# Solve the systemx + 3y + 2z = 73x + 5y + 3z = 22x + 7y + 4z = 1

### 3 Answers | Add Yours

We have to solve the system of equations:

x + 3y + 2z = 7… (1)

3x +5y + 3z =2… (2)

2x+ 7y + 4z =1… (3)

Now (3) - 2*(1)

=> 2x +7y + 4z - 2x- 6y – 4z = 1 - 14

=> y = -13

Substituting y= -13 in (1) and (2) we get the two equations:

x + 2z = 46… (4)

3x +3z = 67… (5)

5-3*(4)

=> 3x + 3z – 3x -6z = 67 -138

=> -3z =-71

=> z = 71/3

Substituting z = 71/3 in x+ 2z = 46

=> x = 46- 142/3

=> x=-4/3

**Therefore x = -4/3, y= -13 and z = 71/3**

(1) x+3y+2z=7 -2x-6y-4z=-14

(3) 2x+7y+4z=1

we have y=-13

Use y=-13 subtitude (1) x+3*(-13)+2z=7

x+2z=46 (5)

Use y=-13 sutitude (2) 3x+5*(-13)+3z=2

3x+3z=67 (6)

Combine (5) and (6) : x+2z=46 -3x-6z=-138

3x+3z=67 3x+3z=67

-3z=-71

z=71/3

(5) x+2z=46 x=46-2z= 46-2*(71/3)=-4/3

the result (-4/3;-13;71/3)

To solve

x + 3y + 2z = 7.............(1)

3x + 5y + 3z = 2...........(2)

2x + 7y + 4z = 1...........(3)

Solution:

(1)*2 - (3) eliminates x and z and gives:

2*3y - 7y = 13.

-y = 13 .

y = -13

Now rewrite the equations (1) and (2) substituting the value y = -13.

Eq(1) becomes:

x+3(-13)+2z = 7. Or x+2z = 46.........(4)

Eq(2) becomes: 3x+5(-13) + 3z = 2 or 3x+3z = 67......(5)

Eq(3)*3 -eq(5) eliminates x and gives: 6z -3z = = 46*3-67 = 71.

Therefore 3z = 71. Or z = 71/3.

Now substite the values y = -3 and z = 71 /3 in (1) and we get:

x+3(-13)+2(71/3) = 7

x = 7+39-142/3 = 46 - 142/3 = (138=142)/3 = -4/3.

Therefore x = -4/3 , y = -13 and z = 71/3.