Question

Which is the solution to this syatem of equation 3x+4y-7z=131x+1y+1z=40x+2y-1z=2

Answer 1

x = 3, y = 1, z = 0

Step-by-step explanation

We have the system of equations:

3x + 4y - 7z = 13 _(1)

x + y + z = 4 ____(2)

2y - z = 2 ______(3)

From (3):

z = 2y - 2

Put that in (1) and (2):

3x + 4y - 7(2y - 2) = 13

3x + 4y - 14y + 14 = 13

3x - 10y = 13 - 14

3x - 10y = -1 ____(4)

and

x + y + 2y - 2 = 4

x + 3y = 4 + 2

x + 3y = 6 _____(5)

Now, we have:

3x - 10y = -1 ____(4)

x + 3y = 6 _____(5)

From (5):

x = 6 - 3y ___(6)

Put that in (4):

3(6 - 3y) - 10y = -1

18 - 9y - 10y = -1

-19y = -1 - 18

-19y = -19

=> y = -19 / -19

y = 1

Put that in (6):

x = 6 - 3(1)

x = 6 - 3

x = 3

Recall that:

z = 2y - 2

=> z = 2(1) - 2

z = 2 - 2

z = 0

So, x = 3, y = 1, z = 0