What is x+y+z=7,2x+3y+z=13,3x-y+2z=12 ?

The solution to x+y+z=7,2x+3y+z=13,3x-y+2z=12 is x= 14/5 ,y= 8/5 ,z= 13/5

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x+y+z=7,2x+3y+z=13,3x-y+2z=12

x + y + z = 7, 2 x + 3 y + z = 13, 3 x - y + 2 z = 12

x = \frac{14}{5}, y = \frac{8}{5}, z = \frac{13}{5}

Solution steps

x + y + z = 7 2 x + 3 y + z = 13 3 x - y + 2 z = 12

Isolate

x

for

x + y + z = 7 : x = 7 - y - z

Substitute

x = 7 - y - z

[2 (7 - y - z) + 3 y + z = 13 3 (7 - y - z) - y + 2 z = 12]

Simplify

[y - z + 14 = 13 - 4 y - z + 21 = 12]

Isolate

y

for

y - z + 14 = 13 : y = z - 1

Substitute

y = z - 1

[- 4 (z - 1) - z + 21 = 12]

Simplify

[- 5 z + 25 = 12]

Isolate

z

for

- 5 z + 25 = 12 : z = \frac{13}{5}

For

y = z - 1

Substitute

z = \frac{13}{5}

y = \frac{13}{5} - 1

\frac{13}{5} - 1 = \frac{8}{5}

y = \frac{8}{5}

For

x = 7 - y - z

Substitute

y = \frac{8}{5}, z = \frac{13}{5}

x = 7 - \frac{8}{5} - \frac{13}{5}

7 - \frac{8}{5} - \frac{13}{5} = \frac{14}{5}

x = \frac{14}{5}

The solutions to the system of equations are:

x = \frac{14}{5}, y = \frac{8}{5}, z = \frac{13}{5}

3 x + y = 21, 2 x - y = 4

3 x - 3 y = 3, 3 x = 4

x = \frac{1}{4}, y = \frac{3}{2}

a + a = 2, a + b = 3 a + b \cdot 2

29 x + 7 y = 15, 5 x - 3 y = 29

What is x+y+z=7,2x+3y+z=13,3x-y+2z=12 ?
The solution to x+y+z=7,2x+3y+z=13,3x-y+2z=12 is x= 14/5 ,y= 8/5 ,z= 13/5