Flat and idk the other one
<span><em />
3x+y=5</span>
<span /><span>2x-2y=22 ...............2(x-y)=22 ⇒ x-y=11</span><span><em /></span>
<span><em>3x+y=5</em></span>
<span /><span><em><u>x-y=11 </u> </em>(+)</span>
<span />3x+x+y-y=11+5
<span>4x=16 /:4</span>
<span /><em>x=4</em>
x-y=11
4-y=11
-y=7 /*(-1)
<em>y=-7</em>
<span><em></em></span>
Well...
2x-6x>-12+8-6
-4x>-10
Attention! ">" change for "<" because we multiply both members for (-1)
x<10/4
It is the same to say (-infinite ; 10/4) or (-infinite ; 5/2) or (-infinite ; 2.5)
Good luck!
M.
A. Factor the numerator as a difference of squares:

c. As

, the contribution of the terms of degree less than 2 becomes negligible, which means we can write

e. Let's first rewrite the root terms with rational exponents:
![\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bx%5Cto1%7D%5Cfrac%7B%5Csqrt%5B3%5Dx-x%7D%7B%5Csqrt%20x-x%7D%3D%5Clim_%7Bx%5Cto1%7D%5Cfrac%7Bx%5E%7B1%2F3%7D-x%7D%7Bx%5E%7B1%2F2%7D-x%7D)
Next we rationalize the numerator and denominator. We do so by recalling


In particular,


so we have

For

and

, we can simplify the first term:

So our limit becomes