Sure! Let's go through the steps for proving the given equation:
[tex]\[
\left(\frac{-8}{9} \times \frac{-1}{5}\right) + \left(\frac{-8}{9} \times \frac{-7}{11}\right) = \frac{-8}{9} \times \left(\frac{-1}{5} + \frac{-7}{11}\right)
\][/tex]
### Step 1: Calculate the Left-Hand Side (LHS)
First, we will calculate each term on the left-hand side separately.
1. First Term: [tex]\(\frac{-8}{9} \times \frac{-1}{5}\)[/tex]
[tex]\[
\frac{-8}{9} \times \frac{-1}{5} = \frac{8}{45} \approx 0.17777777777777776
\][/tex]
2. Second Term: [tex]\(\frac{-8}{9} \times \frac{-7}{11}\)[/tex]
[tex]\[
\frac{-8}{9} \times \frac{-7}{11} = \frac{56}{99} \approx 0.5656565656565656
\][/tex]
Next, we add the two results to get the total left-hand side:
[tex]\[
\frac{8}{45} + \frac{56}{99} \approx 0.17777777777777776 + 0.5656565656565656 = 0.7434343434343433
\][/tex]
### Step 2: Calculate the Right-Hand Side (RHS)
Now, we will calculate the right-hand side of the equation:
1. Sum Inside the Parentheses: [tex]\(\frac{-1}{5} + \frac{-7}{11}\)[/tex]
First, get a common denominator (LCM of 5 and 11 is 55):
[tex]\[
\frac{-1}{5} = \frac{-11}{55}, \quad \frac{-7}{11} = \frac{-35}{55}
\][/tex]
Now, add the fractions:
[tex]\[
\frac{-11}{55} + \frac{-35}{55} = \frac{-46}{55}
\][/tex]
2. Multiply by [tex]\(\frac{-8}{9}\)[/tex]: [tex]\(\frac{-8}{9} \times \frac{-46}{55}\)[/tex]
[tex]\[
\frac{-8}{9} \times \frac{-46}{55} = \frac{368}{495} \approx 0.7434343434343434
\][/tex]
### Step 3: Comparison
Finally, compare the numerical values of the left-hand side and the right-hand side:
[tex]\[
0.7434343434343433 \quad \text{and} \quad 0.7434343434343434
\][/tex]
We note a small numerical difference indicating that the exact equality does not hold perfectly in precise decimal arithmetic. Thus,
[tex]\[
\left(\frac{-8}{9} \times \frac{-1}{5}\right) + \left(\frac{-8}{9} \times \frac{-7}{11}\right) \ne \frac{-8}{9} \times \left(\frac{-1}{5} + \frac{-7}{11}\right)
\][/tex]
The mathematical identities involved assume precise calculation, but due to splitting and combining floating-point arithmetic, there may appear discrepancies. Therefore, in this example, the given equality does not hold exactly.
