Given (x/a) + (y/b) + (z/c) = 1, how do you solve for a?

May 27, 2017

color(blue)(a=(-bcx)/(cy+bz-bc)

Explanation:

$\left(\frac{x}{a}\right) + \left(\frac{y}{b}\right) + \frac{z}{c} = 1$

multiply L.H.S. and R.H.S. by abc

$\therefore \frac{x \left({\cancel{a}}^{\textcolor{b l u e}{1}} b c\right)}{\cancel{a}} ^ \textcolor{b l u e}{1} + \frac{y \left(a {\cancel{b}}^{\textcolor{b l u e}{1}} c\right)}{\cancel{b}} ^ \textcolor{b l u e}{1} + \frac{z \left(a b {\cancel{c}}^{\textcolor{b l u e}{1}}\right)}{\cancel{c}} ^ \textcolor{b l u e}{1} = a b c$

$\therefore b c x + a c y + a b z = a b c$

$\therefore a c y + a b z - a b c = - b c x$

$\therefore a \left(c y + b z - b c\right) = - b c x$

:.color(blue)(a=(-bcx)/(cy+bz-bc)

May 27, 2017

$a = \frac{- b c x}{c y + b z - b c}$

Explanation:

A good strategy is to get rid of the fractions first. You can easily do this by multiplying both sides by $a \cdot b \cdot c$.

$a b c \left(\frac{x}{a}\right) + a b c \left(\frac{y}{b}\right) + a b c \left(\frac{z}{c}\right) = a b c \left(1\right)$

$b c x + a c y + a b z = a b c$

Next, collect all the terms with $a$ together on one side.

$a c y + a b z - a b c = - b c x$

Factor out the common $a$ on the left.

$a \left(c y + b z - b c\right) = - b c x$

Divide both sides by the expression without $a$ on the left.

$a = \frac{- b c x}{c y + b z - b c}$