If the system of linear equation has infinite n.o of solution find the value of k - 2x + 3y -(k-5) =0 and 12x + ky -12 = 0
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For infinite solution we have a1/a2=b1/b2 =c1/c2
In given equation
k - 2x + 3y -(k-5) =0,
12x + ky -12 = 0
Compare these eq with standard eq a1x+b1y+c1=0 a2x+b2y+c2=0
a1=k-2,b1=3,c1=-(k-5),
a2=12,b2=k,c2=-12
Now
a1/a2= b1/b2
=> K-2/12= 3/k
=> k(k-2)= 12*3
=> k^2-2k-36= 0
=>
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