Question

Find the values of k for which the system 2x+ky=12x+ky=1,3x-5y=73x-5y=7 will have a (i) unique solution and (ii) no solution .Is there is a value of k for which the system has infinitely many solutions

Answer 1

Given :

The system of linear equation as

2 x + k y = 1

3 x - 5 y = 7

To Find :

The value of k for which system have

( i ) Unique solution

( ii ) No solution

( iii ) infinitely many solution

Solution :

Since for the system of equation

  $a_1x + b_1y + c_1$ = 0

  $a_2x+b_2y+c_2$ = 0

For Unique solution

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

For No solution

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

For infinite solution

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

Now,

The system of linear equation as

2 x + k y = 1

3 x - 5 y = 7

i.e  $a_1$ = 2          $b_1$ = k            $c_1$ = -1

     $a_2$ = 3          $b_2$ = -5           $c_2$ = -7

( i ) Unique solution

 i.e  $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}\neq \dfrac{c_1}{c_2}$

Or,   $\dfrac{2}{3} \neq \dfrac{k}{-5}\neq \dfrac{-1}{-7}$

∴     k $\neq$ $\dfrac{-10}{3}$  and   k $\neq$ $\dfrac{-5}{7}$

( ii ) No solution

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

Or,  $\dfrac{2}{3} = \dfrac{k}{-5}\neq \dfrac{-1}{-7}$

∴     k = $\dfrac{-10}{3}$  and  k $\neq$ $\dfrac{-5}{7}$

( iii ) Infinite many solution

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

Or,  $\dfrac{2}{3} = \dfrac{k}{-5}= \dfrac{-1}{-7}$

∴     k = $\dfrac{-10}{3}$  and  k = $\dfrac{-5}{7}$

Hence, The value of k

( i )  Unique solution

     k $\neq$ $\dfrac{-10}{3}$  and   k $\neq$ $\dfrac{-5}{7}$

( ii ) No solution

      k = $\dfrac{-10}{3}$  and  k $\neq$ $\dfrac{-5}{7}$

( iii ) Infinite many solution

  k = $\dfrac{-10}{3}$  and  k = $\dfrac{-5}{7}$         Answer

Answer 2

No, there is no value of k for which the system has infinitely many solutions.

Step-by-step explanation:

2x + ky = 1 → (equation 1)

3x - 5y = 7 → (equation 2)

Now, the given system is:

2x + ky = 1 ⇒ a₁x + b₁y - c₁ = 0

Where,

a₁ = 2; b₁ = k; c₁ = -1

3x - 5y = 7 ⇒ a₂x + b₂y - c₂ = 0

Where,

a₂ = 3; b₂ = -5; c₂ = -7

(i) Unique solution:

a₁a₂ = b₁b₂ ⇒ 2 × 3 ≠ -k × -5

a₁/a₂ = b₁/b₂ ⇒ 2/3 ≠ -k/-5 ⇒ k ≠ 10/3

The given equation will have a unique solution.

(ii) No solution:

a₁a₂ = b₁b₂ ≠ c₁c₂ ⇒ 2 × 3 = k × -5 ≠ -1 × -7 ⇒ k = -10/3

a₁/a₂ = b₁/b₂ ≠ c₁/c₂ ⇒ 2/3 = k/-5 ≠ -1/-7 ⇒ k ≠ -5/7

The given equation will have no solution when k = -10/3.