Question

Write the value of k for which the following pair of linear equations has no solution :- 4x +y=11;1x+3y=5 ?

Answer 1

Correct Question : Write the value of k for which the following pair of linear equations has no solution :- 4x +y=11; kx+3y=5 ?

Equations :

4x + y = 11

kx + 3y = 5

When there is no solution then :

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

$\frac{a1}{a2} = \frac{b1}{b2} \\ \\ \frac{4}{k} = \frac{1}{3}$

Cross Multiply the terms.

k = 12

Therefore, the value of k is 12.

Answer 2

• 4x + y = 11

• kx + 3y = 5

» We have to find the value of k if the equation has no solution.

Means..

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

Here.. $a_{1}$ = 4

$a_{2}$ = k

$b_{1}$ = 1

$b_{2}$ = 3

=> $\dfrac{4}{k}$ = $\dfrac{1}{3}$

Cross-multiply them..

=> k = 4(3)

=> k = 12

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$\textbf{k = 12}$ if equation has no solution..

___________ $\bold{[ANSWER]}$

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✡ More information :

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

For unique solutions

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

For infinitely many solutions

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

For no solution

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