The answer to the above system of equations is
x = -1
y = 1
z = 3
Step-by-step explanation:
[tex]3x - y + z = - 1 \\ 2x + 3y + z = 4 \\ 5x + 4y + 2z = 5[/tex]
Let's multiply the first equation by 2 then subtract the 3rd equation from the first:
[tex]2(3x - y + z) = - 1 \times 2\\ 6x - 2y + 2z = - 2 \\ - (5x + 4y + 2z) = 5 \\ \\ 6x - 2y + 2z = - 2 \\ - 5x - 4y - 2z = - 5 \\ = x - 6y = - 7[/tex]
Now let's subtract the second equation from the first
[tex]3x - y + z = - 1 \\ - (2x + 3y + z ) = 4\\ \\ 3x - y + z = - 1 \\ - 2x - 3y - z = - 4 \\ = x - 4y = - 5[/tex]
Now let's subtract our two new equations with two variables
[tex] x - 6y = - 7 \\ x - 4y = - 5[/tex]
[tex]x - 6y = - 7 \\ - (x - 4y) = - 5 \\ \\ x - 6y = - 7 \\ - x + 4y = 5 \\ = - 2y = - 2 \\ \frac{ - 2y}{ - 2} = \frac{ - 2}{ - 2} \\ y = 1[/tex]
Let's now substitute y equals 1 into one of the new equations x - 6y = -7
[tex] x - 6y = - 7 \\ x - 6(1) = - 7 \\ x - 6 = - 7 \\ x = - 7 + 6 \\ x = - 1[/tex]
Now that we have answers for x and y we can substitute them into any of the first three original equations above
[tex]3x - y + z = - 1 \\ 3( - 1) - 1 + z = - 1 \\ - 3 - 1 + z = - 1 \\ - 4 + z = - 1 \\ z = - 1 + 4 \\ z = 3[/tex]
Now let's check by plugging in our values x = -1, y = 1 and
z = 3 into any of the other 3 equations
[tex]5x + 4y + 2z = 5 \\ 5( - 1) + 4(1) + 2(3) = 5 \\ - 5 + 4 + 6 = 5 \\ - 1 + 6 = 5 \\ 5 = 5[/tex]
