Q3. Find the value of k for which the lines of pair of equations 3x+2ky=2 and 2x+5y= -1 are parallel.​

Question

Q3. Find the value of k for which the lines of pair of equations 3x+2ky=2 and 2x+5y= -1 are parallel.​

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Alice 7 months 2021-10-13T18:01:06+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-10-13T18:02:28+00:00

    Answer :

    k = 15/4

    Note:

    ★ A linear equation is two variables represent a straight line .

    ★ The word consistent is used for the system of equations which consists any solution .

    ★ The word inconsistent is used for the system of equations which doesn’t consists any solution .

    ★ Solution of a system of equations : It refers to the possibile values of the variable which satisfy all the equations in the given system .

    ★ A pair of linear equations are said to be consistent if their graph ( Straight line ) either intersect or coincide each other .

    ★ A pair of linear equations are said to be inconsistent if their graph ( Straight line ) are parallel .

    ★ If we consider equations of two straight line

    ax + by + c = 0 and a’x + b’y + c’ = 0 , then ;

    • The lines are intersecting if a/a’ ≠ b/b’ .

    → In this case , unique solution is found .

    • The lines are coincident if a/a’ = b/b’ = c/c’ .

    → In this case , infinitely many solutions are found .

    • The lines are parallel if a/a’ = b/b’ ≠ c/c’ .

    → In this case , no solution is found .

    Solution :

    Here ,

    The given linear equations are ;

    3x + 2ky = 2

    2x + 5y = -1

    These equations can be rewritten as ;

    3x + 2ky – 2 = 0

    2x + 5y + 1 = 0

    Clearly , we have ;

    a = 3 , b = 2k , c = -2

    a’ = 2 , b’ = 5 , c’ = 1

    For the given lines to be parallel ,

    a/a’ = b/b’ ≠ c/c’

    3/2 = 2k/5 ≠ -2/1

    Clearly , 3/2 ≠ -2/1

    Thus ,

    => 3/2 = 2k/5

    => 2k/5 = 3/2

    => k = (3/2)×(5/2)

    => k = 15/4

    Hence , k = 15/4 .

    0
    2021-10-13T18:02:54+00:00

    3/2=2k/5

    3*5=4k

    15=4k

    15/4=k

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