Question

Determine the value of k for which the following pair of linear equations represents a pair of parallel
lines on the graph.
(2k-1)x +(k-2)y =5
(k +2)x+ y = 3

Answer 1

Answer: -1 , 3

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Answer 2

$\huge\bf\underline \green{Solution:-}$

Given equations:-

• (2k-1)x +(k-2)y - 5 = 0
• (k +2)x+ y -3 = 0

The given Linear equations are of the form

$\: \: \: \: \: \: \: \: \: \: \: \: \: \bigstar \bf \: a_1x+b_1y+c_1 = 0 \: \\ \bf \: and \: \: \: \bigstar \: a_2x+b_2y+c_2 = 0$

where,

• a1 = (2k-1) , a2 = (k+2)

• b1 = (k-2) , b2 = 1

• c1 = -5 , c2 = -3

$\therefore \rm \frac{a_1}{a_2} = \frac{(2k - 1)}{(k + 2)} \\ \\ \: \: \: \: \rm \frac{b_1}{b_2} = \frac{k - 2}{1} \\ \\ \: \: \: \: \: \rm \frac{c_1}{c_2} = \frac{ - 5}{ - 3}$

The lines are parallel on the graph so the given equations have no Solution.

Then,

$\bf \: \frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

$\rm\frac{2k - 1}{k + 2}=\frac{k - 2}{1}\neq\frac{5}{3}$

$\rm : \implies\frac{2k - 1}{k + 2}=\frac{k - 2}{1} .........(i)\\\\ \rm : \implies \frac{k-2}{1}\neq\frac{5}{3}........(ii)\\\\ \rm\:From\:eq.\:(i)\\ \\ \rm : \implies \: 2k - 1 =( k - 2)(k + 2) \\ \\ \rm : \implies \: 2k - 1 = {k}^{2} - 4 \\ \\ \rm : \implies \: {k}^{2} = 2k + 3 \\ \\ \rm : \implies {k}^{2} - 2k - 3 = 0$

$\rm : \implies \: {k}^{2} - 3k + k - 3 \\ \\ \rm : \implies \: k(k - 3) + 1(k - 3) \\ \\ \rm : \implies \: (k - 3)(k + 1) \\ \\ \bf : \implies \: k = 3 \: \: or \: \: k = - 1$

When k = 3 or -1 then clearly $\frac{k-2}{1}\neq\frac{5}{3}$

hence, the given system of equations has no Solution when k = 3 or -1