How do you solve for G in #AL+G/EB=R+A #?

1 Answer
Apr 29, 2017

Answer:

See the entire solution process below:

Explanation:

First, subtract #color(red)(AL)# from each side of the equation to isolate the term containing the #G# while keeping the equation balanced:

#AL + G/EB - color(red)(AL) = R + A - color(red)(AL)#

#AL - color(red)(AL) + G/EB = R + A - AL#

#0 + G/EB = R + A - AL#

#G/EB = R + A - AL#

Now, multiply each side of the equation by #color(red)(E)/color(blue)(B)# to solve for #G# while keeping the equation balanced:

#color(red)(E)/color(blue)(B) * G/EB = color(red)(E)/color(blue)(B)(R + A - AL)#

#cancel(color(red)(E))/cancel(color(blue)(B)) * G/color(red)(cancel(color(black)(E)))color(blue)(cancel(color(black)(B))) = color(red)(E)/color(blue)(B)(R + A - AL)#

#G = color(red)(E)/color(blue)(B)(R + A - AL)#

Or

#G = (color(red)(E)(R + A - AL))/B#

Or

#G = (ER + EA - EAL)/B#

Or

#G = (ER)/B + (EA)/B - (EAL)/B#