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+1 vote
4.6k views
in Mathematics by (20 points)
edited by

Find the value of k will the system of equation x +2y=5 3x+ky-15=0 have unique solution.

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2 Answers

+1 vote
by (323k points)

The given system of equations:  

x + 2y = 5  

⇒ x + 2y - 5 = 0 ….(i)  

3x + ky - 15 = 0 …(ii)  

These equations are of the forms:  a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0  

where, a1 = 1, b1= 2, c1 = -5 and a2 = 3, b2 = k, c2 = -15

For a unique solution, we must have:

 

Thus for all real values of k other than 6, the given system of equations will have a unique solution.

+1 vote
by (50 points)
reshown by

given the equation system:

x+2y=5

x+2y-5=0. -(1)

Also,

   3x+ky-15=0. -(2)

Now,

a1 =1, a= 3;  b1=2 , b= k;  c= -5 , c2 = -15

For a unique solution, it should be:

aby a is not equal to b​​​​​​by b2

therefore, 1 by 3 is not equal to  2 by k

Again,

k x 1 not equal to  3 x 2

.. k is not equal to 6

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