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2 years ago

3x+4y-3z=4;5x-y+2z=3;x+2y-z=-2 Solving Systems in three variables

Mathematics

1 answer:

Harlamova29_29 [7]2 years ago

6 0

Answer: x = 3, y = 1, z = 2

Step-by-step explanation:

EQ 1: x - y - z = 0

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

y = 1

EQ 2: 2x - 3y + 2z = 7 → 1(2x - 3y + 2z = 7) → 2x - 3y + 2z = 7

EQ 3: -x + 2y + z = 1 → -2( -x + 2y + z = 1) → -2x + 4y + 2z = 2

y + 4z = 9

y = 1 ⇒ 1 + 4z = 9

4z = 8

z = 2

Input y = 1 and z = 2 into one of the equations to solve for x:

EQ 1: x - y - z = 0

x - (1) - (2) = 0

x - 3 = 0

x = 3

Check:

EQ 2: 2x - 3y + 2z = 7

2(3) - 3(1) + 2(2) = 7

6 - 3 + 4 = 7

3 + 4 = 7

7 = 7 $\checkmark$

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3x+4y-3z=4;5x-y+2z=3;x+2y-z=-2 Solving Systems in three variables​

3x+4y-3z=4;5x-y+2z=3;x+2y-z=-2 Solving Systems in three variables