For what value of k will the pair of linear equations 2x 5y 0 and KX 10y 0has a non zero solution?

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Hint: We will use the concept that for unique solution $\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$ and for many solution we have $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$.

Complete step-by-step answer:

It is given in the question that we have to find the value of k for which the system of equations $3x+5y=0$ and $kx+10y=0$ has a non-zero solution. We know that for a non-zero solution we have two cases. Let us see the cases.Case 1. When we have a unique solution.In this case, the ratio of constants in the equation is as follows, $\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$.In our given equation, $3x+5y=0$ and $kx+10y=0$, the constant values are,$\begin{align} & {{a}_{1}}=3,{{b}_{1}}=5,{{c}_{1}}=0 \\ & {{a}_{2}}=k,{{b}_{2}}=10,{{c}_{2}}=0 \\ \end{align}$So, on putting the values of ${{a}_{1}},{{b}_{1}},{{c}_{1}},{{a}_{2}},{{b}_{2}},{{c}_{2}}$, we get,$\begin{align} & \dfrac{3}{k}\ne \dfrac{5}{10} \\ & 30\ne 5k \\ & k\ne 6 \\ \end{align}$Case 2. When we have many solutions.In this case, the ratio of constants in the equation is as follows, $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$.On putting the values of ${{a}_{1}},{{b}_{1}},{{c}_{1}},{{a}_{2}},{{b}_{2}},{{c}_{2}}$, we get,$\begin{align} & \dfrac{3}{k}=\dfrac{5}{10} \\ & 5k=30 \\ & k=6 \\ \end{align}$Thus, from these two cases, we get the value of k as equal to 6 and not equal to 6. But, we will consider the value of k as 6, because case 2 satisfies our condition when we put the value of k as 6 in the equation, we will get many non-zero solutions.Thus, option (c) is the correct answer.Note: Many times, the students only solve case 1, that is the case for a unique solution, and when they do not get any option similar to their obtained answer, they may think that the question is wrong, but actually they have solved only half the question. So, it is recommended that the students solve the whole question to get the correct answer.

Find the value of k for which the system 3x+5y=0,kx+10y=0 has a nonzero solution.

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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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Page 2

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.

Page 3

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.

Page 4

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.