Question

3×3 Systems Elimination by Addition

-5x- 2y+20z=-28
2x-5y+15z=28
-2x-2y-5z=-12

Please show all of the steps clearly so I will be able to do it on my own with practice.

Answer 1

-5x- 2y+20z=-28
2x-5y+15z=28
-2x-2y-5z=-12

Add last two equations to obtain

-7y +10z = 16

Multiply first equation by 2 and second eqaution by 5 to obtain

-10x- 4y+40z=-56
10x-25y+75z=140

Add both

-29y +115z=84

Now we have

-7y +10z = 16

-29y +115z=84

Multiply first eqaution by 29 and second equation by -7 to obtain

-203y +290z=464

203y-805z = -588

Add both

-515z=-124

z =124/515

Substitute z =124/515 in -7y +10z = 16

-7y +10 *124/515 =16

y= -200/103

Finally substitute y= -200/103,z =124/515 in -5x- 2y+20z=-28

-5x -2(-200/103) +20*124/515 = -28

x= 756/103

Hence solution is

x = 756/103, y = -200/103, z = 124/515