Question

For what value of k, the linear pair of equations 2x-3y=1 ; kx+5y=7 has a unique solution

Answer 1

2x-3y-1=0 --(1)
kx+5y-7=0---(2)
compare (1) and (2) with a1xb1y+c1=0 and a2x+b2y+c2 =0

a1=2; b1=-3; c1=-1

a2=k; b2=5; c2=-7

given (1) and (2) has unique solution
therefore

a1/a2 ≠ b1/b2
2/k ≠-3/5
k/2≠-5/3
k≠-10/3
u can take k all real values other than -10/3

Answer 2

Answer:

The value of k is all real number except $k=-\frac{10}{3}$    

Step-by-step explanation:

Given : Equation $2x-3y=1,kx+5y=7$

To find : The value of k for which the following system of equation has unique solution.

Solution :  

When the system of equation is in form $ax+by+c=0, dx+ey+f=0$ then the condition for unique solutions is  

$\frac{a}{d}\neq\frac{b}{e}$

Comparing, a=2,b=-3,d=k and e=5

Substituting the values in the condition,

$\frac{2}{k}\neq\frac{-3}{5}$

Cross multiply,

$2\times 5\neq-3\times k$

$10\neq-3k$

$k\neq-\frac{10}{3}$

Therefore, The value of k is all real number except $k=-\frac{10}{3}$