For what value of a the given two system of equations have no solutions 3x 5y 22 ax 20y 40?

Find the value of k for which the system of equations 3x + 5y = 0 and kx + 10y = 0 has infinite nonzero solutions.

The given system is3x + 5y = 0                          ……(i)kx + 10y = 0                        ……(ii)This is a homogeneous system of linear differential equation, so it always has a zero solution i.e., x = y = 0.But to have a non-zero solution, it must have infinitely many solutions.For this, we have`(a_1)/(a_2) = (b_1)/(b_2)``⇒ 3/k = 5/10 = 1/2`⇒ k = 6

Hence, k = 6.

Concept: Pair of Linear Equations in Two Variables

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The given system iskx – y – 2 = 0                   ……(i)6x – 2y – 3 = 0                 ……(ii)Here, `a_1 = k, b_1 = -1, c_1 = -2, a_2 = 6, b_2 = -2 and c_2 = -3`For the system, to have a unique solution, we must have`(a_1)/(a_2) ≠ (b_1)/(b_2)``⇒ k/6 ≠ (−1)/(−2) = 1/2`⇒ k ≠ 3

Hence, k ≠ 3.

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The given system is2x + 3y – 5 = 0                         ……(i)4x + ky – 10 = 0                    ……(ii)Here,` a_1 = 2, b_1 = 3, c_1 = -5, a_2 = 4, b_2 = k and c_2 = -10`For the system, to have an infinite number of solutions, we must have`(a_1)/(a_2) = (b_1)/(b_2) = (c_1)/(c_2)``⇒ 2/4 = 3/k = (−5)/(−10)``⇒ 1/2 = 3/k= 1/2`⇒ k = 6

Hence, k = 6.

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The given system is2x + 3y – 1 = 0                    ……(i)4x + 6y – 4 = 0                   ……(ii)Here,` a_1 = 2, b_1 = 3, c_1 = -1, a_2 = 4, b_2 = 6 and c_2 = -4`Now,`(a_1)/(a_2) = 2/4 = 1/2``(b_1)/(b_2) = 3/6 = 1/2``(c_1)/(c_2) = (−1)/(−4) = 1/4`

Thus, `(a_1)/(a_2) = (b_1)/(b_2) ≠ (c_1)/(c_2 )` and therefore the given system has no solution.