For what values of $k$ the set of equations $2 x - 3 y + 6 z - 5 t = 3, y - 4 z + t = 1, 4 x - 5 y + 8 z - 9 T = k$ has infinite number of solutions

Open in App

Solution

Given a system of equations has infinitely many solutions

$2 x - 3 y + 6 z - 5 t = 3$ ....(1)
$y - 4 z + t = 1$ .....(2)
$4 x - 5 y + 8 z - 9 T = k$ ....(3)
Subtracting (2) from (3), we get

$4 x - 6 y + 12 z - 10 t = k - 1$

$\Rightarrow 2 x - 3 y + 6 z - 5 t = \frac{k - 1}{2}$ .....(4)
From (1) and (4), we get
$\frac{k - 1}{2} = 3$

$\Rightarrow k = 7$

Suggest Corrections

Similar questions

2 x - 3 y + 6 z - 5 t = 3

y - 4 z + t = 1

4 x - 5 y + 8 z - 9 t = k

x - 2 y + 3 z = - 1

x - 3 y + 4 z = 1

- 2 x + 4 y - 6 z = k

x + y - 2 z = 0, 2 x - 3 y + z = 0, x - 5 y + 4 z = k

k

(k + 2) x + 10 y = k

k x + (k + 3) y = k - 1

Join BYJU'S Learning Program