determine the value of k for which the following pair of linear equations has infinite solutions.(k-3)X+3y=k, kx+ KY=12
Answers
Answer:
k = 6
Step-by-step explanation:
Given Equations are:
(i)
(k - 3)x + 3y = k
Here, a₁ = (k - 3), b₁ = 3, c₁ = -k
(ii)
kx + ky = 12
Here, a₂ = k, b₂ = k, c₂ = -12
For infinite solution, we have
a₁/a₂ = b₁/b₂ = c₁/c₂
⇒ (k - 3)/k = 3/k = k/12
Case 1:
(k - 3)/k = 3/k
⇒ k(k - 3) = 3k
⇒ k² - 3k = 3k
⇒ k² = 6k
⇒ k = 6
Case 2:
3/k = 6/12
⇒ 36 = 6k
⇒ k = 6
Case 3:
(k - 3)/k = k/12
⇒ 12k - 36 = k²
⇒ k² - 12k + 36 = 0
⇒ (k - 6)² = 0
⇒ k = 6
For k = 6, there are infinitely many solutions.
Hope it helps!
The Given pair of linear equation is :
(k - 3 ) x + 3y = k
kx + ky = 12
We can write these Equations as :
(k - 3 ) x + 3y - k = 0……….(1)
kx + ky - 12 = 0 ………….(2)
On comparing with General form of a pair of linear equations in two variables x & y is:
a1x + b1y + c1 = 0
and a2x + b2y + c2= 0
a1= k-3 , b1= -3, c1= -k
a2= k , b2= k , c2= - 12
a1/a2= k-3 /k , b1/b2= 3/k , c1/c2= -k/-12= k/12
Given: A pair of linear equations has a infinite solution, if
a1/a2 = b1/b2 = c1/c2
k-3 /k =3/k= k/12
I II III
Taking the first two terms
a1/a2 = b1/b2
k-3 /k =3/k
k - 3 = 3
k = 3 + 3
k = 6
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