Question

Solve the simultaneous equations4x+3y=14x^2+3xy+y^2=2​

Answer 1

Notice that the second equation,

4x ² + 3xy + y ² = 2

can be written as

x (4x + 3y) + y ² = 2

and the first equation says 4x + 3y = 1, so this reduces to

x + y ² = 2

Solve the first equation for x :

4x + 3y = 1

4x = 1 - 3y

x = (1 - 3y)/4

Substitute this into the reduced second equation to get a quadratic equation in y, which happens to be easily factorized and solved:

(1 - 3y)/4 + y ² = 2

1 - 3y + 4y ² = 8

4y ² - 3y - 7 = 0

(4y - 7) (y + 1) = 0

4y - 7 = 0   or   y + 1 = 0

y = 7/4   or   y = -1

Solve for x :

x = (1 - 3 (7/4))/4   or   x = (1 - 3 (-1))/4

x = -17/16   or   x = 1

So the two solutions are (x, y) = (-17/16, 7/4) and (x, y) = (1, -1).