Solve the system of equations 2x+4y+3z=6 5x+8y+6z=4

Answer:Solution:

{x, y, z} = {-8,52/7,-18/7}

System of Linear Equations entered:

[1] 2x + 4y + 3z = 6

[2] 5x + 8y + 6z = 4

[3] 4x + 5y + 2z = 0

Solve by Substitution:

/ / Solve equation [3] for the variable z

[3] 2z = - 4x - 5y

[3] z = - 2x - 5y/2

/ / Plug this in for variable z in equation [1]

[1] 2x + 4y + 3• (-2x-5y/2) = 6

[1] - 4x - 7y/2 = 6

[1] - 8x - 7y = 12

/ / Plug this in for variable z in equation [2]

[2] 5x + 8y + 6• (-2x-5y/2) = 4

[2] - 7x - 7y = 4

/ / Solve equation [2] for the variable y

[2] 7y = - 7x - 4

[2] y = - x - 4/7

/ / Plug this in for variable y in equation [1]

[1] - 8x - 7• (-x - 4/7) = 12

[1] - x = 8

/ / Solve equation [1] for the variable x

[1] x = - 8

/ / By now we know this much:

x = - 8

y = - x-4/7

z = - 2x-5y/2

/ / Use the x value to solve for y

y = - (-8) - 4/7 = 52/7

/ / Use the x and y values to solve for z

z = - 2 (-8) - (5/2) (52/7) = - 18/7

Solution:

{x, y, z} = {-8,52/7,-18/7}