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

2 Answers
May 27, 2017

Answer:

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

Explanation:

#(x/a)+(y/b)+z/c=1#

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

#:.(x(cancela^color(blue)1bc))/cancela^color(blue)1+(y(acancelb^color(blue)1c))/cancelb^color(blue)1+(z(abcancelc^color(blue)1))/cancelc^color(blue)1=abc#

#:.bcx+acy+abz=abc#

#:.acy+abz-abc=-bcx#

#:.a(cy+bz-bc)=-bcx#

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

May 27, 2017

Answer:

#a=(-bcx)/(cy+bz-bc)#

Explanation:

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

#abc (x/a)+abc (y/b)+abc (z/c)=abc (1)#

#bcx+acy+abz=abc#

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

#acy+abz-abc=-bcx#

Factor out the common #a# on the left.

#a(cy+bz-bc)=-bcx#

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

#a=(-bcx)/(cy+bz-bc)#