Question

Find the possible values of k if the pairs of linear equations 3x - ky = 5 and
2x+ 2y = 15 will have a unique solution.

Answer 1

☯GIVEN :

• A pair of linear equations : 3x -ky = 5 and 2x +2y = 15

☯TO FIND :

• The value of k so that the equations have a unique solution.

➲SOLUTION :

We know that,

⟼ If a pair of equations is consistent and has a unique solution.

• $\pink{\bf{ \dfrac{a}{d} \ne \dfrac{b}{e}}}$

⟶ Here,

• a = 3

• b = -k

• d = 2

• e = 2

✞ Substituting the given values :

$: \longrightarrow \sf \dfrac{3}{2} \neq \dfrac{ - k}{2}$

• By Cross Multiplication :

$: \longrightarrow \sf3 \times 2\ne 2 \times (- k)$

$: \longrightarrow \sf 6\ne- 2k$

$: \longrightarrow \sf \cancel{\dfrac{6}{ - 2}} \ne k$

$: \longrightarrow \underline{\huge{\boxed{ \purple{\bf{k\neq -3}}}}}$

$\huge{\red{\therefore}}$ The value of k can be any real number except -3.

Answer 2

$\: \: \: \: \: \: \: \: \: \: \cal\huge\red{\mid{\overline{\underline{Question } \red{:-}}}\mid}$

⠀

⠀

Find the possible values of k if the pairs of linear equations 3x - ky = 5 and 2x+ 2y = 15 will have a unique solution.

⠀

⠀

$\: \: \: \: \: \: \: \cal\huge\red{\mid{\overline{\underline{ \: \: \: Solution } \red{:-}}}\mid}$

⠀

The given pair of linear equations is,

$\: \: \: \: \: \: \: \: \: \rm{3x - ky -5 \: = \: 0}$

$\: \: \: \: \: \: \: \: \: \rm{2x + 2y - 15 \: = \: 0}$

⠀

$\rm{ \: \: Where, \: a_1 =3, \: b_1 = - k , c_1 = -5 }$

$\rm{ \: and \: \: \: a_2 =2 , b_2 =2 , c_2 = - 15}$

⠀⠀

⠀⠀For an unique solution, we must have

ㅤ

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm\purple{\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}}$

⠀

$\: \: \: \: \rm{i.e. \: \: \: \: \: \: \: \: \dfrac{3}{2}} \: \ne \: \dfrac{ - k}{2}$

$\: \: \: \: \: \rm{ \implies \: k \: \ne \: -3}$

⠀⠀

⠀ So, the given pair of linear equations will have an unique solution for real values of k other than -3.