Question

for what value of k will be the following pair of linear equations have no solution 2 x+3y=a,6x + (k- 2) y=(3k-2)​

Answer 1

Answer:

k = 11

Explanation:

Given pair of linear equations:

2x+3y=9 ---(1)

6x+(k-2)y=3k-2 ----(2)

Compare these equations with

a1x+b1y=c1 and a2x+b2y=c2

we get

\begin{gathered}a_{1}=2, b_{1}=3\\a_{2}=6,b_{2}=k-2\end{gathered}

a

1

=2,b

1

=3

a

2

=6,b

2

=k−2

Now ,

\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}

a

2

a

1

=

b

2

b

1

/* Given linear equations have no solution */

\implies \frac{2}{6}=\frac{3}{k-2}⟹

6

2

=

k−2

3

Do the cross multiplication, we get,

\implies 2(k-2)=6\times 3⟹2(k−2)=6×3

\implies k-2 = \frac{6\times3}{2}⟹k−2=

2

6×3

\implies k-2=9⟹k−2=9

\implies k = 9+2⟹k=9+2

\implies k = 11⟹k=11

Therefore,

k = 11

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