Solve the following system of equations and explain what the solution means geometrically:
1) x-y+z=10
2) 2x+3y-2z=-21
3) 0.5x+0.4y+0.25z=-0.5
Im pretty sure you need to use substituion.
solve for z in terms of x and y then
put then in another equation and solve for x in terms of y then
put it in the last equation and solve for x
from there you can find y and z. the solution (x,y,z) should be the point at which the planes intersect...i think.
hope this helps
i tried and failed otherwise i would show you how
sorry about that im working on it now. I tried substition and combinations and here is the result (im off somewhere but i believe this should work)...
$\displaystyle
x-y+z=10$
$\displaystyle 2x+3y-2z=-21$
$\displaystyle 10x+8y+5z=-10
$
multiply by -1 for first equation and 2 for the second:
-1 1 -1 = -10
4 3 -4 = -42
10 8 5 = -10
+_________________
13x + 12y = -41
same thing only different
-2 2 -2 = -20
-8 -12 8 = 84
10 8 5 = -10
_____________________
-2y + 11z = 74
from those 2 equation we just found find what x and z equal both in terms of y and you get:
$\displaystyle x=\frac{-12y-62}{13}$
$\displaystyle z=\frac{54+2y}{11}$
dang i thought it would work...sorry...this comes out to be y=-5.6 something. Good luck.
CAS means Computer Algebra System, Computer algebra system - Wikipedia, the free encyclopedia
So not pencil and paper.
Without CAS, you can use row-reduction. First do it with just a 2x2 matrix then do this one. I did the first row reductions below quick so need to check them:
$\displaystyle \left[\begin{array}{cccc} 1 & -1 & 1 & 10 \\
2 & 3 & -2 & 21 \\
10 & 8 & 5 & 10
\end{array}
\right]
$
Now: first row times -2, add to second and first row times -10 and add to third gives:
$\displaystyle \left[\begin{array}{cccc} 1 & -1 & 1 & 10 \\
0 & 5 & -4 & 1 \\
0 & 18 & -5 & -110
\end{array}
\right]
$
Second row times 1/5, add to first and second times 18/5 add to third gives:
$\displaystyle \left[\begin{array}{cccc} 1 & 0 & 1/5 & 51/5 \\
0 & 5 & -4 & 1 \\
0 & 0 & 27/5 & -568/5
\end{array}
\right]
$
Keep doing it until you get ones on the diagonal in the first three columns (divide the second row by 5 right).
I did it by reducing it down. Steps are shown below:
$\displaystyle \left[\begin{array}{cccc} 1 & -1 & 1 & 10 \\
2 & 3 & -2 & -21 \\
10 & 8 & 5 & -10
\end{array}
\right]
$
$\displaystyle \left[\begin{array}{cccc} 1 & -1 & 1 & 10 \\
0 & 5 & -4 & -41 \\
10 & 8 & 5 & -10
\end{array}
\right]
$
$\displaystyle \left[\begin{array}{cccc} 1 & -1 & 1 & 10 \\
0 & 1 & -4/5 & -41/5 \\
10 & 8 & 5 & -10
\end{array}
\right]
$
$\displaystyle \left[\begin{array}{cccc} 1 & 0 & 1/5 & 9/5 \\
0 & 1 & -4/5 & -41/5 \\
10 & 8 & 5 & -10
\end{array}
\right]
$
$\displaystyle \left[\begin{array}{cccc} 1 & 0 & 1/5 & 9/5 \\
0 & 1 & -4/5 & -41/5 \\
0 & 8 & 3 & -28
\end{array}
\right]
$
$\displaystyle \left[\begin{array}{cccc} 1 & 0 & 1/5 & 9/5 \\
0 & 1 & -4/5 & -41/5 \\
0 & 0 & 47/5 & 188/5
\end{array}
\right]
$
So, 47z = 188 sets z = 4.
Then:
1) x + 1/5z = 9/5 equates to x = 1
2) y - 4/5z = -41/5 equates to y = -5