Use the 5-step procedure to solve by substitution: x - y + 2z = -7, y + z = 1, and x - 2y - 3z = 0

Answer 1

Use the 5-step procedure to solve by substitution: x - y + 2z = -7, y + z = 1, and x - 2y - 3z = 0

System of three simultaneous equations in three unknowns x,y and z (although one equation only has 2 unknowns, which makes life easier).

(1) x - y + 2z = -7

(2) y + z = 1

(3) x - 2y - 3z = 0

rearrange (2) to get y = 1 - z which can be substituted in (1) and (3) to eliminate one variable.

x - y + 2z = -7 becomes x - (1-z) + 2z = -7

=> x - 1 + z + 2z = -7

=> x +3z = -6

x - 2y - 3z = 0 becomes x - 2(1-z) - 3z = 0

=> x -2 + 2z - 3z = 0

=> x - z = 2

So now we have a system of two simultaneous equations in two unknowns:

(4) x +3z = -6

(5) x - z = 2

single x in both, so a subtraction will eliminate it:

x +3z = -6

-

x - z = 2

=

0x + 4z = -8

=> z = -2

From equation (2) we already know that y = 1 - z

=> y = 1 - (-2)

=> y = 3

Finally we need x, so we substitute both y and z into (1)

x - y + 2z = -7 becomes x - (3) + 2(-2) = -7

=> x - 3 - 4 = -7

=> x = 0

Hence our solution is x= 0, y= 3, z= -2

As always, double-check your solution by substituting in one of the equations, I'll use (3).

x - 2y - 3z = 0 becomes 0 - 2(3) - 3(-2) = 0

=> 0 - 6 + 6 = 0

=> 0 = 0 which is true

Answer 2

Per eq 3

x = z+5

eq 1 becomes: z+5 - y + 2z = 7

3z - y = 2

eq 2 becomes: 2(z+5) + y + z =8

2z + 10 + y + z = 8

3z + y = -2

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Then 3z = y+2 and 3z = -2 - y

so y+2 = -2 - y

2y = -4

y = -2

then 3z + -2 = -2

3z = 0

z=0

this forces x=5

{x=5, y = -2, z=0} is the solution