Saved Bookmarks
| 1. |
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. |
|
Answer» 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. |
|