Q. 46 - Precalculus Enhanced with Graphing Utilities

Question

Solve the system. Use any method you wish.

$\left\{\begin{array}{l}\frac{1}{x^4}-\frac{1}{y^4}=1\\ \frac{1}{x^4}+\frac{1}{y^4}=4\end{array}\right.$

Step-by-Step Solution

Answer

The solution of system of equations are $\left\{-{\left(\frac{2}{5}\right)}^{\frac{1}{4}},-{\left(\frac{2}{3}\right)}^{\frac{1}{4}}\right\},\left\{-{\left(\frac{2}{5}\right)}^{\frac{1}{4}},{\left(\frac{2}{3}\right)}^{\frac{1}{4}}\right\},\left\{{\left(\frac{2}{5}\right)}^{\frac{1}{4}},-{\left(\frac{2}{3}\right)}^{\frac{1}{4}}\right\},\left\{{\left(\frac{2}{5}\right)}^{\frac{1}{4}},{\left(\frac{2}{3}\right)}^{\frac{1}{4}}\right\}.$

Step 1. Given information.

Consider the given question,

$\left\{\begin{array}{l}\frac{1}{x^4}-\frac{1}{y^4}=1 ...... \left(i\right)\\ \frac{1}{x^4}+\frac{1}{y^4}=4 ...... \left(ii\right)\end{array}\right.$

Multiply equation (i) by $x^4{y}^4$,

$$\begin{gathered} \frac{x^4{y}^4}{x^4}-\frac{x^4{y}^4}{y^4}=x^4{y}^4 \\ {y}^4-x^4=x^4{y}^4 ...... \left(iii\right) \end{gathered}$$

Multiply equation (ii) by $x^4{y}^4$,

$$\begin{gathered} \frac{x^4{y}^4}{x^4}+\frac{x^4{y}^4}{y^4}=4x^4{y}^4 \\ {y}^4+x^4=4x^4{y}^4 ...... \left(iv\right) \end{gathered}$$

Step 2. Add equations (iii) and (iv).

Add equations (iii) and (iv),

$$\begin{gathered} \left(y^4-x^4\right)+\left(y^4+x^4\right)=x^4{y}^4+4x^4{y}^4 \\ 2y^4=5x^4{y}^4 \\ {x}^4=\frac{2}{5} \\ x=\pm {\left(\frac{2}{5}\right)}^{\frac{1}{4}} \end{gathered}$$

Substitute $x={\left(\frac{2}{5}\right)}^{\frac{1}{4}}$ in equation (iii),

$$\begin{gathered} y^4-{\left\{{\left(\frac{2}{5}\right)}^{\frac{1}{4}}\right\}}^4={\left\{{\left(\frac{2}{5}\right)}^{\frac{1}{4}}\right\}}^4{y}^4 \\ {y}^4-\frac{2}{5}=\frac{2}{5}{y}^4 \\ {y}^4=\frac{2}{3} \\ y={\left(-\frac{2}{3}\right)}^{\frac{1}{4}},{\left(\frac{2}{3}\right)}^{\frac{1}{4}} \end{gathered}$$

Step 3. Substitute x = - 2 5 1 4 in equation (iii) and solve the equation.

Substitute $x=-{\left(\frac{2}{5}\right)}^{\frac{1}{4}}$in equation (iii),

$$\begin{gathered} y^4-{\left\{-{\left(\frac{2}{5}\right)}^{\frac{1}{4}}\right\}}^4={\left\{-{\left(\frac{2}{5}\right)}^{\frac{1}{4}}\right\}}^4{y}^4 \\ {y}^4-\frac{2}{5}=\frac{2}{5}{y}^4 \\ {y}^4=\frac{2}{3} \\ {y}^4={\left(-\frac{2}{3}\right)}^{\frac{1}{4}},{\left(\frac{2}{3}\right)}^{\frac{1}{4}} \end{gathered}$$

Therefore, the solution sets are $\left\{-{\left(\frac{2}{5}\right)}^{\frac{1}{4}},-{\left(\frac{2}{3}\right)}^{\frac{1}{4}}\right\},\left\{-{\left(\frac{2}{5}\right)}^{\frac{1}{4}},{\left(\frac{2}{3}\right)}^{\frac{1}{4}}\right\},\left\{{\left(\frac{2}{5}\right)}^{\frac{1}{4}},-{\left(\frac{2}{3}\right)}^{\frac{1}{4}}\right\},\left\{{\left(\frac{2}{5}\right)}^{\frac{1}{4}},{\left(\frac{2}{3}\right)}^{\frac{1}{4}}\right\}$.