stuck with equations *cries*

[fintuck]

please help.. questions are:


Solve these system equations:
A. x^2 + y^2 = 29
x - y =3

B. x - 2y = -1
x^2 - 2xy + 4y^2 = 43

[dglienna]

please help.. questions are:


Solve these system equations:
A. x^2 + y^2 = 29
x - y =3

B. x - 2y = -1
x^2 - 2xy + 4y^2 = 43


The first one is easy. X=5 Y=2
Did it in my head!

[lec87]

For the equation
x - 2y = -1
x^2 - 2xy + 4y^2 = 43

By rearranging the second equation to give x^2 - x(2y) + (2y)^2 = 43, and substituting 2y=x+1 into the equation, 2 solutions are found:
x=6 and y=3.5
or x=-7 and y=-3

[croxley]

The first one is easy. X=5 Y=2
Did it in my head!

Ah, but if you do it analytically, you'll get the other answer y=-5, x=-2

x-y=3 means x=y+3

substitute that into x^2 + y^2=29 gives

(y+3)^2 + y^2=29

and y^2 + 6y + 9 + y^2 = 29
so 2y^2 + 6y - 20 =0

which gives y=2, x=5 and y=-5, x=-2

[Something Else]

please help.. questions are:


Solve these system equations:
A. x^2 + y^2 = 29
x - y =3

B. x - 2y = -1
x^2 - 2xy + 4y^2 = 43
Uhh, this IS chit chat, right???

Well, ok. seriouslyfor now

A:

x - y = 3 ==> x = y + 3

Substitute into first equation:

(y + 3)^2 + y^2 = 29 ...
y^2 + 6y + 9 + y^2 - 29 = 0 ...
2*y^2 + 6*y - 20 = 0
y^2 + 3*y - 10 = 0...
:Method 1: Since -10 = (-1)* 2*5
and 3 = 5 - 2 then...
(y + 5)*(y - 2) = 0

Therefore...
y = -5, OR y = 2, which makes:
x = -2 OR x = 5

METHOD #2:
:chitchat:
the force, LOOK! USE THE FORCE!!!
well, then, how about the quadratic method instead???

B...
Maybe later!