Solve the system by changing the variable (xy) ^ 2-3xy = 18 4x + y = 1

(xy) ^ 2 – 3xy = 18; 4x + y = 1.

Let us express from the second equation y:

4x + y = 1; y = 1 – 4x.

Substitute the expressed value y = 1 – 4x into the first equation:

(x (1 – 4x)) ^ 2 – 3x (1 – 4x) = 18.

Let’s introduce a new variable, let x (1 – 4x) = a.

The equation turns out:

a ^ 2 – 3a – 18 = 0.

Let’s select the roots of the quadratic equation using Vieta’s theorem: x1 + x2 = 3; x1 * x2 = -18.

The roots are 6 and (-3), that is, a = 6 and a = -3.

Let’s go back to replacing x (1 – 4x) = a.

1) a = 6; x (1 – 4x) = 6; x – 4x ^ 2 – 6 = 0; -4x ^ 2 + x – 6 = 0;

multiply by (-1): 4x ^ 2 – x + 6 = 0.

We solve the quadratic equation using the discriminant:

a = 4; b = -1; c = 6;

D = b ^ 2 – 4ac; D = (-1) ^ 2 – 4 * 4 * 6 = 1 – 96 = -95 (D <0, no roots).

2) a = -3; x (1 – 4x) -3; x – 4x ^ 2 + 3 = 0; -4x ^ 2 + x + 3 = 0;

multiply by (-1): 4x ^ 2 – x – 3 = 0.

We solve the quadratic equation using the discriminant:

a = 4; b = -1; c = -3;

D = b ^ 2 – 4ac; D = (-1) ^ 2 – 4 * 4 * (-3) = 1 + 48 = 49 (√D = 7);

x = (-b ± √D) / 2a;

x1 = (1 – 7) / (2 * 4) = -6/8 = -3/4.

x2 = (1 + 7) / 8 = 8/8 = 1.

Let’s calculate the value of y: y = 1 – 4x;

x1 = -3/4; y1 = 1 – 4 * (-3/4) = 1 + 3 = 4.

x2 = 1; y2 = 1 – 4 * 1 = 1 – 4 = -3.

Answer: (-3/4; 4) and (1; -3).