Answer :
Certainly! Let's verify the given equation step-by-step to show that both sides are equal.
Given:
[tex]\[ \left(\frac{-5}{3} + \frac{3}{5}\right) + \frac{-5}{8} = \frac{-5}{3} + \left(\frac{3}{5} + \frac{-5}{8}\right) \][/tex]
First, let's calculate the left-hand side (LHS):
### Step 1: Calculate [tex]\(\left(\frac{-5}{3} + \frac{3}{5}\right)\)[/tex]
To add the fractions [tex]\(\frac{-5}{3}\)[/tex] and [tex]\(\frac{3}{5}\)[/tex], we need a common denominator. The least common multiple (LCM) of 3 and 5 is 15. So, we rewrite the fractions with this common denominator:
[tex]\[ \frac{-5}{3} = \frac{-5 \times 5}{3 \times 5} = \frac{-25}{15} \][/tex]
[tex]\[ \frac{3}{5} = \frac{3 \times 3}{5 \times 3} = \frac{9}{15} \][/tex]
Now, add these fractions:
[tex]\[ \frac{-25}{15} + \frac{9}{15} = \frac{-25 + 9}{15} = \frac{-16}{15} \][/tex]
### Step 2: Add [tex]\(\frac{-16}{15}\)[/tex] and [tex]\(\frac{-5}{8}\)[/tex]
Next, we need a common denominator to add [tex]\(\frac{-16}{15}\)[/tex] and [tex]\(\frac{-5}{8}\)[/tex]. The LCM of 15 and 8 is 120. Rewriting the fractions:
[tex]\[ \frac{-16}{15} = \frac{-16 \times 8}{15 \times 8} = \frac{-128}{120} \][/tex]
[tex]\[ \frac{-5}{8} = \frac{-5 \times 15}{8 \times 15} = \frac{-75}{120} \][/tex]
Now, add these fractions:
[tex]\[ \frac{-128}{120} + \frac{-75}{120} = \frac{-128 - 75}{120} = \frac{-203}{120} \][/tex]
Thus, the LHS is:
[tex]\[ \left(\frac{-5}{3} + \frac{3}{5}\right) + \frac{-5}{8} = \frac{-203}{120} \][/tex]
### Step 3: Calculate the right-hand side (RHS)
[tex]\[ \frac{-5}{3} + \left(\frac{3}{5} + \frac{-5}{8}\right) \][/tex]
### Step 4: Calculate [tex]\(\left(\frac{3}{5} + \frac{-5}{8}\right)\)[/tex]
Using the same common denominator as before (LCM of 5 and 8 is 40):
[tex]\[ \frac{3}{5} = \frac{3 \times 8}{5 \times 8} = \frac{24}{40} \][/tex]
[tex]\[ \frac{-5}{8} = \frac{-5 \times 5}{8 \times 5} = \frac{-25}{40} \][/tex]
Now add these fractions:
[tex]\[ \frac{24}{40} + \frac{-25}{40} = \frac{24 - 25}{40} = \frac{-1}{40} \][/tex]
### Step 5: Add [tex]\(\frac{-5}{3}\)[/tex] and [tex]\(\frac{-1}{40}\)[/tex]
We need a common denominator (LCM of 3 and 40 is 120):
[tex]\[ \frac{-5}{3} = \frac{-5 \times 40}{3 \times 40} = \frac{-200}{120} \][/tex]
[tex]\[ \frac{-1}{40} = \frac{-1 \times 3}{40 \times 3} = \frac{-3}{120} \][/tex]
Now, add these fractions:
[tex]\[ \frac{-200}{120} + \frac{-3}{120} = \frac{-200 - 3}{120} = \frac{-203}{120} \][/tex]
Thus, the RHS is:
[tex]\[ \frac{-5}{3} + \left(\frac{3}{5} + \frac{-5}{8}\right) = \frac{-203}{120} \][/tex]
### Conclusion
Both the left-hand side and right-hand side are equal:
[tex]\[ \left(\frac{-5}{3} + \frac{3}{5}\right) + \frac{-5}{8} = \frac{-5}{3} + \left(\frac{3}{5} + \frac{-5}{8}\right) = \frac{-203}{120} \][/tex]
This verifies the given equation. Therefore, the final verification is:
[tex]\[ \frac{-203}{120} = \frac{-203}{120} \][/tex]
Hence, the verification is true.
