When k is the value, the solution of the equations 3x-y = 4, x + 3Y = 0, the solution of the equations 3x-2y-4 + | K + 1| (x + Y-2 / 3) = 0

When k is the value, the solution of the equations 3x-y = 4, x + 3Y = 0, the solution of the equations 3x-2y-4 + | K + 1| (x + Y-2 / 3) = 0

Solve the equations
3x-y=4,
x+3y=0
The results are as follows
x=6/5,y=-2/5
By substituting 3x-2y-4 + | K + 1 | (x + Y-2 / 3) = 0, we get the following formula:
2/5+|k+1|*2/15=0
|k+1|=3
K = 2, or, k = - 4

Solve the equations y + 2 = 1-x.3x + 2Y = - 3

y+2=1-x
So y = - 1-x
Substituting 3x + 2Y = - 3
3x+2(-1-x)=-3
3x-2-2x=-3
x=-1
y=-1-x=0

Solve the system of equations & nbsp; & nbsp; & nbsp; (1) x = y − 23x + 2Y = − 1 (2) x + y = 3x − y = 5

(1) X = y − 2 & nbsp; ① 3x + 2Y = − 1 & nbsp; ②, substituting ① into ② to get: 3 (Y-2) + 2Y = - 1, the solution is: y = 1, substituting y = 1 into ① to get: x = 1-2 = - 1, the solution of the equation system is x = − 1y = 1. (2) x + y = 3 & nbsp; ① x − y = 5 & nbsp; ②, ① + ② to get: 2x = 8, the solution is: x = 4, substituting x = 4 into ① to get: 4 + y = 3, the solution of the equation system is x = 4Y = − 1