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# Solving systems of equations algebraically

Summary:

Solving systems of equations algebraically typically involves using either the substitution method, where one equation is solved for one variable and then substituted into the other equation, or the elimination method, where equations are added or subtracted to eliminate one variable, making it possible to solve for the remaining variable(s).

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)

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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)