What is 3x+3y+z=3,3y-z=2 ?

The solution to 3x+3y+z=3,3y-z=2 is x=(-2z+1)/3 ,y=(2+z)/3

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3x+3y+z=3,3y-z=2

3 x + 3 y + z = 3, 3 y - z = 2

x = \frac{- 2 z + 1}{3}, y = \frac{2 + z}{3}

Solution steps

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

Isolate

y

for

3 y - z = 2 : y = \frac{2 + z}{3}

Substitute

y = \frac{2 + z}{3}

[3 x + 3 \cdot \frac{2 + z}{3} + z = 3]

Simplify

[3 x + 2 + 2 z = 3]

Isolate

x

for

3 x + 2 + 2 z = 3 : x = \frac{- 2 z + 1}{3}

Expressing

y

in terms of

z

y = \frac{2 + z}{3}

The solutions to the system of equations are:

x = \frac{- 2 z + 1}{3}, y = \frac{2 + z}{3}

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

3 x - 4 y = 6, x + 2 y = 8

b \cdot m + n + r = 3, 2 m - n + r = 2, 3 m + 3 n - r = 5

3 m - 2 n = - 2, 6 m + 4 n = - 20

2 x + 3 y = 10, 3 x - y = 4

What is 3x+3y+z=3,3y-z=2 ?
The solution to 3x+3y+z=3,3y-z=2 is x=(-2z+1)/3 ,y=(2+z)/3