Need solving three-variable systems by substitution algebra 2 problem x-4y+z=6 2x+5y-z=7 2x-y-z=1

Call :   x - 4y + z = 6     (e1) == > z = 6 - x + 4y       2x + 5y - z = 7      (e2)       2x - y    - z = 1      (e3)
sub z from e1 in e2  you have :      2x + 5y - 6 + x - 4y = 7  ==>  3x + y = 13  ==> 3x = 13 - y   (*)sub z from e1 in e3 you have:      2x - y - 6 + x - 4y = 1    ==>   3x - 5y = 7      (**)
sub 3x from (*) in (**), you have     13 - y - 5y = 7   ==>  6 = 6y  ==> y = 1sub y = 1 into (*), you have:     3x = 13 - 1 = 12    ==> x = 4
sub both y = 1 and x = 4 into (e1), you have:     4 - 4(1) + z = 6  ==> z = 6
so answer :   x = 4, y = 1, z =6  (you can use these values to check other equations to see if they come out all right)

Answer:

the image is given below

step-by-step explanation:

since the options are not given so we draw it from the equation

the given equation is

y = (x-1) (x+4)

y = y² + 3x - 4

the above equation is a quadratic equation which will have two roots. if we draw the graph for it, it will look like this