Question

Find the value of k so that the system of linear equations will have infinite number of solutions. X+(k+2)y=4 (2k - 1)x +25y =6k +2 ​

Answer 1

Step-by-step explanation:

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Answer 2

k = 3

The given linear equations are

x + ( k + 2 ) y = 4

= x + ( k + 2 ) y - 4 = 0 ----- ( 1 )

( 2k - 1 ) x + 25y = 6k + 2

= ( 2k - 1 ) x + 25y - ( 6k + 2 ) = 0  ----- ( 2 )

Comparing equation ( 1 ) with a₁x + b₁y + c₁ = 0,

a₁ = 1

b₁ = k + 2

c₁ = - 4

Comparing equation ( 2 ) with a₂x + b₂y + c₂ = 0,

a₂ = 2k - 1

b₂ = 25

c₂ = - ( 6k + 2 )

For two linear equations to have infinite solutions,

a₁ / a₂ = b₁ / b₂ = c₁ / c₂

∴ a₁ / a₂ = b₁ / b₂

= 1 / ( 2k - 1 ) = ( k + 2 ) / 25

=25 = ( k + 2 ) ( 2k - 1 )

= 25 = k ( 2k - 1 ) + 2 ( 2k - 1 )

=25 = 2k² - k + 4k - 2

=25 = 2k² + 3k - 2

=2k² + 3k - 2 - 25 = 0

=2k² + 3k - 27 = 0

=2k² - 6k + 9k - 27 = 0

=2k ( k - 3 ) + 9 ( k - 3 ) = 0

=( k - 3 ) ( 2k + 9 ) = 0

=( k - 3 ) = 0 OR ( 2k + 9 ) = 0

= k - 3 = 0 OR 2k + 9 = 0

=k = 3 OR 2k = - 9

=k = 3 OR k = - 9 / 2

Therefore, the values of k= 3 and k= - 9 / 2.

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