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Filed on Sat Feb 03, 2007 3:11 pm, by shinwoo

$x^{2}-63x+k=0$

$1$

$a+b$

$a,b,c$
$a(b+c)=152, b(c+a)=162$
$c(a+b)=170$
$abc$

$a-7b+8c=4$
$8a+4b-c=7$

$a^{2}-b^{2}+c^{2}$

Ah...

Wed Dec 27, 2006 7:33 pm, by shinwoo

This girl asked me another question....

$\sqrt{-1}$

Trying some more LaTeX

Wed Dec 27, 2006 7:19 pm, by shinwoo

so .....a friend of mine asked me for help on this problem....

To Cindy
okay, so Al said something about systems of equations.
It only means that the number of the variables equal the number of equations. Means it is solvable...

When we are dealing with systems of equation problems, we have couple ways of solving them. One way is to make the values (in this case, the right hand sides of the equations) equal and substitue with other equations. But since we have a nice set of equations that can cancel out nicely, we will use another method. watch and be amazed by this....

we have....

$2x+y+z=12$
$x+3y+4z=9$
$-5x+4y-z=-7$

the first equation and the last equation are the clues.

the Z's!!!!!

one is poitive with the coefficient of 1 and the other is negative with the coefficient of -1.

if we add the two equations, we get....
$-3x+5y=5$

now we have one equation, simpified, but it still has two variable.... therefore we need to find another equation with the same variables (x and y)

so let's force some equations.....

take the last equation again, and multiply it by 4.
We are doing this so the coefficient of the variable z of the last equation is oppositely charged with the coefficient of the varialbe z of the second equation.(so we can cancel them out like the last time we did it)

so

$4*(-5x+4y-z=-7)$
$-20x+16y-4z=-28$

that looks nice now...(although the numbers got a little bigger)

subtract $-20x+16y-4z=-28$ from our second equation, $x+3y+4z=9$

we get.....
$-19x+19y=-19$

this is magical.....

so our big picture is

$x-y=1$
$x=y+1$

substitue in our equation, $-3x+5y=5$ , we get,

$-3-3y+5y=5$
$y=4$

from our previous equation, we can also get the value of $x$

once we get $x$ and $y$ (which we actually got) we can easily get $z$ by substituting in any of the given equations..

whew.....

by the way, if you had a graphics calculator, you could have solved this in two seconds without going through all these.....

(you use matrices)

well, explaing all these wasted my energy, so I'll do the second one on some other time...

from your very friendly moderator