if `a : b = e : d`, show that : `3a + 2b : 3a - 2b = 3c + 2d : 3c - 2d`.

Answer

To prove that if \( a : b = e : d \), then \( 3a + 2b : 3a - 2b = 3e + 2d : 3e - 2d \), we can follow these steps:

### Step 1: Express the given ratio
We start with the given ratio:
\[ \frac{a}{b} = \frac{e}{d} \]
This implies that:
\[ \frac{a}{b} = k \quad \text{(for some constant } k\text{)} \]
Thus, we can express \( a \) and \( b \) in terms of \( k \):
\[ a = kb \quad \text{and} \quad e = kd \]

### Step 2: Substitute \( a \) and \( e \) in the expression
We need to evaluate the left-hand side:
\[ \frac{3a + 2b}{3a - 2b} \]
Substituting \( a = kb \):
\[ \frac{3(kb) + 2b}{3(kb) - 2b} = \frac{(3k + 2)b}{(3k - 2)b} \]

### Step 3: Simplify the expression
Since \( b \) is common in both the numerator and the denominator, we can cancel it out (assuming \( b \neq 0 \)):
\[ \frac{3k + 2}{3k - 2} \]

### Step 4: Express the right-hand side
Now, we evaluate the right-hand side:
\[ \frac{3e + 2d}{3e - 2d} \]
Substituting \( e = kd \):
\[ \frac{3(kd) + 2d}{3(kd) - 2d} = \frac{(3k + 2)d}{(3k - 2)d} \]

### Step 5: Simplify the right-hand side
Similarly, we can cancel \( d \) from both the numerator and the denominator (assuming \( d \neq 0 \)):
\[ \frac{3k + 2}{3k - 2} \]

### Step 6: Conclude the proof
Since both sides simplify to the same expression:
\[ \frac{3k + 2}{3k - 2} = \frac{3k + 2}{3k - 2} \]
We have shown that:
\[ 3a + 2b : 3a - 2b = 3e + 2d : 3e - 2d \]

### Final Result
Thus, we conclude that:
\[ 3a + 2b : 3a - 2b = 3e + 2d : 3e - 2d \]