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

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Answer 1 · 2015-12-29T20:48:28

$11x+y-5z=1$ $..........(i)$
$7x+13y+z=1$ $............(ii)$
$8x-y-z=11$ $.............(iii)$

Adding $(ii)$ and $(iii)$ we get
$15x+12y=12$
$implies 5x+4y=4$ $..........(iv)$

Multiply $(ii)$ by $5$ and add in $(i)$

$implies 35x+65y+5z=5$
$and 11x+y-5z=1$

By addition we have

$46x+66y=6$

$implies 23x+33y=3$ $...............(v)$

Multiply $(iv)$ by $33$ and $(v)$ by $4$ and subtract (v) from (iv)

$implies 165x+132y=132$
$and 92x+132y=12$

By subtraction we have
$73x=120$
$implies x=120/73$

Put $x=120/73$ in $(iv)$

$implies 5(120/73)+4y=4$

$implies 600+292y=292$

$implies 292y=-308$

$implies y=-308/292=-77/73$

$implies y=-77/73$

Put $x=120/73$ and $y=-77/73$ in $(iii)$

$implies 8(120/73)-(-77/73)-z=11$

$implies z=8*120/73+77/73-11$

$implies z=960/73+77/73-11=(960+77-803)/73=234/73$

$implies z=234/73$

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