
David W. answered 10/23/15
Tutor
4.7
(90)
Experienced Prof
The elimination method on the first two equations gets rid of the z, as you said.
x + 2y +z = 6 [use eq1]
4x +2y -z = 19 [use eq2]
--------------------
5x +4y = 25 [eq4]
Now, notice that the third equation is the difference of squares -- (a2-b2) = (a+b)(a-b):
((x+2y) + z)((x+2y) - z) = 24 (from eq3)
(x+2y+z)(6) = 24 (see first equation)
x + 2y + z = 4
x + 2y - z = 6 [use eq1]
----------------------- (elimination method; add equations)
2x + 4y = 10 [eq5]
5x + 4y = 25 [use eq4]
--------------------- (elimination method; subtract top from bottom)
3x = 15
x = 5
2(5) + 4y = 10 [use line above and eq5]
4y = 0
y = 0
Checking (very important:
From eq1, z=-1.
Is 5 +2(0)-(-1) = 6 ? yes
Is 4(5)+2(0)+(-1)=19 ?
20 - 1 = 19 ? yes
Is ((5+2(0))2 -(-1)2 =24 ?
55 - 1 = 24 ? yes