Answer :

Given:

Equation 1: 2x + 3y = 7

Equation 2: (a + b)x + (2a –b)y = 21

Both the equations are in the form of :

a_{1}x + b_{1}y = c_{1} & a_{2}x + b_{2}y = c_{2} where

a_{1} & a_{2} are the coefficients of x

b_{1} & b_{2} are the coefficients of y

c_{1} & c_{2} are the constants

__For the system of linear equations to have infinitely many solutions we must have__

………(i)

According to the problem:

a_{1} = 2

a_{2} = (a + b)

b_{1} = 3

b_{2} = (2a – b)

c_{1} = 7

c_{2} = 21

Putting the above values in equation (i) we get:

On cross multiplication and solving the above equalities we get two sets of linear equation with the variables a & b.

⇒ 7(a + b) = 15*2 ⇒ 7a + 7b = 42 .......(ii)

⇒ 7(2a – b) = 15*3 ⇒ 14a – 7b = 63 ………(iii)

Equation (ii) and (iii) are two linear equations with a and b as variables. To solve this two set of linear equations we use the elimination technique.

In the elimination technique one variable is eliminated by equating it’s coefficient with the other equation. From equation (ii) and (iii) we first eliminate the variable b and find the value of a. Since the coefficient of b are equal but of opposite signs so we add equations (ii) and (iii) On adding we get

(14 + 7)a = 42 + 63 ⇒ 21a = 105 ⇒ a = 5

Putting the value of a in equation (ii) we get

7*5 + 7b = 42 ⇒ 7b = 42 – 35 ⇒ b = 1

__The__ __value of a & b for which the system of equations has infinitely many solutions is a = 5 & b = 1.__

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