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K.marie
How do you do this linear system? {3x+y+2z=3 X-2y+3z=1 4x-8y+12z=7
is it matrices or which topic
Look at equation 1 where you see the term 1y or just y isolate that y to get 3x+y+2z=3 y=3-3x-2z
now go to the other 2 equations and replace y with 3-3x-2z
let @jim_thompson5910 take care of that
so equation 2 goes from x-2y+3z=1 to x-2(3-3x-2z)+3z=1 x-6+6x+4z+3z = 1 7x-6+7z = 1 7x+7z = 1+6 7x+7z = 7 7(x+z) = 7 7(x+z)/7 = 7/7 x+z = 1
Now focus on Equation 3 4x-8y+12z=7 4x-8(3-3x-2z)+12z=7 4x-24+24x+16z+12z=7 28x-24+28z=7 28x+28z=7+24 28x+28z=31
After doing all that, things boil down to these 2 equations x+z = 1 28x+28z=31
use any method to solve that new system of equations for x and z once you have x and z, you can use them both to find y