Find the values of a and b for which the following pair of equations have infinitely many solutions:
`(i) 2x + 3y = 7 and 2ax + ay =28-by`
(ii) ` 2x + 3y =7, (a-b) x + (a +b) y= 3a + b-2`
(iii) `2x - (2a + 5) y =5, (2b +1) x - 9y =15`

Answer

To find the values of \( a \) and \( b \) for which the given pairs of equations have infinitely many solutions, we will use the condition that for two linear equations to have infinitely many solutions, the ratios of their coefficients must be equal.

### (i) Given Equations:
1. \( 2x + 3y = 7 \)
2. \( 2ax + ay = 28 - by \)

**Step 1:** Rewrite the second equation in standard form:
\[ 2ax + ay + by = 28 \]
This can be rearranged to:
\[ 2ax + (a + b)y = 28 \]

**Step 2:** Set up the ratios:
\[ \frac{2}{2a} = \frac{3}{a + b} = \frac{7}{28} \]

**Step 3:** Simplify the third ratio:
\[ \frac{7}{28} = \frac{1}{4} \]

**Step 4:** From the first ratio:
\[ \frac{2}{2a} = \frac{1}{4} \implies 2 = \frac{2a}{4} \implies 2 = \frac{a}{2} \implies a = 4 \]

**Step 5:** Substitute \( a = 4 \) into the second ratio:
\[ \frac{3}{4 + b} = \frac{1}{4} \]
Cross-multiplying gives:
\[ 3 \cdot 4 = 1 \cdot (4 + b) \implies 12 = 4 + b \implies b = 8 \]

**Final Values for (i):**
\[ a = 4, \quad b = 8 \]

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### (ii) Given Equations:
1. \( 2x + 3y = 7 \)
2. \( (a - b)x + (a + b)y = 3a + b - 2 \)

**Step 1:** Set up the ratios:
\[ \frac{2}{a - b} = \frac{3}{a + b} = \frac{7}{3a + b - 2} \]

**Step 2:** From the first ratio:
\[ \frac{2}{a - b} = \frac{7}{3a + b - 2} \]
Cross-multiplying gives:
\[ 2(3a + b - 2) = 7(a - b) \]
Expanding both sides:
\[ 6a + 2b - 4 = 7a - 7b \]
Rearranging gives:
\[ 6a + 2b + 7b = 7a + 4 \implies 6a + 9b = 7a + 4 \implies a - 9b = 4 \quad \text{(Equation 1)} \]

**Step 3:** From the second ratio:
\[ \frac{3}{a + b} = \frac{7}{3a + b - 2} \]
Cross-multiplying gives:
\[ 3(3a + b - 2) = 7(a + b) \]
Expanding both sides:
\[ 9a + 3b - 6 = 7a + 7b \]
Rearranging gives:
\[ 9a - 7a + 3b - 7b = 6 \implies 2a - 4b = 6 \implies a - 2b = 3 \quad \text{(Equation 2)} \]

**Step 4:** Solve the system of equations (1) and (2):
1. \( a - 9b = 4 \)
2. \( a - 2b = 3 \)

Subtracting the second from the first:
\[ (-9b + 2b) = 4 - 3 \implies -7b = 1 \implies b = -\frac{1}{7} \]
Substituting \( b \) back into Equation 2:
\[ a - 2(-\frac{1}{7}) = 3 \implies a + \frac{2}{7} = 3 \implies a = 3 - \frac{2}{7} = \frac{21}{7} - \frac{2}{7} = \frac{19}{7} \]

**Final Values for (ii):**
\[ a = \frac{19}{7}, \quad b = -\frac{1}{7} \]

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### (iii) Given Equations:
1. \( 2x - (2a + 5)y = 5 \)
2. \( (2b + 1)x - 9y = 15 \)

**Step 1:** Set up the ratios:
\[ \frac{2}{2b + 1} = \frac{-(2a + 5)}{-9} = \frac{5}{15} \]

**Step 2:** Simplify the third ratio:
\[ \frac{5}{15} = \frac{1}{3} \]

**Step 3:** From the first ratio:
\[ \frac{2}{2b + 1} = \frac{1}{3} \]
Cross-multiplying gives:
\[ 2 \cdot 3 = 1 \cdot (2b + 1) \implies 6 = 2b + 1 \implies 2b = 5 \implies b = \frac{5}{2} \]

**Step 4:** Substitute \( b = \frac{5}{2} \) into the second ratio:
\[ \frac{-(2a + 5)}{-9} = \frac{1}{3} \]
Cross-multiplying gives:
\[ -(2a + 5) = -3 \implies 2a + 5 = 3 \implies 2a = 3 - 5 \implies 2a = -2 \implies a = -1 \]

**Final Values for (iii):**
\[ a = -1, \quad b = \frac{5}{2} \]

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### Summary of Results:
1. For (i): \( a = 4, \quad b = 8 \)
2. For (ii): \( a = \frac{19}{7}, \quad b = -\frac{1}{7} \)
3. For (iii): \( a = -1, \quad b = \frac{5}{2} \)

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