# algebra

posted by on .

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
2x-y+z=3
6x+7y-z=17

2)
5x+8y+6z=14
2x+5y+z=12
-3x+6y-12z=20

3)
x^2+2xy-2y^2=-11
x^2+xy-2y^2=-9

• algebra - ,

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 - ,

look over the equations and decide which variable would be the "easier" one to eliminate.
In the first I would choose the z

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
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)