Given: 2a+b=3 and substituting for a from eq. 1:
2a+b=3 <==> 2/b+b=3 (eq. 2)
Multiply eq. 2 by b: b(2/b+b)=3b <==> 2+b^2=3b or
b^2-3b+2=0 (eq. 3)
Factoring eq. 3 results in: (b-2)(b-1)=0
By the laws of multiplication, either b-2=0 or b-1=0
Therefore b=2 or b=1
If b=2: a(2)=1 so a=1/2 and 4a^2+b^2 = 4(1/2)^2*2^2 = 5
If b=1: a(1) = 1 so a=1 and 4a^2+b^2 = 4(1)^2*1^2 = 5
The answer is 5:
Given: ab=1 <==> a=1/b (eq. 1)
Given: 2a+b=3 and substituting for a from eq. 1:
2a+b=3 <==> 2/b+b=3 (eq. 2)
Multiply eq. 2 by b: b(2/b+b)=3b <==> 2+b^2=3b or
b^2-3b+2=0 (eq. 3)
Factoring eq. 3 results in: (b-2)(b-1)=0
By the laws of multiplication, either b-2=0 or b-1=0
Therefore b=2 or b=1
If b=2: a(2)=1 so a=1/2 and 4a^2+b^2 = 4(1/2)^2*2^2 = 5
If b=1: a(1) = 1 so a=1 and 4a^2+b^2 = 4(1)^2*1^2 = 5
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