Question

If the pair of linear equations 7x + 21y = k + 3 and
2x + y = k has no solution, then which of the
following cannot be the value of K?​

Answer 1

Answer:

Appropriate Question :-

• The pair of linear equation 7x + 21y = k + 3 and 2x + ky = k has no solution, then which of the following cannot be the value of k ?

Given :-

• The pair of linear equation 7x + 21y = k + 3 and 2x + ky = k has no solution.

To Find :-

• What is the value of k.

Solution :-

As we know the system of equation :

$\dashrightarrow \sf\bold{a_1x + b_1y =\: c_1}$

$\dashrightarrow \sf\bold{a_2x + b_2y =\: c_2}$

Then, it has no solution, if

$\longmapsto \sf\bold{\green{\dfrac{a_1}{a_2} =\: \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}}}$

The pair of linear equation :

$\implies$ 7x + 21y - (k + 3) = 0

$\implies$ 2x + ky - k = 0

where,

• $\sf a_1$ = 7
• $\sf b_1$ = 21
• $\sf c_1$ = - (k + 3)
• $\sf a_2$ = 2
• $\sf b_2$ = k
• $\sf c_2$ = - k

According to the question,

$\longrightarrow \sf \boxed{\bold{\pink{\dfrac{a_1}{a_2} =\: \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}}}}$

$\longrightarrow \sf \dfrac{7}{2} =\: \dfrac{21}{k} \neq \dfrac{\cancel{-} (k + 3)}{\cancel{-} k}$

$\longrightarrow \sf 7k =\: 21(2) \neq \dfrac{k + 3}{k}$

$\longrightarrow \sf 7k =\: 42 \neq \dfrac{k + 3}{k}$

$\longrightarrow \sf k =\: \dfrac{\cancel{42}}{\cancel{7}} \neq \dfrac{k + 3}{k}$

$\longrightarrow \sf k =\: 6 \neq \dfrac{k + 3}{k}$

$\therefore$ The value of k is 6.