# How do you solve the system 4x+4y+z=24, 2x-4y+z=0, and 5x-4y-5z=12?

Jan 15, 2018

Add the equations to eliminate variables and solve.

#### Explanation:

$4 x + 4 y + z = 24$
$2 x - 4 y + z = 0$
$5 x - 4 y - 5 z = 12$

Choose two equations that will easily cancel out.
$4 x + 4 y + z = 24$
$2 x - 4 y + z = 0$
The y's cancel out, leaving you with $6 x + 2 z = 24$. This can be simplified to $3 x + z = 12$.

Choose another two equations to get another binomial.
$4 x + 4 y + z = 24$
$5 x - 4 y - 5 z = 12$
Again, the y's cancel out, leaving you with $9 x - 4 z = 36$.

Use substitution, making $3 x + z = 12$ into $z = 12 - 3 x$. Plug this into $9 x - 4 z = 36$ to get $9 x - 4 \left(12 - 3 x\right) = 36$. Distribute to get the equation $9 x - 48 + 12 x = 36$. Combine like terms, making the equation $21 x = 84$. Divide both sides by $21$ to get $x = 4$.

You can then use this value in the equation $z = 12 - 3 x$. Plug it in to get $z = 12 - 3 \left(4\right)$, which simplifies to $z = 0$.

With these two values you can use any of the original equations to find the remaining value. $2 \left(4\right) - 4 y + \left(0\right) = 0$. Simplify to get $8 - 4 y = 0$. Add $4 y$ to both sides to get $8 = 4 y$, and divide both sides by $4$ to get $y = 2$.

The final values are $x = 4$, $y = 2$ and $z = 0$.
This can be checked by plugging these values into the two remaining original equations.