Question

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 1

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

Answer 2

Answer:

Value of α is all real numbers other tha +6 and -6.

Step-by-step explanation:

Given:

System of equations,

αx + 3y = α - 3

12x + αy = α

System of equation has unique solution.

To find: Value of α

When a system of equation has unique solution then we have following relationship,

$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$

we have ,

$a_1=\alpha\:,\:a_2=12\:,\:b_1=3\:,\:b_2=\alpha$

Consider,

$\frac{\alpha}{12}\neq\frac{3}{\alpha}$

$\alpha^2\neq3\times12$

$\alpha^2\neq36$

$\alpha\neq\pm\sqrt{36}$

$\alpha\neq\pm6$

Therefore, Value of α is all real numbers other tha +6 and -6.