Question

Solve the system. Use any method you wish.

$\left\{\begin{array}{l}\frac{1}{x^4}+\frac{6}{y^4}=6\\ \frac{2}{x^4}-\frac{2}{y^4}=19\end{array}\right.$

Short answer

There is no solution to the two equations.

Step-by-step solution

Step 1. Given information.

Consider the given question,

$\left\{\begin{array}{l}\frac{1}{x^4}+\frac{6}{y^4}=6......\left(i\right)\\ \frac{2}{x^4}-\frac{2}{y^4}=19......\left(ii\right)\end{array}\right.$

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

$$\begin{gathered} \frac{x^4{y}^4}{x^4}+\frac{6x^4{y}^4}{y^4}=6x^4{y}^4 \\ {y}^4+6x^4=6x^4{y}^4......\left(iii\right) \end{gathered}$$

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

localid="1646932051255" role="math" $$\begin{gathered} \frac{2x^4{y}^4}{x^4}-\frac{2x^4{y}^4}{y^4}=19x^4{y}^4 \\ 2y^4-2x^4=19x^4{y}^4......\left(iv\right) \end{gathered}$$

Step 2. Multiply equation (iii) by 2.

Multiply equation (iii) by $-2$,

$-2y^4-12x^4=-12x^4{y}^4......\left(v\right)$

Add equations (v) and (iv),

$$\begin{gathered} \left(2y^4-2x^4\right)+\left(-2y^4-12x^4\right)=19x^4{y}^4-12x^4{y}^4 \\ -14x^4=7x^4{y}^4 \\ {y}^4=-\frac{1}{2} \end{gathered}$$

Taking square root of the two sides isn't possible.

Hence, there is no solution.