find the value of k such that the following pair of equations have no solution .
1- kx- 8y = k+2
2- 6x - 3ky = 9
REMEMBER THAT IT IS FROM THE CHAPTER OF ( LINEAR EQUATION IN TWO VARIABLES )....
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Answers
Answer:
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Step-by-step explanation:
The value of k is 6 for which the system of following equations has a unique solution.
Step-by-step explanation:
Given : Equations 2x-y = 5, Kx-8y = 72x−y=5,Kx−8y=7
To find : The value of k for which the following system of equation has a unique solution.
Solution :
When the system of equation is in form a_1x+b_1y+c_1=0,\ a_2x+b_2y+c_2=0a
1
x+b
1
y+c
1
=0, a
2
x+b
2
y+c
2
=0 then the condition for a unique solutions is
\frac{a_1}{a_2}\neq\frac{b_1}{b_2}
a
2
a
1
≠
b
2
b
1
Comparing, a_1=2,b_1=-1,c_1=-5,a_2=K,b_2=-8,c_1=-7a
1
=2,b
1
=−1,c
1
=−5,a
2
=K,b
2
=−8,c
1
=−7
Substituting the values,
\frac{2}{K}\neq\frac{-1}{-8}
K
2
≠
−8
−1
2\times -8=-1\times K2×−8=−1×K
-K=-16−K=−16
K=16K=16
Therefore, the value of k is 6 for which the system of following equations has a unique solution.
☜☆☞UR ANSWER ☜☆☞
K= 16, K= (-16)