Q. 43 - Precalculus Enhanced with Graphing Utilities

Question

Solve the system. Use any method you wish.

$\left\{\begin{array}{l}\frac{5}{x^2}-\frac{2}{y^2}+3=0\\ \frac{3}{x^2}+\frac{1}{y^2}=7\end{array}\right.$

Step-by-Step Solution

Answer

The solution of system of equations are $\left(1,\frac{1}{2}\right),\left(1,-\frac{1}{2}\right),\left(-1,\frac{1}{2}\right),\left(-1,-\frac{1}{2}\right)$.

Step 1. Given information.

Consider the given question,

$\left\{\begin{array}{l}\frac{5}{x^2}-\frac{2}{y^2}+3=0 ...... \left(i\right)\\ \frac{3}{x^2}+\frac{1}{y^2}=7 ...... \left(ii\right)\end{array}\right.$

Multiply equation (i) by $x^2{y}^2$,

$$\begin{gathered} \frac{5x^2{y}^2}{x^2}-\frac{2x^2{y}^2}{y^2}+3x^2{y}^2=0 \\ 5y^2-2x^2=-3x^2{y}^2 ...... \left(iii\right) \end{gathered}$$

Multiply equation (ii) by $x^2{y}^2$,

$$\begin{gathered} \frac{3x^2{y}^2}{x^2}+\frac{x^2{y}^2}{y^2}=7x^2{y}^2 \\ 3y^2+x^2=7x^2{y}^2 ...... \left(iv\right) \end{gathered}$$

Step 2. Multiply equation (iv) by 2 .

Multiply equation (iv) by $2$,

$6y^2+2x^2=14x^2{y}^2 ...... \left(v\right)$

Add equation (iii) and (v),

$$\begin{gathered} \left(5y^2-2x^2\right)+\left(6y^2+2x^2\right)=14x^2{y}^2-3x^2{y}^2 \\ 11y^2=11x^2{y}^2 \\ {x}^2=1 \\ x=-1,1 \end{gathered}$$

Step 3. Substitute the values of x in equation (iv).

Substitute $x=-1$ in equation (iv),

$$\begin{gathered} 3y^2+{\left(-1\right)}^2=7{\left(-1\right)}^2{y}^2 \\ 3y^2+1=7y^2 \\ 1=4y^2 \\ y=-\frac{1}{2},\frac{1}{2} \end{gathered}$$

Substitute $x=1$ in equation (iv),

$$\begin{gathered} 3y^2+1^2=7{\left(1\right)}^2{y}^2 \\ 3y^2+1=7y^2 \\ 1=4y^2 \\ y=-\frac{1}{2},\frac{1}{2} \end{gathered}$$

Therefore, the solution sets are $\left(1,\frac{1}{2}\right),\left(1,-\frac{1}{2}\right),\left(-1,\frac{1}{2}\right),\left(-1,-\frac{1}{2}\right)$.