Question

18.
For what value of k the following system of linear equations has no solution?
x-ky + 5 = 0; 3x - 6y + 12 = 0​

Answer 1

GIVEN :

Equations :

x - ky + 5 = 0

3x - 6y + 12 = 0

We know that,

When linear equations have no solution.

Then,

$\frac{a1}{a2} = \frac{b1}{b2} \neq \: \frac{c1}{c2}$

From the above equations,

a1 = 1 ; a2 = k

b1 = 3 ; b2 = 6

$\frac{a1}{a2} = \frac{b1}{b2} \\ \\ \frac{1}{3} = \frac{k}{6}$

Cross Multiply the terms.

3 × k = 1 × 6

3k = 6

k = 6/3

k = 2

Therefore, the value of k is 2.

Answer 2

x - ky + 5 = 0

3x - 6y + 12 = 0

__________ [GIVEN EQUATIONS]

• We have to find the value of k when the given equations has no solution.

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Means..

$\dfrac{a_{1}}{a_{2}}$ = $\dfrac{b_{1}}{b_{2}}$ ≠ $\dfrac{c_{1}}{c_{2}}$

Here

$a_{1}$ = 1

$a_{2}$ = 3

$b_{1}$ = k

$b_{2}$ = 6

=> $\dfrac{1}{3}$ = $\dfrac{k}{6}$

Cross-multiply them..

=> 6 = 3k

=> 3k = 6

=> k = $\dfrac{6}{3}$

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$\huge{\bold{k\:=\:2}}$

_____________ $\bold{[ANSWER]}$

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More information :

• $\dfrac{a_{1}}{a_{2}}$ ≠ $\dfrac{b_{1}}{b_{2}}$

For unique solutions

• $\dfrac{a_{1}}{a_{2}}$ = $\dfrac{b_{1}}{b_{2}}$ = $\dfrac{c_{1}}{c_{2}}$

For infinitely many solutions

• $\dfrac{a_{1}}{a_{2}}$ = $\dfrac{b_{1}}{b_{2}}$ ≠ $\dfrac{c_{1}}{c_{2}}$

For no solutions

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