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
$$\frac{4}{5x + 20} = \frac{8}{x + 4} - \frac{3}{5}$$

a. what is/are the value or values of the variable that make(s) the denominators zero?
x = -4
(simplify your answer. use a comma to separate answers as needed.)
b. 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 ∅.

Explanation:

Part b: Solving the Rational Equation

Step 1: Factor the Denominator

First, factor the denominator \(5x + 20\) as \(5(x + 4)\). So the equation becomes:

$$ \frac{4}{5(x + 4)}=\frac{8}{x + 4}-\frac{3}{5} $$

Step 2: Find the Least Common Denominator (LCD)

The denominators are \(5(x + 4)\), \(x + 4\), and \(5\). The LCD is \(5(x + 4)\).

Step 3: Multiply Each Term by the LCD

Multiply each term by \(5(x + 4)\) to eliminate the denominators:

$$ 5(x + 4)\cdot\frac{4}{5(x + 4)} = 5(x + 4)\cdot\frac{8}{x + 4}-5(x + 4)\cdot\frac{3}{5} $$

Simplify each term:

• Left term: \(4\) (the \(5(x + 4)\) cancels out)
• Middle term: \(5(x + 4)\cdot\frac{8}{x + 4}=5\times8 = 40\) (the \(x + 4\) cancels out)
• Right term: \(5(x + 4)\cdot\frac{3}{5}=(x + 4)\times3 = 3x + 12\) (the \(5\) cancels out)

So the equation simplifies to:

$$ 4 = 40-(3x + 12) $$

Step 4: Simplify the Right Side

Simplify \(40-(3x + 12)\):

$$ 4 = 40 - 3x - 12 $$

$$ 4 = 28 - 3x $$

Step 5: Solve for \(x\)

Subtract \(28\) from both sides:

$$ 4 - 28 = -3x $$

$$ -24 = -3x $$

Divide both sides by \(-3\):

$$ x = \frac{-24}{-3}=8 $$

Step 6: Check the Restriction

We know from part (a) that \(x
eq - 4\). The solution \(x = 8\) does not make any denominator zero, so it is valid.

Answer:

A. The solution set is \(\{8\}\)