How do you solve this system of equations: #6x + y = 7, y + z = 2 , and 4x - z = 4#?

2 Answers
Jun 29, 2018

Answer:

#color(maroon)(x = 1/2, y = 4, z = -2#

Explanation:

#6x + y = 7 " Eqn (1)"#

#y + z = 2 " Eqn (2)"#

#4x - z = 4 " Eqn (3)"#

#z = 4x - 4 " from Eqn (3)"#

Substituting the value of z in terms of x in Eqn (2),

#y + 4x - 4 = 2 " or " 4x + y = 6 " Eqn (4)"#

Subtracting Eqn (4) from (1),

#6x + y - 4x - y = 7 - 6#

#2x = 1 ' or x = (1/2)#

Substituting value of x in Eqn (3),

#4 * 1/2 - 4 = z " or " z = -2#

Substituting value of z in Eqn (2),

#y + (-2) = 2 " or " y = 4#

Jun 29, 2018

Answer:

#x=1/2#, #y=4# and #z=-2#

Explanation:

#6x+y-(4x-z)-(y+z)=7-4-2#

#2x=1#, so #x=1/2#

Hence,

#6-1/2+y=7# or #3+y=7#, so #y=4#

Thus,

#4+z=2#, so #z=-2#