Answer:
[tex]\large \boxed{\sf \ \ x=\pm8 \ \ or \ \ x=\pm2 \ \ }[/tex]
Step-by-step explanation:
Hello, please consider the following.
First of all, we can assume that x and y are different from 0, as we cannot divide by 0.
And from the first equation we can write y as a function of x as below.
[tex]y=\dfrac{16}{x}[/tex]
And then, we replace it in the second equation to get.
[tex]\dfrac{x}{\frac{16}{x}}+\dfrac{\frac{16}{x}}{x}=\dfrac{17}{4}\\\\<=> \dfrac{x^2}{16}+\dfrac{16}{x^2}=\dfrac{17}{4}\\\\\text{*** We multiply by }16x^2 \text{ both sides ***}\\\\x^4+16*16=\dfrac{17*16}{4}x^2\\\\x^4-68x^2+3600=0\\\\\text{*** The product of the zeroes is 3600 = 64*4 and the sum is 64+4=68 ***}\\\\\text{*** So we can factorise *** }\\\\x^4-64x^2-4x^2+3600=x^2(x^2-64)-4(x^2-64)=(x^2-64)(x^2-4)=0\\\\x^2=64=8^2 \ \ or \ \ x^2=4\\\\x=\pm8 \ \ or \ \ x=\pm2\\[/tex]
Hope this helps.
Do not hesitate if you need further explanation.
Thank you
