Question

find the value of k so that the following system of equations had no solution 3x-y-5=0 and 6x-2y-k=0

Answer 1

Answer:

The value of k = 10.

Step-by-step explanation:

Given two equations, 3x – y – 5 = 0 (i.e) 3x - y = 5    …. (1)

6x - 2y – k = 0 (i.e) 6x - 2y = k    …. (2)

Divide equation (2) by 2, it becomes, 3x – y = $\left(\frac{k}{2}\right)$ … (3)

From equation (1) and (3) LHS are equal, hence consider RHS

$\left(\frac{k}{2}\right)$ = 5, k = 10 for the problem to have solution,  

Condition for no solution be k ≠ 10.  

Answer 2

Answer: Equations (i) and (ii) will have no solution if k ≠ -10.

Step-by-step explanation:

The given system of equations:

3x - y - 5 = 0 ….(i)

And, 6x - 2y + k = 0 ….(ii)  

These equations are of the following form:

$a_{1}x+ b_{1}y+c_{1}=0, a_{2}x+b_{2}y+c_{2}=0$

where, $a_{1}=3, b_{1}=-1, c_{1}=-5$ and $a_{2}=6, b_{2}=-2, c_{2}=k$

In order that the given system has no solution, we must have:

$\frac{a_{1}}{a_{2}} =\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

As $\frac{3}{6} =\frac{-1}{-2} \neq \frac{-5}{k}$

⇒ $\frac{-1}{-2} \neq \frac{-5}{k}$

⇒$k\neq -10$

Hence, equations (i) and (ii) will have no solution if k ≠ -10.

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