Question

solve the following system of equations.\\(\

$$\begin{cases}x^2 + y^2 = 10\\\\y + x = -4\\end{cases}$$

\\)if there is more than one solution, enter additional solutions with the \or\ button.\
if there is no real solution, use the
o solution\ button.\
\\((x,y) = (\square,\square)\\)

Explanation:

Step1: Isolate $y$ from linear equation

$y = -4 - x$

Step2: Substitute $y$ into quadratic equation

$x^2 + (-4 - x)^2 = 10$

Step3: Expand and simplify the equation

$x^2 + (16 + 8x + x^2) = 10$
$2x^2 + 8x + 6 = 0$
$x^2 + 4x + 3 = 0$

Step4: Factor the quadratic equation

$(x + 1)(x + 3) = 0$

Step5: Solve for $x$ values

$x + 1 = 0 \implies x = -1$
$x + 3 = 0 \implies x = -3$

Step6: Find corresponding $y$ values

For $x=-1$: $y = -4 - (-1) = -3$
For $x=-3$: $y = -4 - (-3) = -1$

Answer:

$(x,y)=(-1,-3)$ or $(x,y)=(-3,-1)$