# How do you solve the following system?: 11x+y-5z=1, 7x+13y+z=1, 8x-y-z=11

Dec 29, 2015

$11 x + y - 5 z = 1$$\ldots \ldots \ldots . \left(i\right)$
$7 x + 13 y + z = 1$$\ldots \ldots \ldots \ldots \left(i i\right)$
$8 x - y - z = 11$$\ldots \ldots \ldots \ldots . \left(i i i\right)$

Adding $\left(i i\right)$ and $\left(i i i\right)$ we get
$15 x + 12 y = 12$
$\implies 5 x + 4 y = 4$$\ldots \ldots \ldots . \left(i v\right)$

Multiply $\left(i i\right)$ by $5$ and add in $\left(i\right)$

$\implies 35 x + 65 y + 5 z = 5$
$\mathmr{and} 11 x + y - 5 z = 1$

By addition we have

$46 x + 66 y = 6$

$\implies 23 x + 33 y = 3$$\ldots \ldots \ldots \ldots \ldots \left(v\right)$

Multiply $\left(i v\right)$ by $33$ and $\left(v\right)$ by $4$ and subtract (v) from (iv)

$\implies 165 x + 132 y = 132$
$\mathmr{and} 92 x + 132 y = 12$

By subtraction we have
$73 x = 120$
$\implies x = \frac{120}{73}$

Put $x = \frac{120}{73}$ in $\left(i v\right)$

$\implies 5 \left(\frac{120}{73}\right) + 4 y = 4$

$\implies 600 + 292 y = 292$

$\implies 292 y = - 308$

$\implies y = - \frac{308}{292} = - \frac{77}{73}$

$\implies y = - \frac{77}{73}$

Put $x = \frac{120}{73}$ and $y = - \frac{77}{73}$ in $\left(i i i\right)$

$\implies 8 \left(\frac{120}{73}\right) - \left(- \frac{77}{73}\right) - z = 11$

$\implies z = 8 \cdot \frac{120}{73} + \frac{77}{73} - 11$

$\implies z = \frac{960}{73} + \frac{77}{73} - 11 = \frac{960 + 77 - 803}{73} = \frac{234}{73}$

$\implies z = \frac{234}{73}$