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Question

# Find the value of k for which each of the following system of equations has no solution: $3x+y=1,\left(2k-1\right)x+\left(k-1\right)y=\left(2k+1\right).$

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Solution

## The given system of equations: 3x + y = 1 ⇒ 3x + y − 1 = 0 ...(i) And, (2k − 1)x + (k − 1)y = (2k + 1) ⇒ (2k − 1)x + (k − 1)y − (2k + 1) = 0 ...(ii) These equations are of the following form: a1x + b1y + c1 = 0, a2x + b2y + c2 = 0 Here, a1 = 3, b1= 1, c1 = −1 and a2 = (2k − 1), b2 = (k − 1), c2 = −(2k + 1) In order that the given system of equations has no solution, we must have: $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$ i.e. $\frac{3}{\left(2k-1\right)}=\frac{1}{\left(k-1\right)}\ne \frac{-1}{-\left(2k+1\right)}$ $\frac{3}{\left(2k-1\right)}=\frac{1}{\left(k-1\right)}$ and $\frac{1}{\left(k-1\right)}\ne \frac{-1}{-\left(2k+1\right)}$ ⇒ 3(k − 1) = 2k − 1 and −(2k + 1) ≠ −(k − 1) ⇒ 3k − 3 = 2k − 1 and −2k − 1 ≠ −k + 1 ⇒ k = 2 and k ≠ −2 Thus, $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$ holds when k is equal to 2. Hence, the given system of equations has no solution when the value of k is 2.

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