Question

Find value of k for which the pair of equations 2x-3y=1 and kx-5y=7 has a unique solutions

Answer 1

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.

___________________________

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 ,

Than,

→ 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 ..

Answer 2

Given: These equations having a unique solution:

• 2x - 3y = 1
• kx  - 5y = 7

To find: The value of k.

Answer:

There are 3 conditions for determining the solutions to an equation(s).

For say, let's consider the two following equations:

$\sf a_1x\ +\ b_1y\ =\ c_1\\\\\sf a_2x\ +\ b_2y\ =\ c_2$

The equation will have no solution (lines will be parallel) if:

$\sf \dfrac{a_1}{a_2}\ =\ \dfrac{b_1}{b_2}\ \neq\ \dfrac{c_1}{c_2} .$

The equation will have a unique solution (lines will intersect) if:

$\sf \dfrac{a_1}{a_2}\ \neq\ \dfrac{b_1}{b_2}.$

And the equation will have infinitely many solutions (lines will coincide) if:

$\sf \dfrac{a_1}{a_2}\ =\ \dfrac{b_1}{b_2}\ =\ \dfrac{c_1}{c_2}.$

Back to the question, it says that the equations have a unique solution. So the second feature will be used.

Let's first obtain $\sf a_1,\ a_2,\ b_1\ and\ b_2.$

$\sf 2x\ -\ 3y\ =\ 1\\\\From\ here,\ \\\\a_1\ =\ 2,\ b_1\ =\ -\ 3.\\\\\\kx\ -\ 5y\ =\ 7\\\\From\ here,\\\\a_2\ =\ k,\ b_2\ =\ -\ 5.$

Now, using the second feature,

$\sf \dfrac{a_1}{a_2}\ \neq \ \dfrac{b_1}{b_2}\\\\Using\ the\ values\ we\ get\ from\ the\ equation,\\\\\dfrac{2}{k}\ \ne\ \dfrac{-\ 3}{-\ 5}\\ \\\\\dfrac{2}{k}\ \ne\ \dfrac{3}{5}\\ \\\sf Cross\ multiplying,\\\\10\ \ne\ \sf 3k\\\\\\\dfrac{10}{3}\ \ne\ k$

Therefore, k can be any value but 10/3.