Question

1. Amanda tells you that because a variable is in the denominator, the equation three over x plus one over three equals five over six becomes unsolvable. Amanda explains, "There is a value for x that makes the denominator zero, and you can't divide by zero." Demonstrate to Amanda how the equation is still solvable and explain your reasoning.
2. When looking at the rational function f of x equals the quantity x minus one times the quantity x plus two times the quantity x plus four all divided by the quantity x plus one times the quantity x minus two times the quantity x minus four, Bella and Edward have two different thoughts. Bella says that the function is defined at x = –1, x = 2, and x = 4. Edward says that the function is undefined at those x values. Who is correct? Justify your reasoning.

Answer 1

Answer: x = 6

Step-by-step explanation:

$\dfrac{3}{x}+\dfrac{1}{3}=\dfrac{5}{6}$

$(6x)\dfrac{3}{x}+(6x)\dfrac{1}{3}=(6x)\dfrac{5}{6}$  multiplied by common denominator

18 + 2x = 5x

      -2x   -2x

18         = 3x

÷3          ÷3

6          =  x

Since "x" is in the denominator, the restriction is that x ≠ 0.  If the solution was x = 0, then the solution would not be valid and would be eliminated as an answer.

Remember that "x" is just an unknown value.  It is possible to add two fractions together and have their sum be a fraction.

For example: $\dfrac{1}{5} + \dfrac{2}{5} = \dfrac{3}{5}$ could be written as $\dfrac{1}{5} + \dfrac{2}{x} = \dfrac{3}{5}$.  When we solve it, we will get x = 5.

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Answer: Edward

Step-by-step explanation:

$f(x)=\dfrac{(x-1)(x+2)(x+4)}{(x+1)(x-2)(x-4)}$

The denominator cannot equal zero, so:

• x + 1 ≠ 0  → x ≠ -1
• x - 2 ≠ 0 → x ≠ 2
• x - 4 ≠ 0 → x ≠ 4

Those x-values are the asymptotes, which is where the function is undefined.