1) 


 
solve the first equation for y:





Substitute in the second equation:





Multiply both sides by 64



Square the binomial:





Possible rational zeros of that are
± fractions whose numerators are
factors of 64 and whose denominators 
are factors of 49.

The factors of 64 are 1,2,4,8,16,32,64

The factors of 49 are 1,7,49

We try the easiest one first in hopes
the person who made up this problem
was kind.

We try x=1, so we divide by x-1

1 | 49 -336  576 -64 
  |      49 -287 289 
  ------------------
    49 -287  289 225

Doesn't leave a 0 remainder.

We try x=2, so we divide by x-2

2 | 49 -336  576 -64 
  |      98 -476 200 
  ------------------
    49 -238  100 136

Doesn't leave a 0 remainder.

We try x=4, so we divide by x-4

4 | 49 -336  576 -64 
  |     196 -560  64 
  ------------------
    49 -140   16   0

Hooray! It leaves a 0 remainder!
So we have factored the left side of
the equation 



as



So we use the zero factor property

 which gives the value 



That does not factor, so we have to use the quadratic
formula:

 











Now we must substitute each of the 3 values for x
into



Substituting 











Therefore on soluton is (4,)

-----

Substituting 





Divide through by 2



Divide both sides by 4



So another solution is 

(x,y) = (,)

---
Substituting 





Divide through by 2



Divide both sides by 4



So another solution is 

(x,y) = (,)

---

2)  {(x+1)^2 - (y-1)^2 = 20
    {x^2 - (y+2)^2 - 24

Sorry, but you left out the equal sign in the second
Re-post it corrected and we can help you.  There must be
an equal sign somewhere in that.  You probably meant for 
one of the - signs to be an = sign but we can't tell 
which one.

Edwin