How do you solve #5(4z+6)-2(z-4)=7z(z+4)-z(7z-2)-48#?

1 Answer
Dec 18, 2016

Answer:

#z = 43/6#

Explanation:

First, expand the terms within parenthesis paying close attention to the terms inside and outside the parenthesis:

#(5*4z) + (5*6) - (2*z) + (2*4) = (7z*z) + (7z*4) - (z*7z) + (z*2) - 48#

#20z + 30 - 2z + 8 = 7z^2 + 28z - 7z^2 + 2z - 48#

Now we can group and combine like terms on each side of the equation:

#20z - 2z + 30 + 8 = 7z^2 - 7z^2 + 28z + 2z - 48#

#(20 - 2)z + 38 = (7 - 7)z^2 + (28 + 2)z - 48#

#18z + 38 = 0z^2 + 30z - 48#

#18z + 38 = 30z - 48#

Next we can isolate the #z# terms on one side of the equation and the constants on the other side of the equation while keeping the equation balanced:

#18z + 38 color(red)( - 18z + 48) = 30z - 48 color(red)( - 18z + 48)#

#18z - 18z + 38 + 48 = 30z - 18z - 48 + 48#

#0 + 38 + 48 = (30 - 18)z - 0#

#86 = 12z#

Finally, we can solve for #z# while keeping the equation balanced:

#86/color(red)(12) = (12z)/color(red)(12)#

#2/2*43/6 = z#

#43/6 = z#