How do you solve for $\frac { h - 2} { h - 3} = \frac { 1} { - c }$ for $h$ ?

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Answer 1 · 2018-03-28T12:29:21

$h=(2c+3)/(c+1)$

Given:
$(h-2)/(h-3)=1/(-c)$

Cross multiplying

$(h-2)(-c)=(h-3))1)$

$-ch+2c=h-3$

$2c+3=h+ch$

$2c+3=(1+c)h$

$h=(2c+3)/(c+1)$

How do you solve for \frac { h - 2} { h - 3} = \frac { 1} { - c } \frac { h - 2} { h - 3} = \frac { 1} { - c } for hh?