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

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Answer 1 · 2017-09-30T09:21:28

#x=1/(2a)-1/(5b)# and #y=1/a+1/(10b)#

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+a)/(10ab)=1/a+1/(10b)#

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

#20abx=10b-4a# or #x=(10b-4a)/(20ab)=1/(2a)-1/(5b)#