How do you solve the system of equations #7x + 5y + 7z = 60#, #15x + 2z = - 195#, and #5z = 75#?

First solve for #z# using the third equation

#z=75/5=15#

Now that we know #z#, we can plug it into the second equation and solve for #x#

#15x+2(15)=-195#

#15x=-225#

#x=-15#

Now we can find #y# using the values of #x# and #z# and plugging them into the first equation

#7(-15)+5y+7(15)=60#

#5y=60#

#y=12#

This question is far easier than it appears at first.

The key idea is that if a a linear equation has #1# variable, there will be one solution, but as soon as there are #2# variable, #2# equations are required and likewise for #3# variables, there must be #3# equations.

We have all of these scenarios presented here.

Solve the #3rd# equation first as it only has #1# variable,

#5z = 75#
#color(blue)(z =15)" "larr# now use this value for #z# in the #2nd# equation:

#15x +2color(blue)(z) = -195#
#15x +2color(blue)((15)) = -195#
#15x +color(blue)(30) = -195#
#15x = -225#
#color(green)(x =-15)" "larr# use this value for #x# in the first equation

#7color(green)(x)+5y+7color(blue)(z)=60#
#7color(green)((-15))+5y+7color(blue)((15))=60#

#-105 +5y +105 = 60#
#5y=60#
#y=12#