How do you solve the system of linear equations: $(a + 2 b) x + (2 a - b) y = 2$ $(a+2b)x+(2a-b)y=2$ and $(a - 2 b) x + (2 a + b) y = 3$ $(a-2b)x+(2a+b)y=3$ ?

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Answer 1 · 2017-09-30T09:21:28

$x = \frac{1}{2 a} - \frac{1}{5 b}$ $x=1/(2a)-1/(5b)$ and $y = \frac{1}{a} + \frac{1}{10 b}$ $y=1/a+1/(10b)$

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

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

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

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

$4 b x - 2 b y = - 1$ $4bx-2by=-1$ ...................(4)

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

$4 a b x + 8 a b y = 10 b$ $4abx+8aby=10b$ ...................(5)

and $4 a b x - 2 a b y = - a$ $4abx-2aby=-a$ ...................(6)

Subtracting (6) from (5), we have

$10 a b y = 10 b + a$ $10aby=10b+a$ or $y = \frac{10 b + a}{10 a b} = \frac{1}{a} + \frac{1}{10 b}$ $y=(10b+a)/(10ab)=1/a+1/(10b)$

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

$20 a b x = 10 b - 4 a$ $20abx=10b-4a$ or $x = \frac{10 b - 4 a}{20 a b} = \frac{1}{2 a} - \frac{1}{5 b}$ $x=(10b-4a)/(20ab)=1/(2a)-1/(5b)$

How do you solve the system of linear equations: (a+2b)x+(2a-b)y=2(a+2b)x+(2a−b)y=2(a+2b)x+(2a-b)y=2 and (a-2b)x+(2a+b)y=3(a−2b)x+(2a+b)y=3(a-2b)x+(2a+b)y=3?