Multiply `(-4)/(3)xy^(3)` by `(6)/(7)x^(2)y`and verify your result for x = 2 and y = 1

Answer

To solve the problem of multiplying \((-4/3)xy^3\) by \((6/7)x^2y\), we will follow these steps:

### Step 1: Write down the expression
We start with the expression:
\[ \left(-\frac{4}{3}xy^3\right) \times \left(\frac{6}{7}x^2y\right) \]

### Step 2: Multiply the coefficients
Next, we multiply the coefficients (the numerical parts) of the two expressions:
\[ -\frac{4}{3} \times \frac{6}{7} = -\frac{4 \times 6}{3 \times 7} = -\frac{24}{21} \]
Now, we can simplify \(-\frac{24}{21}\):
\[ -\frac{24}{21} = -\frac{8}{7} \quad (\text{dividing numerator and denominator by 3}) \]

### Step 3: Multiply the variables
Now, we multiply the variables. When multiplying variables, we add the exponents of like bases:
- For \(x\): \(x^1 \times x^2 = x^{1+2} = x^3\)
- For \(y\): \(y^3 \times y^1 = y^{3+1} = y^4\)

### Step 4: Combine the results
Putting it all together, we get:
\[ -\frac{8}{7} x^3 y^4 \]

### Step 5: Verify the result for \(x = 2\) and \(y = 1\)
Now, we will substitute \(x = 2\) and \(y = 1\) into our expression:
\[ -\frac{8}{7} (2^3) (1^4) = -\frac{8}{7} \times 8 \times 1 = -\frac{64}{7} \]

### Final Result
Thus, the final result of the multiplication is:
\[ -\frac{8}{7} x^3 y^4 \]
And when verified with \(x = 2\) and \(y = 1\), we get:
\[ -\frac{64}{7} \]

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