4a+4b-c=-4

5a+2b+c=-8

8a-4b-4c=-10

Can you please help me with this problem?

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This is a system of three equations with three variables and can be solved by substitution.

`4a+4b-c=-4`      (1)

`5a+2b+c=-8`     (2)

`8a-4b-4c=-10`  (3)

Solve (1) for c.

`c=4a+4b+4`        (4)

Substitute (4) for c in (2) and simplify.

`5a+2b+4a+4b+4=-8`

`9a+6b=-12`

`3a+2b=-4`           (5)

Substitute (4) for c in (3) and simplify.

`8a-4b-4(4a+4b+4)=-10`

`8a-4b-16a-16b-16=-10`

`-8a-20b=6`        (6)

Solve the system of two equations (5) and (6) with two variables by elimination.  Multiply (5) by 10 and add to (6).

`30a+20b=-40`

`-8a-20b=6`

--------------------------

`22a=-34`

Solve for a.

`a=-34/22=-17/11`

Substitute `-17/11` for a in (6).

`-8(-17/11)-20b=6`

Solve for b.

`-20b=6-136/11`

`b=70/((11)(20))=7/22`

Substitute `-17/11` for a and `7/22` for b in (4).

`c=4(-17/11)+4(7/22)+4=-68/11+14/11+44/11=-10/11`

Verify by substituting `-17/11` for a, `7/22` for b and `-10/11` for c in (2).

`5(-17/11)+2(7/22)+(-10/11)=-8`

`-85/11+7/11-10/11=-8`

`-88/11=-8`

`-8=-8`

Thus a=1.55, b=0.32 and c=-0.91

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