# How do you solve the following linear system: 2x-y+3z= 7 , 5x-4y-2z= 3 , 4x+y-2z=-4 ?

$x = \frac{17}{81}$
$y = - \frac{110}{81}$
$z = \frac{47}{27}$

#### Explanation:

Let $2 x - y + 3 z = 7$ 1st equation
Let $5 x - 4 y - 2 z = 3$ 2nd equation
Let $4 x + y - 2 z = - 4$ 3rd equation

Use 3rd equation so that $y$ in terms of $x$ and $z$

$y = - 4 + 2 z - 4 x$ 3rd equation

Use now 1st and 3rd
$2 x - y + 3 z = 7$ 1st equation
$2 x - \left(- 4 + 2 z - 4 x\right) + 3 z = 7$

$2 x + 4 - 2 z + 4 x + 3 z = 7$
$2 x + 4 x - 2 z + 3 z = 7 - 4$
$6 x + z = 3$ Let this be the 4th equation

Use 2nd and 3rd equations:

$5 x - 4 y - 2 z = 3$ 2nd equation

$5 x - 4 \left(- 4 + 2 z - 4 x\right) - 2 z = 3$
$5 x + 16 - 8 z + 16 x - 2 z = 3$
$5 x + 16 x - 8 z - 2 z = 3 - 16$
$21 x - 10 z = - 13$ Let this be the 5th equation

Solve for $x$ and $z$ using 4th and 5th equations

From 4th equation: $z = 3 - 6 x$ insert in 5th equation

$21 x - 10 z = - 13$ 5th equation

$21 x - 10 \left(3 - 6 x\right) = - 13$

$21 x - 30 + 60 x = - 13$

$81 x = 30 - 13$
$81 x = 17$
$x = \frac{17}{81}$

Go back to the 4th equation , use $x = \frac{17}{81}$

From 4th equation: $z = 3 - 6 x$
$z = 3 - 6 \cdot \frac{17}{81}$

$z = \frac{47}{27}$
Go back to the 3rd equation and use $x = \frac{17}{81}$ and $z = \frac{47}{27}$

$y = - 4 + 2 z - 4 x$ 3rd equation
$y = - 4 + 2 \cdot \frac{47}{27} - 4 \cdot \frac{17}{81}$
$y = - \frac{110}{81}$

The solution set:

$x = \frac{17}{81}$
$y = - \frac{110}{81}$
$z = \frac{47}{27}$

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