**Solve the system of equations algebraically x^2+2y^2=10 3x^2-y^2=9**

We have to solve the system of equations:

x^2+2y^2=10...(1)

3x^2-y^2=9 ...(2)

(1) - 2*(2)

=> x^2 + 2y^2 - 3x^2 - 2y^2 = 10 - 18

=> -2x^2 = -8

=> x^2 = 4

Substitute in (2)

3*4 - y^2 = 9

=> y^2 = 3

x^2 = 4

=> x = -2 and 2

y^2 = 3

=> y = sqrt 3 and -sqrt 3

**The required solutions are (2, sqrt 3),(2 , -sqrt 3), (-2 , sqrt 3),
(-2, -sqrt 3)**

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

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)**

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