Question

\left \{ {{x^2+y^2=28} \atop {x+y=4}} \right.

Answer 1

$\left \{ {{x^2+y^2=28} \atop {y=4-x}} \right. \\ \left \{ {{y=4-x} \atop {x^2-4x-6=0}} \right. \\ \left \{ {{x= 2+ \sqrt{10} } } \atop {y=2-\sqrt{10}}} \right. ---or--- \left \{ {{x=2-\sqrt{10}} \atop {y=2+\sqrt{10}}} \right.$

Answer 2

The given system of equations is:

$$\left\{ \begin{array}{ll} x^2 + y^2 &= 28 \\ x + y &= 4 \end{array} \right.$$

To solve this system, you can use substitution or elimination method. Here, I'll demonstrate the substitution method.

From the second equation, you can express $y$ in terms of $x$:

$y = 4 - x$

Substitute this expression for $y$ into the first equation:

$x^2 + (4 - x)^2 = 28$

Now, expand and simplify the equation:

$x^2 + (16 - 8x + x^2) = 28$

Combine like terms:

$2x^2 - 8x - 12 = 0$

Divide the entire equation by 2 to simplify it:

$x^2 - 4x - 6 = 0$

Now, you can solve this quadratic equation using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For this equation, $a = 1$, $b = -4$, and $c = -6$. Plug these values into the quadratic formula:

$x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(-6)}}{2(1)}$

$x = \frac{4 \pm \sqrt{64}}{2}$

$x = \frac{4 \pm 8}{2}$

This gives two possible values for $x$:

1. When $x = 6$:

$y = 4 - x = 4 - 6 = -2$

2. When $x = -2$:

$y = 4 - x = 4 - (-2) = 6$

Therefore, the solutions to the system of equations are $(6, -2)$ and $(-2, 6)$.