How do you solve this system of equations: #6x - y + z = 12, 2x + z = 7, and y - 2z = 4#?

1 Answer
Dec 1, 2017

x = 2.875, y = 6.5, z = 1.25

Explanation:

There are two main ways to arrive at these solutions: solving by substitution and using matrices. Socratic doesn't have a matrix display, so let's consider the solving by substitution method.
Since the 2nd and 3rd equations only have 2 variables, and both contain z, solve these equations for the other variable:
#2x + z = 7#
#2x = 7 - z#
#color(red)(x = frac{7-z}{2})#

#y - 2z = 4#
#color(blue)(y = 2z + 4)#

Since both of these equations are now in terms of z, plug them into the first equation:
#6color(red)x-color(blue)y+z = 12#
#6(color(red)(frac{7-z}{2}))-(color(blue)(2z+4))+z = 12# -- notice you can solve for z!
#""^3cancel(6)(frac{7-z}{""^1cancel(2)})-(2z+4)+z = 12#

#(21-3z)-(2z+4)+z = 12#

#21-3z - 2z-4+z = 12#

#-4z+17 = 12#

#-4z = -5#

#color(green)(z = frac{5}{4} = 1.25)#

Plug this value in to the next 2 equations to solve for x and y:

#color(red)(x = frac{7-color(green)(1.25)}{2}=frac{5.75}{2}=2.875#

#color(blue)(y = 2(color(green)(1.25)) + 4 = 2.5 + 4 = 6.5#

Matrices are quicker, btw, if you can use a calculator capable of handling matrices, but this way practices your algebra skills! :)