Systems of equations

skipjack

Forum Staff
Dec 2006
21,478
2,470
From the first and third equations (eliminating y), 4x = 5z - 23,
so z = 21/(1 - 2a) implies x = (105/(1 - 2a) - 23)/4 = (23a + 41)/(2(1 - 2a))
and y = (9 - 2x - z)/2 = -(53 + 41a)/(2(1 - 2a)).
 
Mar 2019
196
1
TTF area
From the first and third equations (eliminating y), 4x = 5z - 23,
so z = 21/(1 - 2a) implies x = (105/(1 - 2a) - 23)/4 = (23a + 41)/(2(1 - 2a))
and y = (9 - 2x - z)/2 = -(53 + 41a)/(2(1 - 2a)).
I see how you eliminate y, but how do you go from 4x=5z-23 to z=21(1-2a)?
 

skipjack

Forum Staff
Dec 2006
21,478
2,470
I explained how to obtain z = 21/(1 - 2a) here.