Respuesta :
Answer:
[tex]x=\pm \sqrt{\frac{y+2}{y-1}}[/tex]
y can take on values in:
[tex](-\infty,-2] \cup (1,\infty)[/tex]
Step-by-step explanation:
Hopefully this is the right equation:
[tex]y=\frac{x^2+2}{x^2-1}[/tex].
Multiply both sides by [tex]x^2-1[/tex]:
[tex]y(x^2-1)=\frac{x^2+2}{x^2-1}(x^2-1)[/tex]
[tex]y(x^2-1)=x^2+2[/tex]
Distribute:
[tex]yx^2-y=x^2+2[/tex]
Put x terms on one side together and non-x terms on opposite side.
Add y on both sides
and subtract x^2 on both sides:
[tex]yx^2-x^2=y+2[/tex]
Factor:
[tex]x^2(y-1)=y+2[/tex]
Divide both sides by (y-1):
[tex]x^2=\frac{y+2}{y-1}[/tex]
Now square root both sides:
[tex]x=\pm \sqrt{\frac{y+2}{y-1}}[/tex]
y cannot be 1 because we don't want to divide by 0.
Now we also need to find when (y+2)/(y-1) is positive or zero because we can't square root negative values if we intend to have just real solutions.
The fraction is zero when y=-2 because that will give us 0 on top.
I'm going to draw a number line and test which intervals gives us positive values for our expression under the square root.
--------(-2)-------(1)----------
Plugging in -3 gives us (-3+2)/(-3-1)=(-1)/(-4)=1/4 so anything before -2 works.
Let's try 0: (0+2)/(0-1)=-2 so nothing between -2 and 1 will work.
Let's try it for 2: (2+2)/(2-1)=4/1=4 so anything after 1 will work.
So y can take on values in:
[tex](-\infty,-2] \cup (1,\infty)[/tex]
