Question

For what value of k, the following system of equations have (i) a unique solution (ii) no solution
2x+ky= 11 3x-5y= 7​

Answer 1

Solution:

We have been given two equations 2x + ky = 11 & 3x - 5y = 7.

∴ We have to find a unique Solution for system of equation.

(i)For Unique solution, We have;

• a1/a2 ≠ b1/b2

Here, a1 = 2 ,b1 = k & a2= 3 , b2 = 5

Substitute obtained values in Equation:

⇒ a1/a2 ≠ b1/b2

⇒ 2/3 ≠ k/-5

⇒ 3k≠ 10

∴ k = -10/3

(ii)For No Solution, We have ;

• a1/a2 = b1/b2 ≠ c1/c2

Here, a1 = 2 ,b1 = k & a2= 3 , b2 = - 5 & c1 = 11 , c2 = 7

Substitute obtained values in Equation:

⇒a1/a2 = b1/b2 ≠ c1/c2

⇒ 2/3 = k/-5 ≠ 11/7

⇒ 2/3 = k/-5

⇒ 3k = -10

⇒ k = -10/3

∴ k = -10/3

Answer 2

LINEAR EQUATIONS :

( i ). For unique solution,

$\mathsf{\dfrac{a_1}{a_2}}$ ≠$\mathsf{\dfrac{b_1}{b_2}}$

Here,

2x + ky - 11 = 0 , 3x - 5y - 7 = 0

$\mathsf{a_1}$ = 2, $\mathsf{a_2}$ = 3, $\mathsf{b_1}$ = k, $\mathsf{b_2}$ = - 5, $\mathsf{c_1}$ = - 11, $\mathsf{c_2}$ = - 7

Now,

$\mathsf{\dfrac{2}{3}}$ ≠ $\mathsf{\dfrac{k}{-5}}$

$\mathsf{k\:=\:{\dfrac{2*(-5)}{3}}}$

$\boxed{\mathsf{k\:=\:{\dfrac{-10}{3}}}}$

( ii ). For no solution,

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

$\mathsf{\dfrac{2}{3}}$ = $\mathsf{\dfrac{k}{-5}}$ ≠$\mathsf{\dfrac{-11}{-7}}$

For $\mathsf{\dfrac{k}{-5}}$ ≠$\mathsf{\dfrac{-11}{-7}}$,

$\boxed{\mathsf{k\:=\:{\dfrac{55}{7}}}}$

For $\mathsf{\dfrac{2}{3}}$ = $\mathsf{\dfrac{k}{-5}}$

$\boxed{\mathsf{k\:=\:{\dfrac{-10}{3}}}}$