How do you solve for a in #ax+ z = aw-y#?

1 Answer
Apr 12, 2018

Answer:

See a solution process below:

Explanation:

First, subtract #color(red)(z)# and #color(blue)(aw)# from each side of the equation to isolate the #a# terms while keeping the equation balanced:

#ax - color(blue)(aw) + z - color(red)(z) = aw - color(blue)(aw) - y - color(red)(z)#

#ax - aw + 0 = 0 - y - z#

#ax - aw = -y - z#

Next, factor an #a# from each term on the left side of the equation:

#a(x - w) = -y - z#

Now, divide each side of the equation by #color(red)(x - w)# to solve for #a# while keeping the equation balanced:

#(a(x - w))/color(red)(x - w) = (-y - z)/color(red)(x - w)#

#(acolor(red)(cancel(color(black)((x - w)))))/cancel(color(red)(x - w)) = (-y - z)/(x - w)#

#a = (-y - z)/(x - w)#

We can then multiply the right side of the equation by a form of #1# to rewrite the expression as:

#a = (-1)/-1 xx (-y - z)/(x - w)#

#a = (-1(-y - z))/(-1(x - w))#

#a = (y + z)/(-x + w)#

#a = (y + z)/(w - x)#