1. help

solve the system by any method

2x+3y+z=9
4x+y-3z= -7
6x+2y-4z=-8

2. Originally Posted by allyssa
solve the system by any method

2x+3y+z=9
4x+y-3z= -7
6x+2y-4z=-8
multiply the first by -2 and add to the 2nd then
multiply the first by -4 and add to the 3rd to get

$-5y-5z=-25 \iff y+z=5$

$-7y-7z=-35 \iff y+z=5$

From here we can see that the system is consistant dependant.

We can parameterize the solutions

so lets let $z=t$then we get

$y+t=5 \iff y=5-t$

$2x+3(5-t)+t=9 \iff 2x+15-3t+t=9 \iff 2x=2t-6 \iff x=t-3$

So now all of the solutions are of the form.

$(t-3,5-t,t)$

I hope this helps.

2x+3y+z=9
4x+y-3z= -7
6x+2y-4z=-8

divide last line by 2

2x+3y+z=9
4x+y-3z= -7
3x+y-2z = -4

play around a little

4x+y-3z+3= 3x+y-2z

4x -3z +3 = 3x - 2z

x -3z +3 = -2z
good stuff:

x = z - 3
or
z = x+3

plug-n-chug:
2x+3y+(x+3)=9
4x+y-3(x+3)= -7
3x+y-2(x+3) = -4

one of these is easily factorable:

3x+3y+3=9
x+y-9= -7
x+y-6 = -4

3(x+y) = 6

x+y = 2
z = x+3

x = 2-y
x = 3-z

2-y = 3-z

-y = 1-z

x = z - 3
y = z - 1
x+y = 2

(z-3) + (z-1) = 2

2z - 4 = 2
z - 2 = 1

z = 3
y = 2
x = 0

There we go! I do not believe all of my steps here were necessary. In fact, a few I see are redundant and/or superfluous. But if you are careful enough, you can just play around with these things until you see solutions. This took me about 20 seconds on paper. One way to see if you are right! Plug our new x y z values into the original! And yes, (0, 2, 3) is certainly a solution for this system of equations.

Also, if you know Gaussian Elimination, you could do that. It is essentially the same process, but with matrices.

Good luck and I hope this helped,
-Andy

4. i think if u use this it will be better.....

z=3
y=2
x=0

5. Originally Posted by abender
x = 2-y
x = 3-z

2-y = 3-z

-y = 1-z
This is incorrect. Earlier, you found that $x = z - 3$, not $x = 3 - z$. So,

$2 - y = z - 3$

$\Rightarrow y = 5 - z$

If you then try to do your substitution into $x + y = 2$, you will see that the $z$'s will cancel because there are, in fact, an infinite number of solutions. The particular solution that you got, $\left(0,\ 2,\ 3\right)$ happens to work by coincidence, since $5 - z = z - 1,\text{ when }z = 3$.

TheEmptySet provided the correct solution. The general solution is $\left(t - 3,\ 5 - t,\ t\right),\text{ where }t\in\mathbb{R}$.

6. Originally Posted by abender
z = 3
y = 2
x = 0

There we go! I do not believe all of my steps here were necessary. In fact, a few I see are redundant and/or superfluous. But if you are careful enough, you can just play around with these things until you see solutions. This took me about 20 seconds on paper. One way to see if you are right! Plug our new x y z values into the original! And yes, (0, 2, 3) is certainly a solution for this system of equations.

-Andy
I just have to say that the above solution is one of the infinite number of solutions. It is contained in the solution I mentioned.

$(t-3,5-t,t)$ it is when t=3 we get $(3-3,5-3,3) \to (0,2,3)$

Note you can plug this in to see that it works as well (for all values of t)

$2x+3y+z=9 \to 2(t-3)+3(5-t)+t=9 \to 2t-6+15-3t+t=9 \to 9=9$

$4x+y-3z= -7 \to 4(t-3)+(5-t)-3(t)=-7 \to 4t-12+5-t-3t=-7 \to -7=-7$

$6x+2y-4z=-8 \to 6(t-3)+2(5-t)-4t=-8 \to 6t-18+10-2t-4t=-8$
$\to -8=-8$

So the solution checks in all three equations.

I hope this clear it up.

Good luck