Solve the system of equations algebraically x^2+y^2=100 x-y=2

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justaguide's profile pic

Posted on

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

substitute x^2 + y^2 = 100

=> 100 - 2xy = 4

=> 2xy = 96

=> xy = 48

substitute x = 2 + y

=> y(2 + y) = 48

=> y^2 + 2y - 48 = 0

=> y^2 + 8y - 6y - 48 = 0

=> y(y + 8) - 6(y + 8) = 0

=> (y - 6)(y + 8) = 0

=> y = 6 and y = -8

for y = 6, x = 8

for y = -8, x = -6

The required solution is (-6, -8) and (8, 6)

hkj1385's profile pic

Posted on

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)

giorgiana1976's profile pic

Posted on

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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