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Question

# Find the value of k for which each of the following system of linear equations has an infinite number of solutions: $\left(k-1\right)x-y=5,\phantom{\rule{0ex}{0ex}}\left(k+1\right)x+\left(1-k\right)y=\left(3k+1\right).$

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Solution

## The given system of equations: (k − 1)x − y = 5 ⇒ (k − 1)x − y − 5 = 0 ...(i) And, (k + 1)x + (1 − k)y = (3k + 1) ⇒ (k + 1)x + (1 − k)y − (3k + 1) = 0 ...(ii) These equations are of the following form: a1x + b1y + c1 = 0, a2x + b2y + c2 = 0 Here, a1 = (k − 1), b1= −1, c1 = −5 and a2 = (k + 1), b2 = (1 − k), c2 = −(3k + 1) For an infinite number of solutions, we must have: $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$ $\mathrm{i}.\mathrm{e}.\frac{\left(k-1\right)}{\left(k+1\right)}=\frac{-1}{-\left(k-1\right)}=\frac{-5}{-\left(3k+1\right)}\phantom{\rule{0ex}{0ex}}$ $⇒\frac{\left(k-1\right)}{\left(k+1\right)}=\frac{1}{\left(k-1\right)}=\frac{5}{\left(3k+1\right)}$ Now, we have the following three cases: Case I: $\frac{\left(k-1\right)}{\left(k+1\right)}=\frac{1}{\left(k-1\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ $⇒{\left(k-1\right)}^{2}=\left(k+1\right)\phantom{\rule{0ex}{0ex}}$ $⇒{k}^{2}+1-2k=k+1\phantom{\rule{0ex}{0ex}}$ $⇒{k}^{2}-3k=0⇒k\left(k-3\right)=0\phantom{\rule{0ex}{0ex}}⇒k=0\mathrm{or}k=3$ Case II: $\frac{1}{\left(k-1\right)}=\frac{5}{\left(3k+1\right)}\phantom{\rule{0ex}{0ex}}$ $⇒3k+1=5\left(k-1\right)\phantom{\rule{0ex}{0ex}}$ $⇒3k+1=5k-5\phantom{\rule{0ex}{0ex}}$ $⇒2k=6⇒k=3$ Case III: $\frac{\left(k-1\right)}{\left(k+1\right)}=\frac{5}{\left(3k+1\right)}\phantom{\rule{0ex}{0ex}}$ $⇒\left(3k+1\right)\left(k-1\right)=5\left(k+1\right)\phantom{\rule{0ex}{0ex}}$ $⇒3{k}^{2}+k-3k-1=5k+5\phantom{\rule{0ex}{0ex}}$ $⇒3{k}^{2}-2k-5k-1-5=0\phantom{\rule{0ex}{0ex}}$ $⇒3{k}^{2}-7k-6=0\phantom{\rule{0ex}{0ex}}$ $⇒3{k}^{2}-9k+2k-6=0\phantom{\rule{0ex}{0ex}}$ $⇒3k\left(k-3\right)+2\left(k-3\right)=0\phantom{\rule{0ex}{0ex}}$ $⇒\left(k-3\right)\left(3k+2\right)=0\phantom{\rule{0ex}{0ex}}$ $⇒\left(k-3\right)=0\mathrm{or}\left(3k+2\right)=0\phantom{\rule{0ex}{0ex}}$ $⇒k=3\mathrm{or}\mathrm{k}=\frac{-2}{3}$ Hence, the given system of equations has an infinite number of solutions when k is equal to 3.

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