Q. 40 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 39–43, solve each system of equations.

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

Step-by-Step Solution

Answer

Solutions of the system of equations $\left\{\begin{array}{l}2xy+y^2=10\\ 3y^2-xy=2\end{array}\right.$are $\left(2\sqrt{2},\sqrt{2}\right)$

Step 1. Given data

The given system of equation is 

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

Step 2. Formation of the single-variable equation

Multiply 2 to both sides of the equation$3y^2-xy=2$

$$\begin{gathered} 2\left(3y^2-xy\right)=2\left(2\right) \\ 6y^2-2xy=4 \end{gathered}$$

so the new system of equation is $\left\{\begin{array}{l}2xy+y^2=10\\ 6y^2-2xy=4\end{array}\right.$

add both equations of the system of equations, 

$$\begin{gathered} \left(2xy+y^2\right)+\left(6y^2-2xy\right)=10+4 \\ 7y^2=14 \\ y=\pm \sqrt{2} \end{gathered}$$

Step 4. Solution of the system of equations

Substitute $y=\sqrt{2}$ in equation $3y^2-xy=2$

$$\begin{gathered} 3y^2-xy=2 \\ 3{\left(\sqrt{2}\right)}^2-x\sqrt{2}=2 \\ x\sqrt{2}=4 \\ x=2\sqrt{2} \end{gathered}$$

Substitute $y=-\sqrt{2}$ in equation $3y^2-xy=2$

$$\begin{gathered} 3y^2-xy=2 \\ 3{\left(-\sqrt{2}\right)}^2-x\left(-\sqrt{2}\right)=2 \\ x\sqrt{2}=-4 \\ x=-2\sqrt{2} \end{gathered}$$

So solutions of the system of equations are $\left(2\sqrt{2},\sqrt{2}\right) \& \left(-2\sqrt{2},-\sqrt{2}\right)$