{x^2 + 3} \)[/tex]
- [tex]\( x \neq 11, x \neq -1 \)[/tex] → [tex]\( y = \frac{x + 1}{x^2 - 10x - 11} \)[/tex]