Question

Hello!! ^o^
Kindly help me out doing this question..
Plz give complete explanation so that I can understand the topic clearly!!

Answer 1

Here is the solution :

Concept :

Let 2 Equations be,
ax + by = c -------(1)
dx + ey = f --------(2)

Now, (1) and (2) are said to have Infinite solutions ,

if (a/d) = (b/e) = (c/f)

Now, Let us find the values,

(a)
[1st EQ] : 2x + 3y = 7,
[2nd EQ] : a(x+y) - b(x-y) = 3a + b - 2,

Simplifying 2nd EQ ,(Because we need to have like terms seperated),

=> ax + ay - bx + by = 3a + b - 2,

=> x(a-b) + y(a+b) = 3a + b - 2,
[Now the 2nd EQ is simplified]

Applying the conceptual formula here,

=> ((2/(a-b)) = ((3/(a+b)) = ((7/(3a+b-2))

=> Cross Multiplying,

=> 2(a+b) = 3(a-b) = ((7/(3a+b-2))

=> Considering L.H.S and R.H.S of "=" 1st Sign,

=> 2a + 2b = 3a - 3b

=> 5b = a -------- (3)

Substituting (3) in Above one,

=> ((2/(5b-b)) = ((3/(5b+b)) = ((7/(15b+b-2))

=> (1/(2b)) = (1/(2b)) = (7/(16b-2))

=> 1/(2b) = 7/(16b-2)

=> Cross Multiplying,

=> 16b - 2 = 14b

=> 2b = 2,

=> b = 1 ,

=> a = 5 ,

Therefore : Solution for (a) is a = 5, b = 1,

Now,

Solution for (b)

=> Given Equations :

[1st EQ ] : 2x - 3y = 7
[2nd EQ] : (a+b)x - (a+b-3)y = 4a + b,

Here in 2nd EQ Like terms are already seperated, So no need to simplify it !.

From Conceptual Formula Again,

=> (2/(a+b)) = ((-3)/(-(a+b-3))) = (7/(4a+b))

=> (2/(a+b)) = (3/(a+b-3)) = (7/(4a+b))

=> Cross Multiplying terms of 1st "=" Sign,

=> 2(a+b-3) = 3(a+b)

=> 2a + 2b - 6 = 3a + 3b,

=> -6 = a + b,

Substituting this in Above one,

=> (2/(-6)) = (3/(-9)) = (7/(3a + a + b))

=> (-1/3) = (-1/3) = (7/(3a - 6))

=> Cross Multiplying the terms of 2nd "=" Sign,

=> (-1)(3a - 6) = (7)*(3)

=> 6 - 3a = 21,

=> -3a = 15,

=> a = -5,

=> b = -1,

Therefore : Answers for (b) are a is -5 and b is -1,


Finally, Therefore : Solutions of (a) and (b) are

(a) :

a = 5,
b = 1,

(b) :

a = -5,
b = -1,

Hope you understand, Have a Great day,
Thanking you , Bunti 360 !.