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4 years ago

Solve the following system of equations.

Mathematics

2 answers:

Ronch [10]4 years ago

6 0

The answer is A. I plugged it into a TI-83 calculator. If you need to know how I did that I will be happy to help

Lilit [14]4 years ago

4 0

Answer:

Option A ( 2, 2, 2)

Step-by-step explanation:

2x + 2y + z = 10 ------------(1)

3x - y + 3z = 10 ------------(2)

2x + 3y - 2z = 6 ------------(3)

Equation (1) - equation (3)

(2x + 2y + z) - (2x + 3y - 2z) = 10 - 6

(2x - 2x) + (2y - 3y) + (z + 2z) = 4

-y + 3z = 4 --------------------(4)

Equation (1) × 3 - equation (2) × 2

3(2x + 2y + 2) - 2(3x - y + 3z) = 30 - 20

6x + 6y + 3z - 6x + 2y - 6z = 10

(6x - 6x) + (6y + 2y) + (3z - 6z) = 10

8y - 3z = 10 ----------------(5)

Now equation (4) × 8 + equation (5)

8(-y + 3z) + (8y - 3z) = 32 + 10

-8y + 2yz + 8y - 3z = 42

21z = 42

⇒ z = 2

Now we put z = 2 in equation (4)

-y + 3 × 2 = 4 ⇒ -y = 4 - 6

y = 2

From equation (1) 2x + 2 × 2 + 2 = 10

2x + 4 + 2 = 10

2x = 10 - 6 = 4 ⇒ x = 2

Option A ( 2, 2, 2) is the answer.

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