Find the value of k for which the pair of linear equations kx + 3y=k-2 and 12x+ky=k
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kx+3y=k-3 ---(1)
12x+ky=k ---(2)
Here, a1=k b1=3 c1= -(k-3)
a2= 12 b2=k c2= -k
Now, a1/a2=b1/b2≠c1/c2
=> a1/a2= b1/b2
=> k/12= 3/k
=> k2= 36
=> k=6
Now, putting value of k=6, the solution automatically satisfy condition i.e.
a1/a2=b1/b2≠c1/c2
SO value of k is 6.
Hope it helps!
12x+ky=k ---(2)
Here, a1=k b1=3 c1= -(k-3)
a2= 12 b2=k c2= -k
Now, a1/a2=b1/b2≠c1/c2
=> a1/a2= b1/b2
=> k/12= 3/k
=> k2= 36
=> k=6
Now, putting value of k=6, the solution automatically satisfy condition i.e.
a1/a2=b1/b2≠c1/c2
SO value of k is 6.
Hope it helps!
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