1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# For what value of k does the pair of equations $5x+2y=2k\mathrm{and}2\left(\mathrm{k}+1\right)\mathrm{x}+\mathrm{ky}=\left(3\mathrm{k}+4\right)$ have an infinite number of solutions? (a) k = 5 (b) k = 4 (c) $k=\frac{2}{3}$ (d) $k=\frac{-2}{3}$

Open in App
Solution

## The correct option is (b). The given system of equations can be written as follows: 5x + 2y − 2k = 0 and 2(k + 1)x + ky − (3k + 4) = 0 The given equations are of the following form: a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 Here, a1 = 5, b1 = 2, c1 = −2k and a2 = 2(k + 1), b2 = k and c2 = −(3k + 4) ∴ $\frac{{a}_{1}}{{a}_{2}}=\frac{5}{2\left(k+1\right)},\frac{{b}_{1}}{{b}_{2}}=\frac{2}{k}\mathrm{and}\frac{{c}_{1}}{{c}_{2}}=\frac{-2k}{-\left(3k+4\right)}=\frac{2k}{\left(3k+4\right)}$ For an infinite number of solutions, we must have: $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$ ∴ $\frac{5}{2\left(k+1\right)}=\frac{2}{k}=\frac{2k}{\left(3k+4\right)}$ ⇒ $\frac{5}{2\left(k+1\right)}=\frac{2}{k}\mathrm{and}\frac{2k}{\left(3k+4\right)}=\frac{2}{k}$ ⇒ 5k = 4(k + 1) ...(i) And, 2(3k + 4) = 2k2 ...(ii) From (i) we have: 5k = 4k + 4 ⇒ 5k − 4k = 4 ⇒ k = 4 From (ii) we have: 6k + 8 = 2k2 ⇒ k2 − 3k − 4 = 0 ⇒ (k − 4)( k + 1) = 0 ⇒ k = 4, −1 Hence, from (i) and (ii), we get: k = 4

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Graphical Solution
MATHEMATICS
Watch in App
Join BYJU'S Learning Program