# Solve the following system of equations: 3x - 2y + z = 1, x + y + z = 4 and 2x + 2y - 3z = 8

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Since the equation was solved for `z` , replace all occurrences of `z` in the other equations with the solution `(-3x+2y+1)` .

`z=-3x+2y+1 `

`x+y+(-3x+2y+1)=4 `

`2x+2y-3(-3x+2y+1)=8 `

Remove the parentheses around the expression `-3x+2y+1` .

`z=-3x+2y+1 `

`x+y-3x+2y+1=4 `

`2x+2y-3(-3x+2y+1)=8`

Combine all similar terms in the polynomial `x+y-3x+2y+1` .

`z=-3x+2y+1`

`-2x+3y+1=4`

`2x+2y-3(-3x+2y+1)=8`

Multiply `-3` by each term inside the parentheses.

`z=-3x+2y+1`

`-2x+3y+1=4`

`2x+2y+9x-6y-3=8`

Combine all similar terms in the polynomial `2x+2y+9x-6y-3` .

`z=-3x+2y+1 `

`-2x+3y+1=4`

`11x-4y-3=8`

Move all terms not containing `x` to the right-hand side of the equation.

`z=-3x+2y+1`

`-2x=3-3y`

`11x-4y-3=8`

Divide each term in the equation by `-2` .

`z=-3x+2y+1`

`x = (3y)/2 - 3/2`

`11x-4y-3=8 `

Since the equation was solved for `x` , replace all occurrences of `x` in the other equations with the solution `((3y)/2 - 3/2).`

`z=-3x+2y+1 `

`x = (3y)/2 - 3/2`

`11 ((3y)/2 - 3/2) - 4y - 3= 8`

Multiply `11` by each term inside the parentheses.

`z=-3x+2y+1`

`x = (3y)/2 - 3/2`

`(33y)/2 - (33)/2 - 4y - 3 = 8`

Combine `(33y)/2 - 4y` into a single expression by finding the least common denominator (LCD). The LCD of `(33y)/2 - 4y` is `2` .

`z=-3x+2y+1 `

`x = (3y)/2 - 3/2`

`(25y)/2 - (33)/2 - 3 = 8`

Combine `-(33)/2 - 3` into a single expression by finding the least common denominator (LCD). The LCD of `-(33)/2 - 3` is `2` .

`z=-3x+2y+1 `

`x = (3y)/2 - 3/2`

`(25y)/2 - (39)/2 = 8`

Move all terms not containing `y` to the right-hand side of the equation.

`z=-3x+2y+1`

`x = (3y)/2 - 3/2`

`(25y)/2 = (55)/2`

Multiply `55 * 2` to get `110` in the numerator.

`z=-3x+2y+1`

`x = (3y)/2 - 3/2`

`25 y = (110)/2`

Reduce the expression by canceling out all common factors from the numerator and denominator.

`z=-3x+2y+1`

`x = (3y)/2 - 3/2`

`25y=55`

Since the equation was solved for `y` , replace all occurrences of `y` in the other equations with the solution `(11/5).`

`z=-3x+2y+1`

`x = (3y)/2 - 3/2`

`y = 11/5`

Since the equation was solved for `y` , replace all occurrences of `y` in the other equations with the solution `(11/5)`

`z = -3x + 2 (11/5) + 1`

`x = (3(11/5))/2 - 3/2`

`y = 11/5`

Multiply `2` by each term inside the parentheses.

`z = -3x + 22/5 + 1`

`x = (3(11/5))/2 - 3/2`

`y = 11/5`

Combine `22/5 + 1` into a single expression by finding the least common denominator (LCD). The LCD of `22/5 + 1` is `5` .

`z = -3x + 27/5`

`x = (3(11/5))/2 - 3/2`

`y = 11/5`

Remove the single term factors from the expression.

`z = 27/5 - 3x`

`x = 33/10 - 3/2`

`y = 11/5`

Remove the parentheses from the numerator.

`z = 27/5 - 3x`

`x = 18/10`

`y = 11/5`

Combine `33/10 - 3/2` into a single expression by finding the least common denominator (LCD). The LCD of `33/10 - 3/2` is `10` .

`z = 27/5 - 3x`

`x = 9 /5`

`y = 11/5`

Since the equation was solved for `x` , replace all occurrences of `x` in the other equations with the solution `(9/5)`

`z = 27/5 - 3(9/5)`

`x = 9/5`

`y = 11/5`

Multiply `-3` by each term inside the parentheses.

`z = 27/5 - 27/5`

`x = 9/5`

`y = 11/5`

Combine all similar expressions.

`z=0 `

`x = 9/5`

`y = 11/5`

This is the solution to the system of equations.

`z=0`

`x = 9/5`

`y = 11/5`

The system of equations:

3x - 2y + z = 1 ...(1)

x + y + z = 4 ...(2)

2x + 2y - 3z = 8 ...(3)

has to be solved for x, y and z.

(3) - 2*(2)

=> 2x + 2y - 3z - 2*(x + y + z) = 8 - 8

=> 2x + 2y - 3z - 2x - 2y - 2z = 0

=> -5z = 0

=> z = 0

Substituting z = 0 in (1) and (3) gives 3x - 2y = 1 and 2x + 2y = 8

Adding the two equations gives `5x = 9 => x = 9/5`

Substituting x = 9/5 in 3x - 2y = 1 gives 3*(9/5) - 2y = 1

=> 2y = 27/5 - 1

=> 2y = 22/5

=> y = 11/5

**The solution of the given set of equations is x = 9/5, y = 11/5 and z = 0**