Suppose y varies as the sum of two quantities of which one vasries directly as x and the other inversely as x. If y=6 when x=4 and `y=3 1/3` when x=3 then the relation between x and y

Options

A. `x=y+4`
B. `y=2x+8/x`
C. `7=2x-8/x`
D. `y=2x-4/x`

Answer

To solve the problem, we need to establish the relationship between \( y \) and \( x \) based on the given conditions.

### Step-by-Step Solution:

1. **Understanding the Variation**:
Since \( y \) varies as the sum of two quantities, one of which varies directly as \( x \) (let's denote it as \( k_1 x \)) and the other varies inversely as \( x \) (let's denote it as \( \frac{k_2}{x} \)), we can express this relationship as:
\[ y = k_1 x + \frac{k_2}{x} \]

2. **Using the First Condition**:
We know that \( y = 6 \) when \( x = 4 \). Substituting these values into the equation gives:
\[ 6 = k_1(4) + \frac{k_2}{4} \]
Simplifying this, we get:
\[ 6 = 4k_1 + \frac{k_2}{4} \]
Multiplying through by 4 to eliminate the fraction:
\[ 24 = 16k_1 + k_2 \quad \text{(Equation 1)} \]

3. **Using the Second Condition**:
Next, we use the second condition where \( y = 3 \frac{1}{3} = \frac{10}{3} \) when \( x = 3 \). Substituting these values gives:
\[ \frac{10}{3} = k_1(3) + \frac{k_2}{3} \]
Multiplying through by 3 to eliminate the fraction:
\[ 10 = 3k_1 + k_2 \quad \text{(Equation 2)} \]

4. **Solving the Equations**:
Now we have two equations:
- \( 24 = 16k_1 + k_2 \) (Equation 1)
- \( 10 = 3k_1 + k_2 \) (Equation 2)

We can subtract Equation 2 from Equation 1:
\[ 24 - 10 = (16k_1 + k_2) - (3k_1 + k_2) \]
This simplifies to:
\[ 14 = 13k_1 \]
Therefore, solving for \( k_1 \):
\[ k_1 = \frac{14}{13} \]

5. **Finding \( k_2 \)**:
Now, substitute \( k_1 \) back into Equation 2 to find \( k_2 \):
\[ 10 = 3\left(\frac{14}{13}\right) + k_2 \]
Simplifying this:
\[ 10 = \frac{42}{13} + k_2 \]
To isolate \( k_2 \):
\[ k_2 = 10 - \frac{42}{13} = \frac{130}{13} - \frac{42}{13} = \frac{88}{13} \]

6. **Final Relation**:
Now we can write the final relation between \( y \) and \( x \):
\[ y = \frac{14}{13}x + \frac{\frac{88}{13}}{x} \]
This can be simplified to:
\[ y = \frac{14x + 88}{13x} \]

### Final Answer:
The relation between \( x \) and \( y \) is:
\[ y = \frac{14x + 88}{13x} \]