Question

Find the value of k so that the following system of equations has no solution's. 3x-y-5=0, 6x-2y-k=0

Answer 1

Given Equation is 3x - y - 5 = 0  = > a1 = 3, b1 = -1, c1 = -5.

Given Equation is 6x - 2y - k = 0 = > a2 = 6, b2 = -2, c2 = -k.

Now,

Given that Equation has no solutions.

= > (a1/a2) = (b1/b2) ≠ (c1/c2)

= > (3/6) = (-1/-2) ≠ (-5/-k)

=. (1/2) = (1/2) ≠ (5/k)

= > (1/2) ≠ (5/k)

= > k ≠ 10.

Therefore, k can take all real values except 10

Hope this helps!

Answer 2

Answer:

k≠10

Step-by-step explanation:

Given Equation is 3x - y - 5 = 0  = > a1 = 3, b1 = -1, c1 = -5.

Given Equation is 6x - 2y - k = 0 = > a2 = 6, b2 = -2, c2 = -k.

Now,  

Given that Equation has no solutions.

=> $\frac{a1}{b1}$= $\frac{b1}{b2}$ ≠ $\frac{c1}{c2}$

= > $\frac{3}{6}$ = $\frac{-1}{-2}$≠ $\frac{-5}{-k}$

=> $\frac{1}{2}$ = $\frac{1}{2}$ ≠ $\frac{5}{k}$

=> $\frac{1}{2}$ ≠ $\frac{5}{k}$

Multiplying 2×5

=> k ≠ 10.

 

Therefore, k can take all real values except 10.

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