Solve the equations 3x^2 + 3y^2 = 36 and x + 2y = 10

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The equations 3x^2 + 3y^2 = 36 and x + 2y = 10 have to be solved.

The equation 3x^2 + 3y^2 = 36 is that of a circle with center (0, 0) and radius `2*sqrt 3` . x + 2y = 10 is a straight line.

Substitute x = 10 - 2y in (1)

=> 3(10 - 2y)^2 + 3y^2 =36

=> 100 + 4y^2 - 40y + y^2 = 12

=> 5y^2 - 40y + 88 = 0

The solutions of this equation are

It is seen that 40^2 < 4*5*88

The quadratic equation does not give real solutions. The imaginary solutions of the equation are `(40 + sqrt (-160))/10`

=> `4 + i*2*sqrt 10/5`

and ` 4 - i*2*sqrt 10/5`

For `y = 4 + i*2*sqrt 10/5` , `x = 10 - 8 - 4*i*sqrt 10/5` = `2 - i*4*sqrt 10/5`

For `y = 4 - i*2*sqrt 10/5` , `x = 2 + 4*i*sqrt 10/5`

The solution of the equations are `(2 - i*4*sqrt 10/5,4 + i*2*sqrt 10/5)` and `( 2 + 4*i*sqrt 10/5, 4 - i*2*sqrt 10/5)`

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