Given:
[tex]\[ \left(\frac{-5}{3} + \frac{3}{5}\right) + \frac{-5}{8} = \frac{-5}{3} + \left(\frac{3}{5} + \frac{-5}{8}\right) \][/tex]
First, let's calculate the left-hand side (LHS):
### Step 1: Calculate [tex]\(\left(\frac{-5}{3} + \frac{3}{5}\right)\)[/tex]
To add the fractions [tex]\(\frac{-5}{3}\)[/tex] and [tex]\(\frac{3}{5}\)[/tex], we need a common denominator. The least common multiple (LCM) of 3 and 5 is 15. So, we rewrite the fractions with this common denominator:
[tex]\[ \frac{-5}{3} = \frac{-5 \times 5}{3 \times 5} = \frac{-25}{15} \][/tex]
[tex]\[ \frac{3}{5} = \frac{3 \times 3}{5 \times 3} = \frac{9}{15} \][/tex]
Now, add these fractions:
[tex]\[ \frac{-25}{15} + \frac{9}{15} = \frac{-25 + 9}{15} = \frac{-16}{15} \][/tex]
### Step 2: Add [tex]\(\frac{-16}{15}\)[/tex] and [tex]\(\frac{-5}{8}\)[/tex]
Next, we need a common denominator to add [tex]\(\frac{-16}{15}\)[/tex] and [tex]\(\frac{-5}{8}\)[/tex]. The LCM of 15 and 8 is 120. Rewriting the fractions:
[tex]\[ \frac{-16}{15} = \frac{-16 \times 8}{15 \times 8} = \frac{-128}{120} \][/tex]
[tex]\[ \frac{-5}{8} = \frac{-5 \times 15}{8 \times 15} = \frac{-75}{120} \][/tex]
Now, add these fractions:
[tex]\[ \frac{-128}{120} + \frac{-75}{120} = \frac{-128 - 75}{120} = \frac{-203}{120} \][/tex]
Thus, the LHS is:
[tex]\[ \left(\frac{-5}{3} + \frac{3}{5}\right) + \frac{-5}{8} = \frac{-203}{120} \][/tex]
### Step 3: Calculate the right-hand side (RHS)
[tex]\[ \frac{-5}{3} + \left(\frac{3}{5} + \frac{-5}{8}\right) \][/tex]
### Step 4: Calculate [tex]\(\left(\frac{3}{5} + \frac{-5}{8}\right)\)[/tex]
Using the same common denominator as before (LCM of 5 and 8 is 40):
[tex]\[ \frac{3}{5} = \frac{3 \times 8}{5 \times 8} = \frac{24}{40} \][/tex]
[tex]\[ \frac{-5}{8} = \frac{-5 \times 5}{8 \times 5} = \frac{-25}{40} \][/tex]
Now add these fractions:
[tex]\[ \frac{24}{40} + \frac{-25}{40} = \frac{24 - 25}{40} = \frac{-1}{40} \][/tex]
### Step 5: Add [tex]\(\frac{-5}{3}\)[/tex] and [tex]\(\frac{-1}{40}\)[/tex]
We need a common denominator (LCM of 3 and 40 is 120):
[tex]\[ \frac{-5}{3} = \frac{-5 \times 40}{3 \times 40} = \frac{-200}{120} \][/tex]
[tex]\[ \frac{-1}{40} = \frac{-1 \times 3}{40 \times 3} = \frac{-3}{120} \][/tex]
Now, add these fractions:
[tex]\[ \frac{-200}{120} + \frac{-3}{120} = \frac{-200 - 3}{120} = \frac{-203}{120} \][/tex]
Thus, the RHS is:
[tex]\[ \frac{-5}{3} + \left(\frac{3}{5} + \frac{-5}{8}\right) = \frac{-203}{120} \][/tex]
### Conclusion
Both the left-hand side and right-hand side are equal:
[tex]\[ \left(\frac{-5}{3} + \frac{3}{5}\right) + \frac{-5}{8} = \frac{-5}{3} + \left(\frac{3}{5} + \frac{-5}{8}\right) = \frac{-203}{120} \][/tex]
This verifies the given equation. Therefore, the final verification is:
[tex]\[ \frac{-203}{120} = \frac{-203}{120} \][/tex]
Hence, the verification is true.
