Question

the following rational equation has denominators that contain variables. for this equation, a. write the value or values of the variable that make a denominator zero. these are the restrictions on the variable. b. keeping the restrictions in mind, solve the equation.
\\(\frac{2}{x + 2} + 5 = \frac{17}{x + 2}\\)

a. what is/are the value or values of the variable that make(s) the denominators zero?
x = -2
(simplify your answer. use a comma to separate answers as needed.)

b. solve the equation. select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the solution set is { }
(type an integer or a simplified fraction. use a comma to separate answers as needed.)
b. the solution set is {x|x is a real number}
c. the solution set is \\(\varnothing\\)

Explanation:

Part b: Solving the Equation

Step 1: Identify the common denominator

The denominators in the equation \(\frac{2}{x + 2}+5=\frac{17}{x + 2}\) are both \(x + 2\) (and we know \(x
eq - 2\) from part a). Multiply each term by \(x + 2\) to eliminate the denominators.

$$ (x + 2)\times\frac{2}{x + 2}+5\times(x + 2)=(x + 2)\times\frac{17}{x + 2} $$

Step 2: Simplify each term

Simplify the left - hand side and the right - hand side:

• For the first term \((x + 2)\times\frac{2}{x + 2}\), the \(x + 2\) terms cancel out, leaving \(2\).
• For the second term \(5\times(x + 2)\), we use the distributive property \(a(b + c)=ab+ac\), so \(5\times(x + 2)=5x+10\).
• For the right - hand side \((x + 2)\times\frac{17}{x + 2}\), the \(x + 2\) terms cancel out, leaving \(17\).

So the equation becomes:

$$ 2+5x + 10=17 $$

Step 3: Combine like terms

Combine the constant terms on the left - hand side: \(2 + 10=12\). So the equation is \(5x+12 = 17\).

Step 4: Solve for \(x\)

Subtract \(12\) from both sides of the equation:

$$ 5x+12-12=17 - 12 $$

$$ 5x=5 $$

Divide both sides by \(5\):

$$ x = 1 $$

We need to check if \(x = 1\) is not a restricted value (we know \(x
eq - 2\), and \(1
eq - 2\)).

Answer:

A. The solution set is \(\{1\}\)