1.

find the value of alpha for which the following system of equation has a unique solution alpha x +3y=alpha - 3 12x + xy = alpha

Answer»

Alfa×x +3y-alfa+3=0
12x+alfa×y-alfa=0
for UNIQUE solution a1÷a2 is not equal TI b1÷b2
a1÷a2=alfa÷12---eq 1
b1÷b2=3÷alfa---eq 2
cross MULTIPLY (of eq 1 and eq 2)
(alfa)^2=36
alfa= 6



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