To divide the fractions, you first need to invert (take the reciprocal of) the second fraction and then multiply.
Here is the division problem:
\[
\frac{\frac{12}{10x} - \frac{40}{4}}{\frac{2x - 8}{1}}
\]
Step 1: Simplify the fractions where possible.
For the numerator:
\[
\frac{12}{10x} - \frac{40}{4}
\]
Simplify \(\frac{40}{4}\):
\[
\frac{40}{4} = 10
\]
Then, the numerator becomes:
\[
\frac{12}{10x} - 10
\]
For the denominator:
\[
\frac{2x - 8}{1}
\]
This is already simplified as \(2x - 8\).
However, let's modify the problem to make it simpler and clearer:
Let's rewrite the original problem without confusion:
\[
\left( \frac{12}{10x} - \frac{40}{4} \right) \div \left( \frac{2x - 8}{1} \right)
\]
Step 2: Simplify the terms:
\[
\frac{12}{10x} = \frac{6}{5x} \quad \text{and} \quad \frac{40}{4} = 10
\]
So you have:
\[
\frac{6}{5x} - 10
\]
The denominator after simplification is:
\[
2x - 8 = 2(x-4)
\]
Step 3: Invert the bottom fraction and multiply:
The problem now looks like this:
\[
\left( \frac{6}{5x} - 10 \right) \times \frac{1}{2(x-4)}
\]
Let's simplify \(\frac{1}{2(x-4)}\) and multiply it by each term in the numerator:
\[
\left( \frac{6}{5x} \times \frac{1}{2(x-4)} \right) - \left( 10 \times \frac{1}{2(x-4)} \right)
\]
This leads to:
\[
\frac{6}{10x(x-4)} - \frac{10}{2(x-4)}
\]
Simplify each term:
\[
\frac{3}{5x(x-4)} - \frac{5}{x-4}
\]
Combine the fractions with a common denominator:
\[
\frac{3 - 5 \cdot 5x}{5x(x-4)} = \frac{3 - 25x}{5x(x-4)}
\]
This is the simplified form:
\[
\frac{3 - 25x}{5x(x-4)}
\]
divide these fractions.
12/10x-40/4/2x-8
invert the bottom fraction and then multiply it by the first fraction
1 answer
1 answer