Question

find value of a and b for which the following equation has an infinitely number of solution 2x-(a-4)y=2b+1 and 4x-(a-1)y=5b-1

Answer 1

Note:

If we consider two linear equations in two variables say;

a1x + b1y + c1 = 0 and

a2x + b2y + c2 = 0

Then;

The condition for infinitely many solutions is given by;

a1/a2 = b1/b2 = c1/c2.

Here;

The given equations are;

2x-(a-4)y=2b+1 OR 2x-(a-4)y-(2b +1)=0

And;

4x-(a-1)y=5b-1 OR 4x-(a-1)y-(5b-1)=0

Clearly, we have;

a1 = 2

a2 = 4

b1 = -(a-4)

b2 = -(a-1)

c1 = -(2b+1)

c2 = -(5b-1)

Thus;

For the given pair of linear equations to have infinitely many solutions,

We have;

=> a1/a2 = b1/b2 = c1/c2

=> 2/4 = -(a-4)/-(a-1) = -(2b+1)/-(5b-1)

=> 1/2 = (a-4)/(a-1) = (2b+1)/(5b-1)

Thus;

=> 1/2 = (a-4)/(a-1)

=> (a-1) = 2(a-4)

=> a - 1 = 2a - 8

=> 2a - a = 8 - 1

=> a = 7

Also;

=> 1/2 = (2b+1)/(5b-1)

=> (5b-1) = 2(2b+1)

=> 5b - 1 = 4b + 2

=> 5b - 4b = 2 + 1

=> b = 3

Thus;

The required values of "a" and "b" are

7 and 3 respectively.

Answer 2

AnswEr :

⠀

To Get Infinite Number of Solution, Pair of Linear Equation should must follow this Condition :

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

We Have Two Different Linear Pair of Equation as :

⇒ 2x - (a - 4)y = 2b + 1

⇒ 2x - (a - 4)y - (2b + 1) = 0 ⠀⠀(¡)

⇒ 4x - (a - 1)y = 5b - 1

⇒ 4x - (a - 1)y - (5b - 1) = 0 ⠀⠀(¡¡)

⋆ We Clearly Get the Values from these Equations :

• a1 = 2 ⠀and, ⠀a2 = 4

• b1 = - (a - 4) ⠀and, ⠀b2 = - (a - 1)

• c1 = - (2b + 1) ⠀and, ⠀c2 = - (5b - 1)

⠀

_________________________________

⋆ So, Now :

$\mathsf{ \dfrac{2}{4} = \dfrac{ \cancel- (a - 4)}{ \cancel - (a - 1)} = \dfrac{ \cancel - (2b + 1)}{ \cancel - (5b - 1)} }$

⋆ Once We Will Take This.

$\implies \mathsf{ \dfrac{2}{4} = \dfrac{(a - 4)}{(a - 1)} }$

$\implies \mathsf{ 2 \times (a - 1) = 4 \times (a - 4) }$

$\implies \mathsf{ 2a - 2 = 4a - 16 }$

$\implies \mathsf{ 16 - 2 = 4a - 2a }$

$\implies \mathsf{ 14 = 2a }$

$\implies \mathsf{ a = \cancel \dfrac{14}{2} }$

$\implies \boxed{\mathsf{ a = 7 }}$

⠀

⋆ Now, We Will Take Another Term.

$\implies \mathsf{ \dfrac{2}{4} = \dfrac{(2b + 1)}{(5b - 1)} }$

$\implies \mathsf{ 2 \times (5b - 1) = 4 \times (2b + 1) }$

$\implies \mathsf{10b - 2 = 8b + 4}$

$\implies \mathsf{10b - 8b = 4 + 2}$

$\implies \mathsf{2b = 6}$

$\implies \mathsf{ b = \cancel \dfrac{6}{2} }$

$\implies \boxed{\mathsf{ b = 3 }}$

⠀

$\therefore$ Value of a = 7 and, b = 3 will give infinite number of Solution for this Pair of Linear Equation.