Question

2x+3y=5 and 4x+ky=8 pair of linear equations has unique solution.then find the value of k

Answer 1

Step-by-step explanation:

2x+3y=5.... (1)

4x+ky=8.....(2)

Multiply equation (1) by 2,we get

4x+6y=10....(3)

Subtracting equation (3) from (1) we get

ky-6y=8-10

y(k-6)= -2

k-6= -2/y

k= -2/y+6

Answer 2

The value of k given by k ≠ 6

Given :

The pair of equations 2x + 3y = 5 and 4x + ky = 8 has unique solution

To find :

The value of k

Concept :

For the given two linear equations

$\displaystyle \sf{ a_1x+b_1y+c_1=0 \: and \: \: a_2x+b_2y+c_2=0}$

Consistent :

One of the Below two condition is satisfied

1. Unique solution :

$\displaystyle \sf{ \: \frac{a_1}{a_2} \ne \frac{b_1}{b_2} }$

2. Infinite number of solutions :

$\displaystyle \sf{ \: \frac{a_1}{a_2} = \frac{b_1}{b_2} = \: \frac{c_1}{c_2}}$

Inconsistent :

No solution

$\displaystyle \sf{ \: \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \: \frac{c_1}{c_2}}$

Solution :

Step 1 of 2 :

Write down the given pair of equations

Here the given pair of linear equations are

2x + 3y = 5 - - - - - (1)

4x + ky = 8 - - - - - (2)

Step 2 of 2 :

Find the value of k

Rewriting the given equation as below

2x + 3y - 5 = 0 - - - - - (1)

4x + ky - 8 = 0 - - - - - (2)

Comparing with the equations

a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 we get

a₁ = 2 , b₁ = 3 , c₁ = - 5 and a₂ = 4 , b₂ = k , c₂ = - 8

Now the given pair of linear equations has unique solution

Thus we get

$\displaystyle \sf{ \: \frac{a_1}{a_2} \ne \frac{b_1}{b_2} }$

$\displaystyle \sf{ \implies } \frac{2}{4} \ne \frac{3}{k}$

$\displaystyle \sf{ \implies } \frac{1}{2} \ne \frac{3}{k}$

$\displaystyle \sf{ \implies }k \ne \: 6$

Hence the value of k given by k ≠ 6

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