Solve the system of equations.
y= x^2/(x-1) +1
y= 1/(x-1)

someone help, thanks!

Question

Solve the system of equations.
y= x^2/(x-1) +1
y= 1/(x-1)

someone help, thanks!

anonymous · Answer

set y1 = y2 and solve for x first

hero · Answer

@TheRealMeeeee , @turnupcity probably needs help with it because the system posted includes fractions.

anonymous · Answer

all i know was to set y1 = y2 and solve for x but don't really know how to do it

hero · Answer

After setting $y_1 = y_2$ you have:

$$\frac{1}{x - 1} = \frac{x^2}{x - 1} + 1$$
Right @turnupcity ?

anonymous · Answer

yeah thats exactly it

hero · Answer

Okay, now since we have two fractions with the same denominator we can add or subtract them right?

anonymous · Answer

yes..

hero · Answer

Okay, well, the next step before we do that is to subtract 1/(x - 1) from both sides:

$$0 = \frac{x^2}{x - 1} - \frac{1}{x - 1} + 1$$

hero · Answer

Hopefully you understood how I got that.

anonymous · Answer

yes so far i do

hero · Answer

Because the next step would be to combine the fractions with like denominators:

$$0 = \frac{x^2 - 1}{x - 1} + 1$$

hero · Answer

At this point you should notice that the numerator has the term $x^2 - 1$ which is a difference of squares. In other words, we can factor it to get $(x + 1)(x - 1)$

anonymous · Answer

okay.. lol how would do that

hero · Answer

Keep in mind that in general, if you have a difference of squares $a^2 - b^2$ it factors to $(a + b)(a - b)$

hero · Answer

Because in general

$a^2 - b^2 = (a + b)(a - b)$

hero · Answer

So what we have is

$$0 = \frac{(x + 1)(x - 1)}{x - 1} + 1$$

anonymous · Answer

so would we multiply (x+1) on the bottom?

anonymous · Answer

or add (x+1)?

hero · Answer

No, you would not. At this point you should see that you have the same number on the top and bottom $(x - 1)$

anonymous · Answer

so it would cancel out?

hero · Answer

Exactly

anonymous · Answer

so it would leave it at y=(x+1)+1?

hero · Answer

No, if you look at my last step you'll notice that the left side was zero. Why did you replace zero with y?

anonymous · Answer

oops didnt see that, so does x =-2?

hero · Answer

Exactly. Now if you plug x back in to the original system, you'll be able to find the value of y

anonymous · Answer

thank you, apperciate it!

hero · Answer

Let me know what you get for y

hero · Answer

If you run into any trouble with it, let me know. I can easily and quickly assist.

anonymous · Answer

i got -.33333333? i might have typed it wrong in my calc

hero · Answer

I was hoping you would do it without the need of a calc.

hero · Answer

To do it without a calc, here's what you would do. First realize that it would be easier to solve for y if we use this equation:

$$y = \frac{1}{x - 1}$$

So plug in x = -2...

$$y = \frac{1}{-2 - 1}$$
$$y = \frac{1}{-3}$$
$$y = -\frac{1}{3}$$

hero · Answer

Basically, for finding y, you shouldn't need a calc if all you're doing is plugging back in.

anonymous · Answer

oh that makes sense, yeah i guess i didnt my calc for that

hero · Answer

So the solution point is 

$$(x,y) = \left(-2, -\frac{1}{3}\right)$$