Let the system of equations x + 5y – z = 1 4x + 3y – 3z = 7 24x + y + z = , have infinitely many solutions. Then the number of the solutions of this system, if x, y, z are integers and satisfy , is : [2025]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Let the system of equations

x + 5y – z = 1

4x + 3y – 3z = 7

24x + y + $λ$ z = $μ$

$λ, μ \in R$ , have infinitely many solutions. Then the number of the solutions of this system, if x, y, z are integers and satisfy $7 \leq x + y + z \leq 77$ , is : [2025]

1 3

2 5

3 6

4 4

Ans.
(1)

For infinitely many solutions, we have, D = 0

$\Rightarrow$ $| \begin{matrix} 1 & 5 & - 1 \\ 4 & 3 & - 3 \\ 24 & 1 & λ \end{matrix} |$ $= 0$

$\Rightarrow 1 (3 λ + 3) - 5 (4 λ + 72) - 1 (4 - 72) = 0$

$\Rightarrow 17 λ = - 289 \Rightarrow λ = - 17$ ... (i)

Now, $D_{1} = 0$

$\Rightarrow$ $| \begin{matrix} 1 & 5 & - 1 \\ 7 & 3 & - 3 \\ μ & 1 & - 17 \end{matrix} |$ $= 0$

$\Rightarrow 1 (- 51 + 3) - 5 (- 119 + 3 μ) - 1 (7 - 3 μ) = 0$

$\Rightarrow - 48 + 595 - 15 μ - 7 + 3 μ = 0 \Rightarrow 12 μ = 540$

$\Rightarrow μ = 45$ ... (ii)

Using (i) and (ii), we get

$x + 5 y - z = 1, 4 x + 3 y - 3 z = 7, 24 x + y - 17 z = 45$

$\Rightarrow z = x + 5 y - 1$

$∴ 4 x + 3 y - 3 x - 15 y + 3 = 7$

$\Rightarrow x - 12 y = 4$

$\Rightarrow x = 4 + 12 y and z = 4 + 12 y + 5 y - 1 = 3 + 17 y$

$∴$ (x, y, z) = (4 + 12k, k, 3 + 17k) ( $∵$ Assume y = k)

Also, $7 \leq 7 + 30 k \leq 77$

$\Rightarrow 0 \leq 30 k < 70$

$\Rightarrow 0 \leq k \leq 2.3 \Rightarrow k = 0, 1, 2$

Thus, there are three possible solutions.

Want Us to Email you About Special Offers & Updates?