whats this question? please and thanks broski!

I am a girl but it doesn't matter!

 

First you have to simplify the first equation.

a+b = 8

  -b     -b

a = 8 - b

Then plug in the value for a wherever it is.

1  +    1  = 1       +    1

a        b     (8-b)       b

 

 

Then Solve!

 

Or you could subtract a and do it the other way.

Thanks Quaz...,  I am going to look at it too :))

 

whats

a + b = 8                       (1)

 

1          1       2

_   +     _   =   _              (2)

a          b       3

 

 

I'm going to simplify (2)

 

\(\frac{1}{a}+\frac{1}{b}=\frac{2}{3}\\ 3ab(\frac{1}{a}+\frac{1}{b})=3ab*\frac{2}{3}\\ 3b+3a=2ab\\ 3(b+a)=2ab\\ \qquad a+b=8\\ \qquad b=8-a\\ 3*8=2*a*(8-a)\\ 24=2(8a-a^2)\\ 12=8a-a^2\\ a^2-8a+12=0\\ (a-2)(a-6)=0\\ a=2 \qquad or \qquad a=6\\~\\ When \;\;a=2\;\;\qquad b=8-2=6\\ When\;\; a=6\;\;\qquad b=8-6=2\\ \)

Either a=2 and b=6    or     b=2 and a=6