SOLUTION: I have been sitting here for hours trying to figure the following problems. I am working on Non-Linear Systems and have to solve the following systems. No matter how I work it I

Algebra ->  Linear-equations -> SOLUTION: I have been sitting here for hours trying to figure the following problems. I am working on Non-Linear Systems and have to solve the following systems. No matter how I work it I       Log On


   

Question 252412: I have been sitting here for hours trying to figure the following problems. I am working on Non-Linear Systems and have to solve the following systems. No matter how I work it I can't seem to come to a solution.
1) {7x-8y=24
{xy^2=1
2) {(x+1)^2 - (y-1)^2 = 20
{x^2 - (y+2)^2 - 24
Thank you - Lori
Found 2 solutions by jim_thompson5910, Edwin McCravy:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Even though these are non-linear equations, we can still use substitution to solve them. I'll do the first one to get you started. If that doesn't help either repost or ask me.


# 1

xy%5E2=1 Start with the second equation.


x=1%2F%28y%5E2%29 Divide both sides by {{y^2}}} to isolate 'x'.


7x-8y=24 Move onto the first equation.


7%281%2F%28y%5E2%29%29-8y=24 Plug in x=1%2F%28y%5E2%29


7-8y%5E3=24y%5E2 Multiply EVERY term by the LCD y%5E2 to clear out the fractions.


-8y%5E3-24y%5E2%2B7=0 Get every term to the left side.


8y%5E3%2B24y%5E2-7=0 Multiply every term by -1.


Now use the rational root theorem to find that y=1%2F2 is a root to the polynomial equation above. In other words, if you plug in y=1%2F2 into 8y%5E3%2B24y%5E2-7, you will get 0. Because of this fact, this means that 2y-1 is a factor of 8y%5E3%2B24y%5E2-7


Now use polynomial long division to find that %288y%5E3%2B24y%5E2-7%29%2F%282y-1%29=4y%5E2%2B14y%2B7. So 8y%5E3%2B24y%5E2-7=%282y-1%29%284y%5E2%2B14y%2B7%29


This tells us that %282y-1%29%284y%5E2%2B14y%2B7%29=0. Since we know that 2y-1=0 gives a root of 1%2F2, we can ignore this equation. So the next step is to solve 4y%5E2%2B14y%2B7=0 for 'y'. Use the quadratic equation to find the next two solutions of y=%28-7%2Bsqrt%2821%29%29%2F4 or y=%28-7-sqrt%2821%29%29%2F4


So the three solutions in terms of 'y' are y=1%2F2, y=%28-7%2Bsqrt%2821%29%29%2F4 or y=%28-7-sqrt%2821%29%29%2F4


From here, plug each solution (in terms of 'y') into x=1%2F%28y%5E2%29 to find the corresponding solution in terms of 'x'.


I skipped a bit of steps (since they're a bit long and I'm out of time for now), so feel free to ask about any step.

Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!


1) 

system%287x-8y=24%2C+xy%5E2=1%29
 
solve the first equation for y:

7x-8y=24
7x-24=8y
%287x-24%29%2F8=y

Substitute in the second equation:

x%28%287x-24%29%2F8%29%5E2=1

x%287x-24%29%5E2%2F64=1

Multiply both sides by 64

x%287x-24%29%5E2=64

Square the binomial:

x%2849x%5E2-336x%2B576%29=64

49x%5E3-336x%5E2%2B576x-64=0

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 

49x%5E3-336x%5E2%2B576x-64=0

as

%28x-4%29%2849x%5E2-140x%2B16%29=0

So we use the zero factor property

x-4=0 which gives the value x=4

49x%5E2-140x%2B16=0

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

 

x+=+%28140+%2B-+sqrt%2816464%29%29%2F98+

x+=+%28140+%2B-+sqrt%28784%2A21%29%29%2F98+

x+=+%28140+%2B-+28sqrt%2821%29%29%2F98+

x+=+%2814%2810+%2B-+2sqrt%2821%29%29%29%2F98+

x+=+%2810+%2B-+2sqrt%2821%29%29%2F7+

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

7x-8y=24

Substituting x=4

7%284%29-8y=24

28-8y=24

4=8y

4%2F8=y

1%2F2=y

Therefore on soluton is (4,1%2F2)

-----

Substituting x+=+%2810+%2B+2sqrt%2821%29%29%2F7+

7%28+++%2810+%2B+2sqrt%2821%29%29%2F7+++++++++%29-8y=24

%2810+%2B+2sqrt%2821%29%29-8y=24%7D%7D%0D%0A%0D%0A%7B%7B%7B-14%2B2sqrt%2821%29=8y

Divide through by 2

-7%2Bsqrt%2821%29=4y

Divide both sides by 4

%28-7%2Bsqrt%2821%29%29%2F4=y

So another solution is 

(x,y) = (%2810+%2B+2sqrt%2821%29%29%2F7,%28-7%2Bsqrt%2821%29%29%2F4)

---
Substituting x+=+%2810+-+2sqrt%2821%29%29%2F7+

7%28+++%2810+-+2sqrt%2821%29%29%2F7+++++++++%29-8y=24

%2810+-+2sqrt%2821%29%29-8y=24%7D%7D%0D%0A%0D%0A%7B%7B%7B-14-2sqrt%2821%29=8y

Divide through by 2

-7-sqrt%2821%29=4y

Divide both sides by 4

%28-7-sqrt%2821%29%29%2F4=y

So another solution is 

(x,y) = (%2810+-+2sqrt%2821%29%29%2F7,%28-7-sqrt%2821%29%29%2F4)

---

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