Question

for what value of k will the following pair of linear equation have no solution 2x + 3y =9 ; 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

$a_{1}=2, b_{1}=3\\a_{2}=6,b_{2}=k-2$

Now ,

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

/* Given linear equations have no solution */

$\implies \frac{2}{6}=\frac{3}{k-2}$

Do the cross multiplication, we get,

$\implies 2(k-2)=6\times 3$

$\implies k-2 = \frac{6\times3}{2}$

$\implies k-2=9$

$\implies k = 9+2$

$\implies k = 11$

Therefore,

k = 11

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Answer 2

Step-by-step explanation:

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