3x + y + z = 63x - y + 2z = 9y + z = 3y + z = 3y = 3 - z (substitute 3 - z for y into the first equation of the system)3x + y + z = 63x + (3 - z) + z = 63x + 3 = 63x = 3x = 1 (substitute 3 - z for y and 1 for x into the second equation of the system)3x - y + 2z = 93(1) - (3 - z) + 2z = 93 - 3 + z + 2z = 93z = 9z = 3 (which yields y = 0)y = 3 - z = 3 - 3 = 0So that solution of the system of the equations is x = 1, y = 0, and z = 3.
x2+7x+12=0 x2+7x=-12 x+Sqaure Root of 7x= 2(SQ Root Symbol)3
A=0 b=0 c=0
-8z + 12 = 12 - 4z-8z +4z + 12 = 12 - 4z +4z-4z + 12 -12 = 12 -12-4z / -4 = 0 / -4z = 0:)
It's the equation of a straight line with a slope of -1.5 and a y-intercept of -3.
N=0.
This could be intended as a trick question; the answer is 0. 0 + 6 = 12 - 6
3x + y + z = 63x - y + 2z = 9y + z = 3y + z = 3y = 3 - z (substitute 3 - z for y into the first equation of the system)3x + y + z = 63x + (3 - z) + z = 63x + 3 = 63x = 3x = 1 (substitute 3 - z for y and 1 for x into the second equation of the system)3x - y + 2z = 93(1) - (3 - z) + 2z = 93 - 3 + z + 2z = 93z = 9z = 3 (which yields y = 0)y = 3 - z = 3 - 3 = 0So that solution of the system of the equations is x = 1, y = 0, and z = 3.
12 plus 0 is 12
If: x+3+6x+9 = 0 then 7x+12 = 0 So: 7x = -12 and x = -12/7
3x + 6y = 363x - 6y = 0 add both equations6x = 36x = 63x - 6y = 03(6) - 6y = 018 - 6y = 018 = 6y3 = yThe solution point is (6, 3).
There are infinitely many possible answers. 1.5 + 2.5 + 18 = 10 + 0 + 12
0
(x + 12)(x - 6) = 0 x = 6 or -12
I come up with 20.
(11, 1, 0) and (0, 1, 11) are the two possible solutions.
x = 3 and y = 0