Solve by Addition/Elimination x-3y+4z=-3 , 3x+y-z=4 , -2x+2y+3z=-5

Problem

{[x−3*y+4*z=−3],[3*x+y−z=4],[−2*x+2*y+3*z=−5])

Solution

1. Label the equations for clarity.

(L_1):x−3*y+4*z=−3

(L_2):3*x+y−z=4

(L_3):−2*x+2*y+3*z=−5

2. Eliminate x from (L_2) by adding −3×(L_1) to (L_2)

−3*(x−3*y+4*z)+(3*x+y−z)=−3*(−3)+4

−3*x+9*y−12*z+3*x+y−z=9+4

10*y−13*z=13

3. Eliminate x from (L_3) by adding 2×(L_1) to (L_3)

2*(x−3*y+4*z)+(−2*x+2*y+3*z)=2*(−3)−5

2*x−6*y+8*z−2*x+2*y+3*z=−6−5

−4*y+11*z=−11

4. Solve the new system of two equations for y and z

(L_4):10*y−13*z=13

(L_5):−4*y+11*z=−11

5. Eliminate y by multiplying (L_4) by 2 and (L_5) by 5 then adding them.

2*(10*y−13*z)+5*(−4*y+11*z)=2*(13)+5*(−11)

20*y−26*z−20*y+55*z=26−55

29*z=−29

z=−1

6. Substitute z=−1 into (L_4) to find y

10*y−13*(−1)=13

10*y+13=13

10*y=0

y=0

7. Substitute y=0 and z=−1 into (L_1) to find x

x−3*(0)+4*(−1)=−3

x−4=−3

x=1

Final Answer

(x,y,z)=(1,0,−1)

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