Find the value of a and b so that following sysyem of equation has infinitely many solutions
`2x - (a - 4) y = 2b +1 `
`4x - (a-1) y = 5 b -1`

Answer

To find the values of \( a \) and \( b \) such that the given system of equations has infinitely many solutions, we need to ensure that the two equations represent the same line. This occurs when the ratios of the coefficients of \( x \), \( y \), and the constant terms are equal.

The given equations are:
1. \( 2x - (a - 4)y = 2b + 1 \)
2. \( 4x - (a - 1)y = 5b - 1 \)

Let's rewrite these equations in the standard form \( Ax + By + C = 0 \).

1. Rearranging the first equation:
\[ 2x - (a - 4)y - (2b + 1) = 0 \]
Here, \( A_1 = 2 \), \( B_1 = -(a - 4) \), and \( C_1 = -(2b + 1) \).

2. Rearranging the second equation:
\[ 4x - (a - 1)y - (5b - 1) = 0 \]
Here, \( A_2 = 4 \), \( B_2 = -(a - 1) \), and \( C_2 = -(5b - 1) \).

For the two lines to be coincident, the following ratios must hold:
\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} \]

Substituting the values we have:
1. From \( \frac{A_1}{A_2} \):
\[ \frac{2}{4} = \frac{1}{2} \]

2. From \( \frac{B_1}{B_2} \):
\[ \frac{-(a - 4)}{-(a - 1)} = \frac{a - 4}{a - 1} \]
Setting this equal to \( \frac{1}{2} \):
\[ \frac{a - 4}{a - 1} = \frac{1}{2} \]
Cross-multiplying gives:
\[ 2(a - 4) = 1(a - 1) \]
Simplifying:
\[ 2a - 8 = a - 1 \]
\[ 2a - a = 8 - 1 \]
\[ a = 7 \]

3. From \( \frac{C_1}{C_2} \):
\[ \frac{-(2b + 1)}{-(5b - 1)} = \frac{2b + 1}{5b - 1} \]
Setting this equal to \( \frac{1}{2} \):
\[ \frac{2b + 1}{5b - 1} = \frac{1}{2} \]
Cross-multiplying gives:
\[ 2(2b + 1) = 1(5b - 1) \]
Simplifying:
\[ 4b + 2 = 5b - 1 \]
\[ 4b - 5b = -1 - 2 \]
\[ -b = -3 \]
\[ b = 3 \]

Thus, the values of \( a \) and \( b \) are:
\[ a = 7, \quad b = 3 \]