Question

Hello friends ‼‼‼‼‼

Here is my question ➡➡➡

Please Solve my problems --

⚫ Find value of k system has infinite solution

♦ Kx-2y+6=0 ,
4x-3y+9=0

♦ 8x+5y=9 ,
Kx+10y =18

♦ 2x-3y=7 ,
(K+2)x -(2x+1)y = 3(2k-1)

♦ 2x+3y=2 ,
(K+2)x + (2k+1)y =2(K-1)

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⚠ No sparm ⚠

Answer 1

Hey mate !!

Here's the answer !!

For an equation to have infinite solutions, it must satisfy the following rule.

a₁ / a₂ = b₁ / b₂ = c₁ / c₂.

Here a₁ and a₂ are coefficients of x in both equations, b₁ and b₂ are coefficients of y in both the equations and c₁ and c₂ are constants of those equations.

Eqn 1 :

kx - 2y = - 6
4x - 3y = - 9

According to the rule,

k / 4 = - 2 / - 3 = - 6 / - 9

k / 4 = - 2 / - 3

=> k = 4 * - 2 / - 3 

=> k = - 8 / - 3 

=> k = 8 / 3.

Eqn 2 :

8x + 5y = 9
kx + 10y = 18

According to the rule,

8 / k = 5 / 10 = 9 / 18

=> 8 / k = 1 / 2 = 1 / 2

=> 8 / k = 1 / 2

=> 8 * 2 = 1 * k

=> k = 16

Eqn 3 :

2x - 3y = 7
( k + 2 )x - ( 2x + 1 )y = 3 (2k - 1 )

According to the rule,

2 / ( k + 2 ) = 7 / 3 ( 2k - 1 )

=> 2 * 3 ( 2k - 1 ) = 7 ( k + 2 )

=> 6 ( 2k - 1 ) = 7k + 14

=> 6k - 6 = 7k + 14

=> 7k - 6k = -14 -6

=> k = - 20

Eqn 4 :

2x + 3y = 2
( k + 2 )x + ( 2k + 1 )y = 2 ( k - 1 )

According to the formula,

2 / k + 2 = 3 / 2k + 1

=> 2 ( 2k + 1 ) = 3 ( k + 2 )

=> 4k + 2 = 3k + 6

=> 4k - 3k = 6 - 2

=> k = 4

Hope my answer helped !!

Cheers !!