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March 30, 2015

March 30, 2015

Posted by **hal** on Monday, April 8, 2013 at 1:38am.

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 -
**Shaniqua**, Monday, April 8, 2013 at 2:36amIf 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 8:44amlook 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)

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