# Solve the following the equations: x + y = z , x + 2y = 3z , x + y = 8.

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Given the equations:

x + y = z............(1)

x + 2y = 3z ..............(2)

x + y = 8 .............(3)

Now , from (1) and (3) we notice that z = 8.

Then we will substitute into (1) and (2).

==> x + y= 8............(3)

==> x + 2y = 3*8

==> x+ 2y = 24 ..............(4)

Now we will use the elimination method to solve for y.

We will subtract (3) from (4)

==> y= 24 - 8 = 16

==> y= 16.

Now we will substitute into (3) to find x.

==> x + y = 8

==> x + 16 = 8

==> x = 8

Then the answer for the system is:

**x = 8, y= 16, and z = 8.**

We have to the following equations to solve for x, y and z:

x + y = z… (1)

x + 2y = 3z… (2)

x + y = 8… (3)

Subtracting (1) from (3),

=> x + y – x – y = 8 – z

=> 8 – z = 0

=> z = 8

So we have (1)

=> x + y = 8

=> x = 8 – y

Substitute this in (2)

8 – y + 2y = 24

=> 8 + y = 24

=> y = 16

x = 8 – y

=> x = 8 – 16

=> x = -8

**Therefore x = -8, y = 16 and z = 8**