How do you solve for y in #k(y+3z)=4(y-5)#?

1 Answer
May 23, 2018

Answer:

See a solution process below:

Explanation:

First, expand the terms in parenthesis on each side of the equation by multiplying each term within the parenthesis by the term outside the parenthesis:

#color(red)(k)(y + 3z) = color(red)(4)(y - 5)#

#(color(red)(k) xx y) + (color(red)(k) xx 3z) = (color(red)(4) xx y) - ((color(red)(4) xx 5)#

#ky + 3kz = 4y - 20#

Next, subtract #color(red)(3kz)# and #color(blue)(4y)# from each side of the equation to isolate the #y# terms while keeping the equation balanced:

#ky - color(blue)(4y) + 3kz - color(red)(3kz) = 4y - color(blue)(4y) - 20 - color(red)(3kz)#

#ky - 4y + 0 = 0 - 20 - 3kz#

#ky - 4y = 20 - 3kz#

The factor out the common term on the left side of the equation:

#kcolor(red)(y) - 4color(red)(y) = 20 - 3kz#

#(k - 4)y = 20 - 3kz#

Now, divide both sides of the equation by #color(red)((k - 4))# to solve for #y# while keeping the equation balanced:

#((k - 4)y)/color(red)((k - 4)) = (20 - 3kz)/color(red)((k - 4))#

#(color(red)(cancel(color(black)((k - 4))))y)/cancel(color(red)((k - 4))) = (20 - 3kz)/(k - 4)#

#y = (20 - 3kz)/(k - 4)#