Hello, Kim!

I'll do the last one . . . it seems to be the messiest one.

I hope you can use my solution as a template for solving the others.

d) .[1] .x² + y² + 4x + 6y - 40 .= .0

. . .[2] .x - y .= .10

We have a quadratic equation and a linear equation.

Solve the linear equation for one of its variables.

. . From [2], we have: .y .= .x - 10 .[3]

Substitute into [1]:

. . . . .x² + (x - 10)² + 4x + 6(x - 10) - 40 .= .0

. .x² + x² - 20x + 100 + 4x + 6x - 60 - 40 .= .0

. . . . . . . . . . . . . . . . . . . . . . .2x² - 10x .= .0

Factor: . . . . . . . . . . . . . . . . . .2x(x - 5) .= .0

Solve the two equations:

. . . 2x .= .0 . .→ . x .= .0

. . x - 5 .= .0 . → . x .= .5

Substitute into [3]:

. . x = 0: . y .= .0 - 10 . → . y .= -10

. . x = 5: . y .= .5 - 10 . → . y .= .-5

Therefore, the solutions are: .(0,-10) and (5,-5)