Number of values of k for which system of equations (k + 1)x + 8y = 4k, kx + (k + 3)y = 3k - 1 has i

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3 and 4 .Determinants and Matrices

medium

The number of values of $k $ for which the system of equations $(k + 1)x + 8y = 4k,$ $kx + (k + 3)y = 3k - 1$ has infinitely many solutions, is

A

$0$

B

$1$

C

$2$

D

Infinite

(IIT-2002)

Solution

(b) For infinitely many solutions, the two equations must be identical

$ \Rightarrow \frac{{k + 1}}{k} = \frac{8}{{k + 3}} = \frac{{4k}}{{3k – 1}}$

$ \Rightarrow (k + 1)(k + 3) = 8k$ and $8\,(3k – 1) = 4k(k + 3)$

$ \Rightarrow {k^2} – 4k + 3 = 0$ and ${k^2} – 3k + 2 = 0$.

By cross multiplication, $\frac{{{k^2}}}{{ – 8 + 9}} = \frac{k}{{3 – 2}} = \frac{1}{{ – 3 + 4}}$

${k^2} = 1$ and $k = 1$; $k = 1$.

Standard 12

Mathematics

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