## Task B-2.6.

Determine all ordered pairs of real numbers $(x, y)$ that are solutions to the system of equations. $\left\{\begin{array}{l}\sqrt[3]{2-x}+\sqrt[3]{7-y}=-1 \\ \sqrt{x+y}=4\end{array}\right.$
## First solution.

First, let's express the unknown $y$ from the equation $\sqrt{x+y}=4$.

$\sqrt{x+y}=4 /^{2}$

$x+y=16$

$y=16-x$

By substituting $y=16-x$ into the above equation, we get the equation

$\sqrt[3]{2-x}+\sqrt[3]{x-9}=-1$.

1 point

$\sqrt[3]{2-x}+\sqrt[3]{x-9}=-1 /{ }^{3}$

$2-x+3 \cdot \sqrt[3]{(2-x)^{2}(x-9)}+3 \cdot \sqrt[3]{(2-x)(x-9)^{2}}+x-9=-1$

$3 \cdot \sqrt[3]{(2-x)^{2}(x-9)}+3 \cdot \sqrt[3]{(2-x)(x-9)^{2}}=6 /: 3$

$\sqrt[3]{(2-x)^{2}(x-9)}+\sqrt[3]{(2-x)(x-9)^{2}}=2$
$\sqrt[3]{(2-x)(x-9)} \cdot(\sqrt[3]{2-x}+\sqrt[3]{x-9})=2$

Notice that we can substitute the value of the expression $\sqrt[3]{2-x}+\sqrt[3]{x-9}=-1$ into the equation.

$\sqrt[3]{(2-x)(x-9)} \cdot(-1)=2$

$\sqrt[3]{(2-x)(x-9)}=-2 / 3$

$-x^{2}+11 x-18=-8$

$x^{2}-11 x+10=0$

1 point

$(x-1)(x-10)=0$

The solutions to the equation are the numbers $x_{1}=1$ and $x_{2}=10$.

1 point

Let's determine the corresponding values of the unknown $y$.

$y_{1}=16-1=15$

$y_{2}=16-10=6$

The solutions to the system of equations are the ordered pairs $(1,15)$ and $(10,6)$.