Question

Ahola...!!!


Solve by cross multiplication method :-


(a+2b)x+(2a-b)y=2 ,
(a-2b)x+(2a+b)y=3 .


All the best...!!!​

Answer 1

SOLUTION

(a+2b)x+(2a-b)y-2=0

(a-2b)x+(2a+b)y-3=0

Here,

a1= a+2b, b1= 2a-b, c1= -2

a2= a-2b, b2= 2a+b, c2= -3

By cross- multiplication, we have

$= > \frac{x}{ - 3(2a - b) - ( - 2)(2a + b)} = \frac{ - y}{3(a + 2b) - ( - 2)(a - 2b)} = \frac{1}{(a + 2b)(2a + b) - (2a - b)(a - 2b)} \\ \\ = > \frac{x}{ - 6a + 3b + 4a + 2b} = \frac{ - y}{ - 3 - 6b + 2a - 4b} = \frac{1}{2 {a}^{2} + ab + 4ab + 2 {b}^{2} - (2 {a}^{2} - 4ab - ab + 2 {b}^{2}) } \\ \\ = > \frac{x}{ - 2a + 5b} = \frac{ - y}{ - a - 10b} = \frac{1}{2 {a}^{2} + ab + 4ab + 2 {b}^{2} - (2a {}^{2} - 4ab - ab + 2b {}^{2} } \\ \\ = > \frac{x}{ - 2a + 5b} = \frac{ - y}{ - (a + 10b)} = \frac{1}{10ab} \\ \\ = > \frac{x}{ - 2a + 5b} = \frac{y}{a + 10b} = \frac{1}{10ab} \\ = > now \\ = > \frac{x}{ - 2a + 5b} = \frac{1}{10ab} \\ \\ = > x = \frac{5b - 2a}{10ab} \\ and \: \\ = > \frac{y}{a + 10b} = \frac{1}{10ab} \\ \\ = > y = \frac{a + 10b}{10ab} \\ \\ \\ = > hence \: x = \frac{5b - 2a}{10ab} \: and \: y = \frac{a + 10b}{10ab}$

Hope it helps ✔️

Answer 2

Explanation:

We have (a+2b)x+(2a−b)y=2 ...................(1)

(a−2b)x+(2a+b)y=3 ...................(2)

Adding the two we get 2ax+4ay=5...................(3)

and subtracting (2) from (1), we get

4bx−2by=−1 ...................(4)

Multiplying (3) by 2b and (4) by a, we get

4abx+8aby=10b ...................(5)

and 4abx−2aby=−a ...................(6)

Subtracting (6) from (5), we have

10aby=10b+a or y=10b+a10ab=1a+110b

Multipying (6) by 4 and adding to (5), we get

20abx=10b−4a or x=10b−4a20ab=12a−15b

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