Question

Determine the value of k in wch the following system of equation has a unique solution 2x-3y=1 kx+5y=7

Answer 1

2x-3y-1=0
kx+5y-7=0
Here
a1=2, b1=-3, c1=-1
a2=k, b1=5,c1=-7
For unique solution
a1/a2≠b1/b2
Now
2/k≠-3/5
-3k≠10
k≠-10/3

Answer 2

Answer:

$\huge \cal \underline \color{magenta}★AɴSᴡᴇʀ★$

Question :- Find value of k for which the pair of equations 2x-3y=1 and kx-5y=7 has a unique solution.. ?

Concept used :-

A linear equation in two variables represents a straight line in 2D Cartesian plane .

If we consider two linear equations in two variables, say ;

a1x + b1y + c1 = 0 and

a2x + b2y + c2 = 0

Then ;

✪ Both the straight lines will coincide if ;

a1/a2 = b1/b2 = c1/c2

In this case , the system will have infinitely many solutions.

✪ Both the straight lines will be parallel if ;

a1/a2 = b1/b2 ≠ c1/c2

In this case , the system will have no solution.

✪ Both the straight lines will intersect if ;

a1/a2 ≠ b1/b2

In this case , the system will have an unique solution.

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Solution :-

⇛2x -3y = 1

→ 2x - 3y - 1 = a1x + b1y + c1 = 0

and ,

→ kx - 5y - 7 = a2x + b2y + c2 = 0

From this we get,

→ a1 = 2, b1 = (-3)

→ a2 = k , b2 = (-5)

Since , The Equations has a unique solution ,

ᴛʜᴇɴ,

→ a1/a2 ≠ b1/b2

putting values we get,

→ 2/(-3) ≠ k/(-5)

→ 2/3 ≠ k/5 ( -ve will cancel)

→ 10 ≠ 3k

→ k ≠ 10/3

Hence, value of k will be except 10/3 , than the Equations will intersect and have an unique solution .

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