# How do you solve the system of equations algebraically x-3z=7, 2x+y-2z=11, -x-2y+9z=13?

Jan 14, 2017

$x = 10$; $y = - 7$; $z = 1$

#### Explanation:

Write out the three equations to line up the unknown values. This will give you a clue how to proceed.

Always be careful of the signs $\left(-\right) \left(+\right)$ as they move across the $\left(=\right)$. Always write the sign with the value to which it refers. $\left(- 3 z\right)$

$x - 3 z = 7$

$2 x + y - 2 z = 11$

$- x - 2 y + 9 z = 13$

Already we can see that it is easy to solve for $x$ in the first equation:

x -3z = 7 >> x = 3z +7

Insert the value for $x$ into the second equation;

$2 \left(3 z + 7\right) + y - 2 z = 11$

$6 z + 14 + - 2 z = 11$

$4 z + y = - 3$

$y = - 4 z - 3$

Substitute the values of $x$ and $y$ into the third equation:

$- \left(3 z + 7\right) - 2 \left(- 4 z - 3\right) + 9 z = 13$

$- 3 z - 7 + 8 z + 6 + 9 z = 13$

$14 z = 14$

$z = 1$

Substitute the value of $z$ into the first equation:

$x - 3 \left(1\right) = 7$

$x = 10$

Substitute the values of $x$ and $z$ into the second equation:

$2 \left(10\right) + y - 2 \left(1\right) = 11$

$20 + y - 2 = 11$

$y = - 20 + 2 + 11$

$y = - 7$

Use either equation that includes all three unknowns to check:

$- x - 2 y + 9 z = 13$

$- \left(10\right) - 2 \left(- 7\right) + 9 \left(1\right) = 13$

$13 = 13$