Question

find the value of 'k'
3x+ y =7
4x+ky =9 has a unique solution​

Answer 1

Explanation:

$\sf{\bold{\blue{\underline{\underline{Given}}}}}$

● pair of equation

• 3x+y=7

• 4x+ky=9

⠀⠀⠀⠀

$\sf{\bold{\red{\underline{\underline{To\:Find}}}}}$

■ Value of "k".??■⠀⠀⠀⠀

$\sf{\bold{\purple{\underline{\underline{Solution}}}}}$

➔ Here it is given that the pair of equations have a unique solution

➔ If a pair of equations have a unique solution we know that,

 ::$\implies$$\boxed{\orange{\dfrac{a_1}{a_2}\neq \dfrac{b_1}{b_2}}}$

where,

a1=3

a2=4

b1=1

b2=k

so..

$\rightsquigarrow{\dfrac{a_1}{a_2}\neq \dfrac{b_1}{b_2}}$

$\rightsquigarrow{\dfrac{3}{4}\neq \dfrac{1}{k}}$

$\rightsquigarrow{3×k}\neq {1×4}$

$\rightsquigarrow{3k}\neq {4}$

$\rightsquigarrow{k}\neq \dfrac{4}{3}$

$\sf{\bold{\green{\underline{\underline{Answer}}}}}$

$\rightsquigarrow{\boxed{\orange{k\neq{\frac{4}{3}}}}}$⠀⠀

$\sf{\bold{\blue{\underline{\underline{Notes:-}}}}}$

$\rightsquigarrow$If a pair of equations:

    a₁x + b₁y + c₁ = 0

    a₂x + b₂y + c₂ = 0

$\rightsquigarrow$has a unique solution and is consistent,

$\red{\dfrac{a_1}{a_2}\neq \dfrac{b_1}{b_2}}$

$\rightsquigarrow$Has infinite number of solutions and is consistent,

$\red{\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2} }$

$\rightsquigarrow$Has no solution and is inconsistent,

$\red{\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq \dfrac{c_1}{c_2} }$

Answer 2

Explanation:

3x +y =7 ( multiple by 4)

4x + ky = 9 ( multiple by 3)

12x + 4y = 28

12x + 3ky = 27

y( 4-3k) = 1 ( by eliminated method)

y = 1/ 4-3k

put value in 3x + y =7

3x +1/4-3k = 7

3x = 7- 1/ 4- 3k

3x = 7( 4-3k) - 1 / 4-3k

3x = 28 - 21k -1 / 4- 3k

3x = 27 - 21 k / 4- 3k

x = 9 - 7 k / 4- 3 k

put the value x and y

3x +y =7

3 * ( 9-7k) / 4- 3k ) + 1/ 4-3k = 7

27 -21k / 4-3k + 1/ 4-3 k = 7

28 -21 k = 7 * ( 4- 3k)

28 - 21 k = 28 - 21 k

k = 0