Question

Solve the system by elimination 3x - 2y + z = 0 6x + 2y +3z = -2 3x - 4y + 5z = 5

Answer 1

Answer:

x = -$\frac{2}{3}$, y = -$\frac{1}{2}$ and z = 1 is the solution of the system of the equations.

Step-by-step explanation:

The system of equation is given as

3x - 2y + z = 0 -----(1)

6x + 2y + 3z = -2 -------(2)

3x - 4y + 5z = 5 ------(3)

Now we add equations 1 from 2

(6x + 2y + 3z) + (3x - 2y + z) = 0 - 2

6x + 3x + 2y - 2y + 3z + z = -2

9x + 4z = -2 -----(4)

Further we multiply equation 1 by 2 and subtract from equation 3

(3x - 4y + 5z) - 2(3x - 2y + z) = 5 - 0

3x - 6x - 4y + 4y + 5z - 2z = 5

-3x + 3z = 5 -----(5)

Now we multiply equation 5 by 3 and add it to equation 4

3(-3x + 3z) + (9x + 4z) = 5×3 - 2

-9x + 9z + 9x + 4z = 15 -2

13z = 13

z = 1

Now we put z = 1 in equation number 5

-3x + 3×1 = 5

-3x + 3 = 5

-3x = 5 - 3

-3x = 2

x = -$\frac{2}{3}$

Now we put the values of x and z in equation 1

3(-$\frac{2}{3}$ - 2y + 1 = 0

-2 - 2y + 1 = 0

-1 - 2y = 0

2y = -1

y = -$\frac{1}{2}$

Therefore, x = -$\frac{2}{3}$, y = -$\frac{1}{2}$ and z = 1 is the solution of the system of the equations.

Answer 2

12x+-4y+9z=3
that's all i got, hope this is right.