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
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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
Quote:
Originally Posted by fintuck
The first one is easy. X=5 Y=2
Did it in my head!
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
Ah, but if you do it analytically, you'll get the other answer y=-5, x=-2Quote:
Originally Posted by dglienna
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
Uhh, this IS chit chat, right???Quote:
Originally Posted by fintuck
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!