For the given pair of equation kx+3y= -7and 2x+6y= -14 to have infinitely many solution ,the value of k should be 1 .Is the statement is true? About the author Amelia
Answer 1st equation = kx + 3y +7 = 0 2nd equation = 2x + 6y +14 = 0 a₁ = k ,b₁ = 3 , c₁ = 7 a₂ = 2 , b₂ = 6 ,c₂ = 14 ⠀⠀⠀⠀⠀⠀⠀⠀⠀According to the Question it is given that the given pair of equation has infinitely many solution . So , the condition for infinitely many solution . [tex]:\implies[/tex] a₁/a₂ = b₁/b₂ = c₁/c₂ = 0 Substitute the value we get [tex]:\implies[/tex] k/2 = 3/6 = 7/14 [tex]:\implies[/tex] k/2 = 1/2 = 1/2 [tex]:\implies[/tex] k/2 = 1/2 or k/2 = 1/2 [tex]:\implies[/tex] k = 2/2 or k = 2/2 [tex]:\implies[/tex] k = 1 or k = 1 Hence, the value of k is1 . So , the given statement is True . Reply
Answer: Given :– For the given pair of equation kx+3y= -7 and 2x+6y= -14 to have infinitely many solution To Find :– Value of k Solution :– We know that a1/a2 = b1/b2 = c1/c2 k/2 = 3/6 = 7/14 k/2 = 1/2 = 7/14 k/2 = 1/2 = 1/2 k × 2/2 × 1 = 1/2 2k/2 = 1/2 2k = 2 k = 1 [tex] \\ [/tex] Reply
Answer
a₁ = k ,b₁ = 3 , c₁ = 7
a₂ = 2 , b₂ = 6 ,c₂ = 14
⠀⠀⠀⠀⠀⠀⠀⠀⠀According to the Question
it is given that the given pair of equation has infinitely many solution . So , the condition for infinitely many solution .
[tex]:\implies[/tex] a₁/a₂ = b₁/b₂ = c₁/c₂ = 0
Substitute the value we get
[tex]:\implies[/tex] k/2 = 3/6 = 7/14
[tex]:\implies[/tex] k/2 = 1/2 = 1/2
[tex]:\implies[/tex] k/2 = 1/2 or k/2 = 1/2
[tex]:\implies[/tex] k = 2/2 or k = 2/2
[tex]:\implies[/tex] k = 1 or k = 1
So , the given statement is True .
Answer:
Given :–
For the given pair of equation kx+3y= -7
and 2x+6y= -14 to have infinitely many solution
To Find :–
Value of k
Solution :–
We know that
a1/a2 = b1/b2 = c1/c2
k/2 = 3/6 = 7/14
k/2 = 1/2 = 7/14
k/2 = 1/2 = 1/2
k × 2/2 × 1 = 1/2
2k/2 = 1/2
2k = 2
k = 1
[tex] \\ [/tex]