Question

2x^2 + y^2 = 19
x + 3y = 0

Answer 1

(-3,1) & (3, -1)

Explanation:

Solving quadratic equation and linear equation:

2x² + y² = 19 -----------------(I)

x + 3y = 0 -----------------(II)

x = -3y --------------(III)

Substitute x = -3y in equation (I) and we can find the value of y.

2(-3y)² + y² = 19

2*9y² + y² = 19

18y² + y² = 19

Combine like terms,

19y² = 19

y² = 19 ÷ 19

y² = 1

y = √1

y = ±1

Substitute y = ±1 in equation (III) and find the value of x.

When y = 1,

x = -3*1

x = -3

When y = -1,

x = -3*(-1)

x = 3

Solutions: (-3, 1) ; (3, -1)