Solve the system of equations algebraically x^2+y^2=100 x-y=2
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We have x^2 + y^2 = 100 and x - y = 2. We have to solve these equations for x and y.
x - y = 2
=> (x - y)^2 =2^2
=> x^2 + y^2 - 2xy = 4
(The entire section contains 85 words.)
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Given:-
(x^2) + (y^2) = 100 ..........(1)
x - y = 2...........(2)
Squaring (2) on both sides we get
(x-y)^2 = (2)^2
or, (x^2) + (y^2) - 2xy = 4
Putting the value of (x^2) + (y^2) from (1) in the above equation we get
100 - 2xy = 4
or, 2xy = 96...........(3)
Now, (1) + (3) gives
(x^2) + (y^2) + 2xy = 100 + 96
or, (x+y)^2 = 196
or, (x+y)^2 = (14)^2
or, (x+y) = 14.........(4)
or, (x+y) = -14..........(5)
Now, (2) + (4) gives
2x = 16
or, x = 8
Thus, y = 6...........(6)
Also, (2) + (5) gives
2x = -12
or, x = -6
Thus, y = -8..........(7)
Hence the two values of (x,y) are (8,6) & (-6,-8)
We'll solve the system using substitution method.
We'll solve for x the 2nd equation:
x = 2 + y (3)
We'll substitute x into the 1st equation:
(2+y)^2+y^2=100
We'll expand the square:
4 + 4y + y^2 + y^2 = 100
We'll combine like terms:
2y^2 + 4y + 4 = 100
We'll divide by 2 and we'll move all terms to one side:
y^2 + 2y + 2 - 50 = 0
y^2 + 2y - 48 = 0
We'll apply quadratic formula:
y1 = [-2 + sqrt(4 + 192)]/2
y1 = (-2+14)/2
y1 = 6
y2 = -8
We'll substitute y1 and y2 into the 3rd equation:
x1 = 2 + y1
x1 = 2 + 6
x1 = 8
x2 = 2 + y2
x2 = 2 - 8
x2 = -6
The solutions of the system are: (8 ; 6) and (-6 ; -8).
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