Question

Solve the system by substitution, solve by elimination. Please do both & explain. Don’t answer if you don’t know. Will mark brainliest. Thank you :)

Answer 1

(1)

[1]   x + 5y - 4z = -10

[2]   2x - y + 5z = -9

[3]   2x - 10y - 5z = 0

Eliminate x by adding -2[1] to [2], and adding -1[2] to [3]:

-2(x + 5y - 4z) + (2x - y + 5z) = -2(-10) + (-9)

-2x - 10y + 8z + 2x - y + 5z = 20 - 9

[4] -11y + 13z = 11

-(2x - y + 5z) + (2x - 10y - 5z) = -(-9) + 0

-2x + y - 5z + 2x - 10y - 5z = 9

[5]  -9y - 10z = 9

Eliminate y by adding -9[4] to 11[5]:

-9(-11y + 13z) + 11(-9y - 10z) = -9(11) + 11(9)

99y - 117z - 99y - 110z = -99 + 99

-227z = 0

z = 0

Plug this solution into either [4] or [5] and solve for y :

[5] -9y - 10(0) = 9

-9y = 9

y = -1

Plug them both into either of [1], [2], or [3] and solve for x :

[1] x + 5(-1) - 4(0) = -10

x - 5 = -10

x = -5

(2)

[1] 2x - y + z = -4

[2] z = 5

[3] -2x + 3y - z = -10

Substitute [2] into [1] and [3]:

2x - y + 5 = -4

[4] 2x - y = -9

-2x + 3y - 5 = -10

[5] -2x + 3y = -5

Solve [4] for y :

2x - y = -9

2x + 9 = y

Substitute this into [5] and solve for x :

-2x + 3(2x + 9) = -5

-2x + 6x + 27 = -5

4x = -32

x = -8

Plug this into either [4] or [5] to solve for y :

[4] 2(-8) - y = -9

-16 - y = -9

y = -5