Find the value for which the system of equation 3x +5y =0 , kx +10y = 0 has non zero solution
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Answered by
291
condition for the system of equations to have non-zero solutions,
a1/a2 = b1/b2 = c1/c2
here, a1 = 3 , a2 = k, b1 = 5 , b2 = 10
3/k = 5/10
3/k = 1/2
k = 6
a1/a2 = b1/b2 = c1/c2
here, a1 = 3 , a2 = k, b1 = 5 , b2 = 10
3/k = 5/10
3/k = 1/2
k = 6
Answered by
84
Answer:
For all real values k other than 6 , the system of equations has non zero solution.
Step-by-step explanation:
Given system of equations:
3x+5y=0 ,
kx+10=0 ,
Compare this with a1x+b1y+c1=0
and
a2x+b2y+c2 = 0, we get
a1 = 3 , b1 = 5, c1 = 0
a2 = k , b2= 10, c2=0
The equations have non zero solution,
Here ,
a1b2-a2b1≠0
=> 3×10-k×5 ≠ 0
=> 30 - 5k ≠ 0
=> -5k ≠-30
On dividing each term by -5, we get
=> k ≠ 6
If plot the lines ,we will notice that they are parallel.
So, for this problem to have non -zero solution , k is any real number other than 6.
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