Chapter 11, Problem 71RE

To determine

Solve each system of equations.

Answer to Problem 71RE

  $\boxed{(2\sqrt{2},\sqrt{2})and(-2\sqrt{2,}\sqrt{2)}}$

Explanation of Solution

Given information:

Solve each system of equations.

  $\{\begin{array}{l}2xy+y^2=10\\ 3y^2-xy=2\end{array}$

Calculation:

Use the elimination method of solve the system of equations.

Multiply both sides of the equation $3y^2-xy=2$ by $2$ so as to make the coefficients of $xy$

negatives of one another.

  $6y^2-2xy=4$

Thus, the system of equation is $\{\begin{array}{l}6y^2-2xy=4\\ 2xy+y^2=10\end{array}$

Add the two equations and solve for $x$ .

  $\{\begin{array}{l}6y^2-2xy=4\\ \begin{array}{c}2xy+y^2=10\end{array}\end{array}$

  $7y^2=14$

  $y^2=2$

  $y=\pm \sqrt{2}$

Hence, $y$ has two values $\sqrt{2}$ and $-\sqrt{2}$ .

Substitute $\sqrt{2}$ and $-\sqrt{2}$ for $y$ in the equation $2xy+y^2=10$ and solve for $x$ .

When $y=\sqrt{2}$

  $x=\frac{10-y^2}{2y}=\frac{10-2}{2\sqrt{2}}=2\sqrt{2}$

When $y=-\sqrt{2}$

  $x=\frac{10-y^2}{2y}=\frac{10-2}{-2\sqrt{2}}=-2\sqrt{2}$

So, we have $x=2\sqrt{2},-2\sqrt{2}$ .

Hence, the solutions are $(2\sqrt{2},\sqrt{2})and(-2\sqrt{2,}\sqrt{2)}$ .