ok...

[speed4baseball]

ok so

\(\displaystyle 1/f=1/d_{0}+1/d_{1}\)
solve for \(\displaystyle d_{1}\)

i isolated \(\displaystyle d_{1}\)
by multiplying it to both sides but i couldnt figure out if i had to divide \(\displaystyle 1/d_{0} or 1/f\)

 

[soroban]

Hello, speed4baseball!

\(\displaystyle \frac{1}{f} \:=\:\frac{1}{d_o}+ \frac{1}{d_1} \qquad \text{Solve for }d_1\)

First, get rid of the fractions . . .

\(\displaystyle \text{Multiply through by }d_0d_1f\!:\quad d_od_1f\bigg[\frac{1}{f} \;=\;\frac{1}{d_o} + \frac{1}{d_1}\bigg] \quad\Rightarrow\quad d_od_1 \;=\;d_1f + d_of\)


. . \(\displaystyle \text{Then: }\;d_od_1 - d_1f \;=\;d_of \quad\Rightarrow\quad d_1(d_o -f) \;=\;d_of\)


. . \(\displaystyle \text{Therefore: }\;d_1 \;=\;\frac{d_of}{d_o-f}\)

 

[BigGlenntheHeavy]

\(\displaystyle \frac{1}{f} \ = \ \frac{1}{d_0}+\frac{1}{d_1}\)

\(\displaystyle \frac{1}{d_1} \ = \ \frac{1}{f}-\frac{1}{d_0}, \ \frac{1}{d_1} \ = \ \frac{d_0-f}{d_0f}\)

\(\displaystyle Taking \ recipocal \ gives: \ d_1 \ = \ \frac{d_0f}{d_0-f}\)

 

[speed4baseball]

i like the way u explain things soroban
thnx