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If equations ( 2 x-3 y=7 ) and  (a+b) x-(4 a+b) y=(a+b-3)  have no solution then a and  b  satisfy the equation.

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Given system of equations is

2x - 3y = 7

(a + b)x - (4a + b)y = a + b -3

system has no solution if \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

\(\Rightarrow\) \(\frac{2}{a + b} = \frac{-3}{-(4a + b)}\) \(\neq \frac{7}{a + b -3}\)

 \(\Rightarrow\) \(\frac{2}{a + b} = \frac{3}{4a + b}\)

\(\Rightarrow\) 2(4a + b) = 3(a + b)

\(\Rightarrow\) 8a + 2b = 3a + 3b

\(\Rightarrow\) 5a - b = 0

\(\Rightarrow\) b = 5a

Also \(\frac{2}{a + b} \neq \frac{7}{a + b - 3}\)

\(\Rightarrow\) 2a + 2b - 6 \(\neq\) 7a + 7b

\(\Rightarrow\) 5a + 5b \(\neq\) -6

\(\Rightarrow\) b + 5b \(\neq\) -6

\(\Rightarrow\) 6b \(\neq\) -6

\(\Rightarrow\) b \(\neq\) -1

Hence, given system has no solution if a and b satisfies the equation b = 5a but b \(\neq\) -1.

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