(1) 5 x -4 y -3 z =3
(2) x + y -z =0
(3) x = 3 y + 1
We can use substitution, when we take the equation x = 3 y + 1, and substitute x in the equations (1) and (2) by 3y+1.
(1) 5 x -4 y -3 z =3
5(3y+1) -4 y -3 z =3 ---> 15y+5-4y-3z=3 -----> 11y - 3z = - 2 (4)
(2) x + y -z =0 -----> (2) (3y + 1) + y -z =0 -----> 4y - z+1 ----> z= 4y+1 (5)
Now, we can take the equation (5) and substitute the value of (z) into the equation (4).
(4) 11y - 3z = - 2
(5) z= 4y+1
(6) 11y - 3z = - 2 ---> 11y - 3(4y+1) = - 2 --->11y -12y -3 = -2 ---> -y= 1----> y= - 1
Take the equation (5) , and now we know y = - 1.
(5) z= 4y+1 =4*(-1)+1 = -4 +1= - 3, ----> z= - 3
Take the equation (3), and we know that y = - 1,
x = 3 y + 1 = 3*(-1) + 1 = -3 +1 = -2, x = -2.
So, x = - 2, y = - 1, z = - 3.
Check:
(1) 5 *(-2) -4*(-1) -3*(-3) =3 ---> -10+4+9=3---> 3=3
(2) (-2) + (-1) -(-3) =0 ---> -2-1+3=0 ----> 0=0
(3) (-2) = 3*(-1) + 1 ----> -2=-3+1 ---> -2 = -2
