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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: $5x+2y=2k,\phantom{\rule{0ex}{0ex}}2\left(k+1\right)x+ky=\left(3k+4\right).$

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Solution

## The given system of equations: 5x + 2y = 2k ⇒ 5x + 2y − 2k= 0 ...(i) And, 2(k + 1)x + ky = (3k + 4) ⇒ 2(k + 1)x + ky − (3k + 4) = 0 ...(ii) These equations are of the following form: a1x + b1y + c1 = 0, a2x + b2y + c2 = 0 Here, a1 = 5, b1= 2, c1 = −2k and a2 = 2(k + 1), b2 = k, c2 = −(3k + 4) For an infinite number of solutions, we must have: $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$ $\frac{5}{2\left(k+1\right)}=\frac{2}{k}=\frac{-2k}{-\left(3k+4\right)}\phantom{\rule{0ex}{0ex}}$ $⇒\frac{5}{2\left(k+1\right)}=\frac{2}{k}=\frac{2k}{\left(3k+4\right)}$ Now, we have the following three cases: Case I: $\frac{5}{2\left(k+1\right)}=\frac{2}{k}$ $⇒2×2\left(k+1\right)=5k⇒4\left(k+1\right)=5k$ $⇒4k+4=5k⇒k=4$ Case II: $\frac{2}{k}=\frac{2k}{\left(3k+4\right)}$ $⇒2{k}^{2}=2×\left(3k+4\right)$ $⇒2{k}^{2}=6k+8⇒2{k}^{2}-6k-8=0$ $⇒2\left({k}^{2}-3k-4\right)=0$ $⇒{k}^{2}-4k+k-4=0$ $⇒k\left(k-4\right)+1\left(k-4\right)=0$ $⇒\left(k+1\right)\left(k-4\right)=0\phantom{\rule{0ex}{0ex}}$ $⇒\left(k+1\right)=0\mathrm{or}\left(k-4\right)=0\phantom{\rule{0ex}{0ex}}$ $⇒k=-1\mathrm{or}k=4$ Case III: $\frac{5}{2\left(k+1\right)}=\frac{2k}{3k+4}\phantom{\rule{0ex}{0ex}}$ $⇒15k+20=4{k}^{2}+4k\phantom{\rule{0ex}{0ex}}$ $⇒4{k}^{2}-11k-20=0\phantom{\rule{0ex}{0ex}}$ $⇒4{k}^{2}-16k+5k-20=0\phantom{\rule{0ex}{0ex}}$ $⇒4k\left(k-4\right)+5\left(k-4\right)=0\phantom{\rule{0ex}{0ex}}$ $⇒\left(k-4\right)\left(4k+5\right)=0\phantom{\rule{0ex}{0ex}}$ $⇒k=4\mathrm{or}\mathrm{k}=\frac{-5}{4}$ Hence, the given system of equations has an infinite number of solutions when k is equal to 4.

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