Question

PLEASE HELP ASAP. SOLVE FOR X AND Y. SHOW ALL WORK

Answer 1

`\left( \begin{array}{l l}x=-3+ 3√(3), & y=-3+2√(3)\\\\x=-3- 3√(3), & y=-3-2√(3)\end{array}\right)`

Explanation:

Given system of equations:

`\begin{cases}x-2y+(x)/(y)=6\\\\x^2-2xy-6y=0 \end{cases}`

Rearrange the second equation to isolate y:

`\implies x^2-2xy-6y=0`

`\implies x^2=2xy+6y`

`\implies x^2=y(2x+6)`

`\implies y=(x^2)/(2x+6)`

Substitute the found expression for y into the first equation:

`\implies x-2\left((x^2)/(2x+6)\right)+(x)/((x^2)/(2x+6))=6`

`\implies x-(x^2)/(x+3)\right)+(2x+6)/(x)-6=0`

`\implies (x^2(x+3))/(x(x+3))-(x^3)/(x(x+3))\right)+((2x+6)(x+3))/(x(x+3))-(6x(x+3))/(x(x+3))=0`

`\implies (x^2(x+3)-x^3+(2x+6)(x+3)-6x(x+3))/(x(x+3))=0`

`\implies x^3+3x^2-x^3+2x^2+12x+18-6x^2-18x=0`

`\implies -x^2-6x+18=0`

Solve using the Quadratic Formula.

Quadratic Formula

`x=(-b \pm √(b^2-4ac) )/(2a)\quad\textsf{when }\:ax^2+bx+c=0`

Therefore:

`\implies x=(-(-6) \pm √((-6)^2-4(-1)(18)) )/(2(-1))`

`\implies x=(6 \pm √(108))/(-2)`

`\implies x=(6 \pm √(36 \cdot 3))/(-2)`

`\implies x=(6 \pm √(36)√(3))/(-2)`

`\implies x=(6 \pm 6√(3))/(-2)`

`\implies x=-3\pm 3√(3)`

Substitute the found values of x into the expression for y:

`\textsf{When }x=-3+3√(3):`

`\implies y=((-3+3√(3))^2)/(2(-3+ 3√(3))+6)`

`\implies y=(36-18√(3))/(6√(3))`

`\implies y=(36)/(6√(3)) -(18√(3))/(6√(3))`

`\implies y=(6√(3))/(√(3)√(3)) -3`

`\implies y=-3+2√(3)`

`\textsf{When }x=-3-3√(3):`

`\implies y=((-3- 3√(3))^2)/(2(-3- 3√(3))+6)`

`\implies y=(36+18√(3))/(-6√(3))`

`\implies y=(36)/(-6√(3))+(18√(3))/(-6√(3))`

`\implies y=-(6)/(√(3))-3`

`\implies y=-(6√(3))/(√(3)√(3))-3`

`\implies y=-3-2√(3)`

Therefore, the solutions are:

`\left( \begin{array}{l l}x=-3+ 3√(3), & y=-3+2√(3)\\\\x=-3- 3√(3), & y=-3-2√(3)\end{array}\right)`