value of k for which the system would have unique solution
k x +2y=5; 3x + y = 1
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kx+2y-5=0
3x+y-1=0
from middle end first rule
2 -5 K 2
2 -1 3 2
x/-2+10 = y/ -15+k = 1/2k-6
x/8 = 1/2k-6
x=8/2k-6
y= -15+k/2k-6
equate x and y
8/2k-6 = -15+k/2k-6
8 = -15 +k
k=8+15=23 is answer
3x+y-1=0
from middle end first rule
2 -5 K 2
2 -1 3 2
x/-2+10 = y/ -15+k = 1/2k-6
x/8 = 1/2k-6
x=8/2k-6
y= -15+k/2k-6
equate x and y
8/2k-6 = -15+k/2k-6
8 = -15 +k
k=8+15=23 is answer
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