Posted by hal on Monday, April 8, 2013 at 1:38am.
Could you show me how to do these math systems and show your work so i understand how to do them.
1)
3x+2y+4z=19
2xy+z=3
6x+7yz=17
2)
5x+8y+6z=14
2x+5y+z=12
3x+6y12z=20
3)
x^2+2xy2y^2=11
x^2+xy2y^2=9

algebra  Shaniqua, Monday, April 8, 2013 at 2:36am
If i remember you just take two out of the three and solve using elimination. Then plug in your found variables to the third equation and youll have your variables! Its a lot of work per problem though

algebra  Reiny, Monday, April 8, 2013 at 8:44am
look over the equations and decide which variable would be the "easier" one to eliminate.
In the first I would choose the z
add the last two equations:
8x + 6y = 0 = 20
reduce to
4x + 3y = 10
2nd times 4 :
8x  4y + 4z = 12
3x + 2y + 4z = 19 , now subtract those two

5x  6y + 0 = 7
now look at the y's, they would be the easier to get rid of ...
double the first: 8x + 6y = 20
keep the 2nd ...: 5x  6y = 7
add them
13x = 13
x = 1
sub into 4x+3y=10
4 + 3y = 10
3y = 6
y = 2
back into the 2nd, the easier looking one ...
2x  y + z = 3
2  2 + z = 3
z = 3
x=1
y=2
z=3
try the second the same way, I would choose to work on the z's to start with
The last one is a lot different:
notice that the x^2 and y^2 terms are the same.
So, let's just subtract them ....
xy = 2
or
y = 2/x
sub that into the first:
x^2 + 2x(2/x)  2(4/x^2) = 11
x^2  4  8/x^2 = 11
x^2  8/x^2 = 7
times x^2
x^4  8 = 7x^2
x^4 + 7x^2  8 = 0
(x^2 + 8)(x^2  1) = 0
x^2 = 8 = 0 > x^2 = 8 > no real solution
(or x = ± 2i√2 for complex numbers)
or
x^2  1 = 0
x^2 = 1
x = ± 1
IF x= +1 , then y = 2/1 = 2  (1 , 2)
IF x = 1 , then y = 2/1 = +2  (1, 2)