Question 162620
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Good, {{{ 10k+15c=360 }}} -------------------------> EQN 1
we get, {{{k=(360-15c)/10}}} ----------------------> EQN 2
But {{{cross(20+1/2c=220)}}}, we'll figure it out oks,
2 cats are gone, so --> {{{k-2}}} right?
1/2 canaries are gone, so --> {{{(1/2)c}}} right?
New value when they're gone, {{{220}}}
Putting into eqn.
{{{10(k-2)+15(1/2)c=220}}} ------------------------> EQN 3
subst, EQN 2 in EQN 3:
{{{10((360-15c)/10-2)+15(1/2)c=220}}}
{{{10((360-15c-20)/10)+7.5c=220}}} Get LCD
{{{cross(10)(360-15c-20)/cross(10)+7.5c=220}}}, 
{{{360-20-220=15c-7.5c}}}, combine similar terms
{{{120=7.5c}}} ---> {{{cross(120)16/cross(7.5)=cross(7.5)c/cross(7.5)}}}
{{{c=16}}}, TOTAL CANARIES BEFORE ESCAPING
For cats, go back EQN 2:
{{{k=(360-15*16)/10=(360-240)/10}}}
{{{k=12}}}, TOTAL CATS BEFORE ESCAPING
In doubt? Go back eqn 1:
{{{10*12+15*16=360}}}
{{{120+240=360}}}
{{{360=360}}}
Go back eqn 2,
{{{10(12-2)+15(1/cross(2))(cross(16)8)=220}}}
{{{10*10+15*8=220}}}
{{{100+120=220}}}
{{{220=220}}}
Thank you,
Jojo</pre>