Find the value of k for which the system if equations x+2y-5=0 and 3x+ky+15=0 has a unique solution
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Thus for all real values of k other than 6, the given system of equations will have a unique solution. Hence, the required value of k is 6.
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The given system of equations can be written as
x+2y−5=0
3x+ky−15=0
This system of equation is of the form
a1x+b1y+c1=0
a2x+b2y+c2=0
where, a1=1,b1=2,c1=−5 and a2=3,b2=k,c2=−15
The given system of equations will have a unique solution, if
31=k2⇒k=6.
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