Question

Solve the system of equations x²y²−12x+6y+32=0 and x²=−8(y−2).

Answer 1

The solution to the system of equations is
`\(x = 2\)` and
`\(y = 3\).`

Step-by-step explanation:

To solve the given system of equations, we'll substitute the expression for
`\(x^2\)` from the second equation into the first equation. The system is:

`\[x^2y^2 - 12x + 6y + 32 = 0\]`

`\[x^2 = -8(y - 2)\]`

Substitute
`\(-8(y - 2)\)` for
`\(x^2\)` in the first equation:

`\[(-8(y - 2))y^2 - 12(-8(y - 2)) + 6y + 32 = 0\]`

Simplify and solve for
`\(y\)`:

`\[8y^3 - 104y + 352 = 0\]`

Factoring this cubic equation, we find that
`\(y = 3\)` is a solution. Substituting
`\(y = 3\)` back into the second equation, we get
`\(x^2 = -8(3 - 2)\)`, which simplifies to
`\(x^2 = -8\)`, and thus
`\(x = 2\)` or
`\(x = -2\).`

So, the solutions to the system are
`\(x = 2\)` and
`\(y = 3\).`

Answer 2

To solve the provided system of equations, one can substitute the expression for x² from the second equation into the first equation. The solution to the system of equations is x = 2 and y = 3.

Step-by-step explanation:

To solve the given system of equations, we'll substitute the expression for from the second equation into the first equation. The system is:

x²y² - 12x + 6y + 32 = 0

x²=−8(y−2)

Substitute x²=−8(y−2) for x² in the first equation:

(-8(y - 2) y² - 12 (-8(y - 2)) + 6y + 32 = 0

Simplify and solve for y:

8y³ - 104y + 352 = 0

Factoring this cubic equation, we find that y = 3 is a solution.

Substituting y = 3 back into the second equation, we get x² -8(3 - 2), which simplifies to x² - 8, and thus x = 2 or x = -2

So, the solutions to the system are x = 2 and y = 3.