Question

If the given system of equation x-4y=6. , 3x+ky=5 are inconsistent ? Then what is the value of k

Answer 1

Answer:

x -4 y = 6

x - 4 y - 6 = 0

3x + ky - 5  = 0

as per the given condition, to make these set of equation inconsistant,

i.e, no solution  a1/b1 = a2/b2 ≠ c1 / c2

the possible value of k to make this equations     ,

    K = -12

Answer 2

Answer:

-12 is the required value of k

Step-by-step explanation:

Explanation:

Given , x - 4y = 6 and 3x + ky = 5

Equation are inconsistent when there is no point of intersection when a system of linear equations represents two parallel lines, and hence no pair of x and y values that fulfil both equations.

∴ Equations $a_1x + b_1y +c_1 = 0$ and $a_2x +b_2y +c2 = 0$ are inconsistent when

$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

Step 1:

We have x- 4y - 6 = 0 and 3x + ky - 5  = 0

⇒$a_1 = 1 , a_2 = 3 , b_1 = -4 ,b_2 = k , c_1 = -6 and c_2 = -5$

Therefore , $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

⇒ $\frac{1}{3} = \frac{-4}{k} \neq \frac{-6}{-5}$

⇒k = -4 × 3 = -12       [By cross multiplication ]

Final answer:

Hence , value of k is -12 .

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