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  • For which value(s) of k will the pair of equations kx+ 3y = k – 3 12x + ky= k have no solution

For which value(s) of k will the pair of equations kx+ 3y = k – 3, 12x + ky= k have no solution?

Answers (1)

Solution:
The given equations are
kx + 3y = k –3
12x + ky = k
In equation
kx + 3y = k – 3
a1 = k, b1 = 3, c1 = –k + 3
In equation
12x + ky = k
a2 = 12, b2 = k, c2 = –k

a_{2}= 12,b_{2}= k,c_{2}= -k
\frac{a_{1}}{a_{2}}= \frac{k}{12},\frac{b_{1}}{b_{2}}= \frac{3}{k},\frac{c_{1}}{c_{2}}= \frac{-k+3}{-k}= \frac{k-3}{k}

We know that if the equations have no solution then
\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}
\frac{k}{12}= \frac{3}{k}\equiv \frac{k-3}{k}
\frac{k}{12}= \frac{3}{k}
k^{2} = 36
k= \pm 36
(k\neq 6 because if k= +6, then \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}} which is not possible in no solution.)

Hence value of k is –6.

Posted by

infoexpert27

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