Chapter 11, Problem 40RE

To determine

To find: The solution of system of equations.

Answer to Problem 40RE

The solution of the system of equations $\left(2\sqrt{2},\sqrt{2}\right)$ and $\left(-2\sqrt{2},-\sqrt{2}\right)$

Explanation of Solution

Given information:

The equations are $2xy+y^2=10$ and $3y^2-xy=2$

Calculation:

Consider the equations.

  $2xy+y^2=10$ …… (1)

  $3y^2-xy=2$ …… (2)

Multiply both sides of Equation (2) by $2$ .

  $6y^2-2xy=4$ …… (3)

Add equation (1) and (3).

  $\begin{array}{l}2xy+y^2+6y^2-2xy=4+10\\ 7y^2=14\\ y^2=2\\ y=\pm \sqrt{2}\end{array}$

Subtract $\sqrt{2}$ for $y$ in equation (1).

  $\begin{array}{l}2x\sqrt{2}+{\left(\sqrt{2}\right)}^2=10\\ 2x=\frac{10-2}{\sqrt{2}}\\ x=2\sqrt{2}\end{array}$

Subtract $-\sqrt{2}$ for $y$ in equation (1).

  $\begin{array}{l}2x\left(-\sqrt{2}\right)+{\left(-\sqrt{2}\right)}^2=10\\ 2x=\frac{10-2}{-\sqrt{2}}\\ x=-2\sqrt{2}\end{array}$

Therefore, the solution of the system of equations $\left(2\sqrt{2},\sqrt{2}\right)$ and $\left(-2\sqrt{2},-\sqrt{2}\right)$