📺 Subscribe Our YouTube Channels: Doubtify JEE | Doubtify Class 10

Search Suggest

Solve Linear System with Parameters and Count Solutions

Learn how to find conditions for infinite solutions in a system of equations and count integer solutions within a given range. This method is useful..

Question:

Let the system of equations

{x+5y−z=14x+3y−3z=724x+y+λz=ÎĽ,λ,ÎĽ∈R\begin{cases} x + 5y - z = 1 \\[1em] 4x + 3y - 3z = 7 \\[1em] 24x + y + \lambda z = \mu,\quad \lambda, \mu \in \mathbb{R} \end{cases}

have infinitely many solutions.
Then, the number of solutions of this system, if x,y,zx, y, z are integers and satisfy

7≤x+y+z≤77,

is:


Question Image

Let the system of equations x + 5y − z = 1, 4x + 3y − 3z = 7, 24x + y + λz = ÎĽ, λ, ÎĽ ∈ R, have infinitely many solutions. Then the number of solutions of this system, if x, y, z are integers and satisfy 7 ≤ x + y + z ≤ 77, is:


Short Solution

Idea:

  1. For infinitely many solutions, the third equation must be a linear combination of the first two.

  2. Find λ\lambda and ÎĽ\mu.

  3. Reduce to two equations in x,y,zx, y, z, find the relation among variables.

  4. Put x+y+z=tx + y + z = t, derive bounds 7≤t≤77

  5. Count integer solutions.


Image Solution

Let the system of equations x + 5y − z = 1, 4x + 3y − 3z = 7, 24x + y + λz = ÎĽ, λ, ÎĽ ∈ R, have infinitely many solutions. Then the number of solutions of this system, if x, y, z are integers and satisfy 7 ≤ x + y + z ≤ 77, is:

Step 1: Check dependence
We have:

E1:x+5y−z=1
E2:4x+3y−3z=7E_2: 4x + 3y - 3z = 7

Let E3:24x+y+λz=μE_3: 24x + y + \lambda z = \mu
For infinite solutions:

E3=pE1+qE2

Matching coefficients:

24=p(1)+q(4)1=5p+3qλ=−p−3qÎĽ=p(1)+q(7)

Solve:

  • From first: p+4q=24p + 4q = 24

  • From second: 5p+3q=15p + 3q = 1

Multiply first by 5:

5p+20q=120

Subtract:

17q=119⇒q=7,p=−4.

Then:

λ=−(−4)−21=4−21=−17,ÎĽ=−4+49=45.

So E3E_3 is consistent.


Step 2: Solve the pair
Take E1E_1 and E2E_2:

x+5y−z=1(1)4x+3y−3z=7(2)

Multiply (1) by 4:

4x+20y−4z=4

Subtract from (2):

−17y+z=3⇒z=17y+3.-17y + z = 3 \Rightarrow z = 17y + 3.From (1):

x+5y−(17y+3)=1⇒x−12y−3=1⇒x=12y+4.


Step 3: Sum constraint

x+y+z=12y+4+y+17y+3=30y+7.

So:

t=30y+7,7≤t≤77.

Subtract 7:

0≤30y≤70⇒0≤y≤7030=73.

So y=0,1,2


Step 4: Solutions

yy
xx
zz
x+y+zx+y+z
0437
1162037
2283767

All lie between 77 and 7777.

✅ Number of solutions = 3


Conclusion & Video Solution

The system has exactly 3 integer solutions satisfying the given condition.

3\boxed{3}

👉 Watch the detailed video explanation here:



Post a Comment

Have a doubt? Drop it below and we'll help you out!