The number of values of k for which the system of equations (k+1)x + 8

Question

The number of values of k for which the system of equations

(k + 1) x + 8 y = 4 k, k x + (k + 3) y = 3 k - 1

has infinitely many solutions, is

Options:

A . 0

B . 1

C . 2

D . Infinite

Answer: Option B
:
B
For infinitely many solutions, the two equations must be identical

\Rightarrow \frac{k + 1}{k} = \frac{8}{k + 3} = \frac{4 k}{3 k - 1} \Rightarrow (k + 1) (k + 3) = 8 k

and 8(3k-1) = 4k(k+3)

\Rightarrow k^{2} - 4 k + 3 = 0

and

k^{2} - 3 k + 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.

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