pfizer

ITOT().
Hitachi Global Website
... : Hitachi, Ltd. [9] .
()6501 ...
[6501]PER ...
SE .
10/1110/15Hitachi Social Innovation Forum 2021 JAPAN 2021927 BD-STX110G.
6501<>
100-8280 66 JR () 1 .
Astemo !? EVAstemo ...

ITOT().
Hitachi Global Website
... : Hitachi, Ltd. [9] .
()6501 ...
[6501]PER ...
SE .
10/1110/15Hitachi Social Innovation Forum 2021 JAPAN 2021927 BD-STX110G.
6501<>
100-8280 66 JR () 1 .
Astemo !? EVAstemo ...
- 6b + 25b = 155 \Rightarrow 19b = 69 \Rightarrow b = \frac{69}{19}
This is exact. So one bead costs $ \frac{69}{19} $ gold coins. But let's double-check the original equations with integer values.
Try solving again using elimination:
Multiply first equation by 5: $ 25t + 15b = 215 $
Multiply second by 3: $ 6t + 15b = 93 $
Subtract:
Then plug into $ 5t + 3b = 43 $:
\cdot \frac{122}{19} + 3b = 43 \Rightarrow \frac{610}{19} + 3b = 43 \Rightarrow 3b = 43 - \frac{610}{19} = \frac{817 - 610}{19} = \frac{207}{19}
\Rightarrow b = \frac{69}{19}
Thus, one bead costs $ \boxed{\frac{69}{19}} $ gold coins.