Question

find the values of k for which the pair of linear equations kx +3y=k2 and 12x+ky=k which has no solution

Answer 1

In equation 1, kx + 3y =2k
here a1=k b1=3 & c1=2k
And equation is 12x + ky = k
here a2=12 b2=k and c2=k
For a equation to have no solutions,
a1/a2 = b1/b2 not equal to c1/c2

Now a1/a2 = b1/b2
so k/12 = 3/k
=> k^2=3×12
=>k = √(12×3)=6

Now b1/b2 not equal to c1/c2
=> 3/k not =2k/k
=>3/k not=2
=>k not=3/2
=> k not equal to 1.5

So the value of k for which the pair of equations have no soln is 6.

HOPE IT HELPS....

Answer 2

• It seems like you are searching for " For which value(s) of k will the pair of equations kx + 3y = k – 3, 12x + ky = k has no solution?

Answer:

Required value of k for which the given pair of linear equations has no solutions is -6.

Step-by-step explanation:

Given:

• $k x+3 y=k-3 and 12 x+k y=k$

• Comparing with  $a_{1} x+b_{1} y+c_{1}=0 and a_{2} x+b_{2} y+c_{2}=0$,

we get

• $\begin{aligned}&a_{1}=k, b_{1}=3 \text { and } c_{1}=-(k-3) \\&a_{2}=12, b_{2}=k \text { and } c_{2}=-k\end{aligned}$

For the pair of linear equations to have no solution,

$\begin{aligned}&\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \\\frac{k}{12}=\frac{3}{k} \neq \frac{-(k-3)}{-k}\end{aligned}$

Now,

$\frac{\mathrm{k}}{12}=\frac{3}{\mathrm{k}}\\ k^{2}=36 \\\\ k=\pm 6\end{aligned}$

And,

$&\frac{3}{k} \neq \frac{k-3}{k} \\\\3 k \neq k(k-3) \\\\ 3 k-k(k-3) \neq 0 \\\\ k(3-k+3) \neq 0 \\\\ k \neq 0 \text { and } k \neq 6\end{aligned}$

Hence, required value of k for which the given pair of linear equations has no solutions is -6.