3 years ago

Solve the system using elimination...

Mathematics

1 answer:

Galina-37 [17]3 years ago

3 0

Answer: x = -1, y = 1, z = 0

Step-by-step explanation:

EQ 1: -2x + 2y + 3z = 0 ⇒ 1(-2x + 2y + 3z = 0) ⇒ -2x + 2y + 3z = 0

EQ 2: -2x - y + 3z = -3 ⇒ -1(-2x - y + 3z = -3) ⇒ 2x + y - 3z = 3

3y = 3

y = 1

EQ 1: -2x + 2y + 3z = 0 ⇒ -2x + 2(1) + 3z = 0 ⇒ -2x + 2 + 3z = 0

EQ 3: 2x + 3y + 3z = 5 ⇒ 2x + 3(1) + 3z = 5 ⇒ 2x + 3 + 3z = 5

5 + 6z = 5

6z = 0

z = 0

EQ 1: -2x + 2y + 3z = 0

-2x + 2(1) + 3(0) = 0

-2x + 2 = 0

-2x = -2

x = -1

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