Question

Q.4 Find the value of k, for which the system of equations x+3y-2 and 2x+ky- 8 has no solution.
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Answer 1

Answer:

• k = 6

Step-by-step explanation:

Given:

• Equations - x + 3y - 2 and 2x + ky - 8

To Find:

• Value of k.

Formula used:

• $\sf{For\;no\;solution,\; \dfrac{a_{1}}{a_{2}} =\dfrac{b_{1}}{b_{2}} \neq \dfrac{c_{1}}{c_{2}}}$

Now, we have

$\implies{\sf{a_{1}=1\;and\;a_{2}=2}}$

$\implies{\sf{b_{1}=3\;and\;b_{2}=k}}$

$\implies{\sf{c_{1}=-2\;and\;c_{2}=-8}}$

Now, put the values in the formula,

$\therefore{\sf{For\;no\;solution,\; \dfrac{a_{1}}{a_{2}} =\dfrac{b_{1}}{b_{2}} \neq \dfrac{c_{1}}{c_{2}}}}$

$\implies{\sf{\dfrac{1}{2} =\dfrac{3}{k} \neq \dfrac{-2}{-8}}}$

$\implies{\sf{\dfrac{1}{2} =\dfrac{3}{k}}}$

$\implies{\sf{k=2\times 3}}$

$\implies{\sf{k=6}}$

Hence, Value of k = 6.

Answer 2

Answer:

given equations are : x+3y=2 , 2x+ky=8

here , by the question it has no solution

x+3y-2=0 , 2x+ ky -8 = 0

• so , the formula is a1/a2 = b1 / b2 =/ c1/c2
• here, a1= 1 , a2=2 ,
• b1 = 3 , b2 = k
• c1 = -2 , c2= -8

½ = 3/k =/ -2/8

so , ½ =3/k

= by cross multiplying , we get

= 1×k =2×3

= k = 6