Question

use the substitution method to solve the following sys. of equations.

x+2y-z=2
3x+y-2z=-15
3x+4y-3z=-2

Answer 1

x = -2y +2 +z
y = -3x +2z -15
x = -2(-3x+2x-15) + 2 +z
x = 6x -4x +30 + 2 +z
x = -32 -z
y = -3(-32-z) + 2z -15
y = 96+3z + 2z -15 = 5z + 81
3(-32-z) + 4(5z+81) -3z = -2
Not sure if there are any mistakes

Answer 2

this is why they invented technology. and matrices

Answer 3

Linear Algebra says x = -13, y = 2 and z = -11. Use as corroboration.

Answer 4

what is a corroboration? lol can u tell im horrible at math?

Answer 5

\[x+2y-z=2 => x+2y-2=z\]
\[3x+y-2z=-15 => 3x+y-2(x+2y-2)=-15\]
\[3x+y-2(x+2y-2)=-15=> x-3y+4=-15 => x-3y=-19\]
\[x-3y=-19 => x=-19+3y\]
\[z=x+2y- 2 \& x=-19+3y => z=-19+3y+2y-2=>z=-19+5y-2\]
\[z=-19+5y-2=>z=-21+5y\]
\[z=-21+5y \& x=-19+3y \& 3x+4y-3z=-2 =>\]
\[3(-19+3y)+4y-3(-21+5y)=-2\]
now lets simplify and solve for y
\[-51+9y+4y+63-15y=-2\]
\[-2y+12=-2 => -2y=-14 => y=7\]
i can't figure out where i made mistake i have y=7

Answer 6

you should tell your teacher solve this using substitution style is crazy talk lol

Answer 7

i can solve it by typing
http://www.wolframalpha.com/input/?i=x%2B2y-z%3D2+and+3x%2By-2z%3D-15+and+3x%2B4y-3z%3D-2

Answer 8

ok i see a mistake i don't know how to multiply

Answer 9

3(-19) =-57 not freaking -51

Answer 10

so those last two lines should be
\[-57+9y+4y+63-15y=-2\]
\[-2y+6=-2 => -2y=-8 => y=4\]

Answer 11

OMG thank you satellite73 and myininaya

Answer 12

\[\text{ if y=4 and }z=-21+5y=-21+5(4)=-21+20=-1\]
\[\text{ if y=4 and } x=-19+3y=-19+3(4)=-19+12=-7\]

Answer 13

yay! :)