Show that the system 2x + 3y = 1, 4x + 6y = 4, has no solutions.

1.	Show that the system 2x + 3y = 1, 4x + 6y = 4, has no solutions.
Answer» Given system of equations are 2x + 3y = 1 and 4x + 6y = 4. Comparing given system of equations with a₁x + b₁y = c₁ and a₂x + b₂y = c₂, we get a₁ = 2, b₁= 3, c₁ = 1 and a₂= 4, b₂ = 6, c₂ = 4. Now, \(\frac{a_1}{a_2}= \frac{2}{4} = \frac{1}{2},\) \(\frac{b_1}{b_2}\) = \(\frac{3}{6} = \frac{1}{2}\) & \(\frac{c_1}{c_2}\) = \(\frac{1}{4}\) Hence, \(\frac{a_1}{a_2} = \frac{b_1}{b_2}\) = \(\frac{1}{2} \neq \frac{1}{4} = \frac{c_1}{c_2}\) Hence, \(\frac{a_1}{a_2} = \frac{b_1}{b_2}\) \(\neq\) \(\frac{c_1}{c_2}\), which is condition of No Solution. Hence, given system of equations has no solution. Hence proved

Answer»

Given system of equations are 2x + 3y = 1 and 4x + 6y = 4.

Comparing given system of equations with a₁x + b₁y = c₁ and a₂x + b₂y = c₂, we get

a₁ = 2, b₁= 3, c₁ = 1 and a₂= 4, b₂ = 6, c₂ = 4.

Now, \(\frac{a_1}{a_2}= \frac{2}{4} = \frac{1}{2},\)

\(\frac{b_1}{b_2}\) = \(\frac{3}{6} = \frac{1}{2}\) & \(\frac{c_1}{c_2}\) = \(\frac{1}{4}\)

Hence, \(\frac{a_1}{a_2} = \frac{b_1}{b_2}\) = \(\frac{1}{2} \neq \frac{1}{4} = \frac{c_1}{c_2}\)

Hence, \(\frac{a_1}{a_2} = \frac{b_1}{b_2}\) \(\neq\) \(\frac{c_1}{c_2}\), which is condition of No Solution.

Hence, given system of equations has no solution.

Hence proved

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