Is the answer not suppose to be pretty?- nonlinear equations

[Paradox]

1.

x^2 + y^2 = 8
16x + 4y^2 = 80

2.

x^2 + y^2 = 25
x - y = 1

These are systems of equations (obviously). The first one, I can't remember what to do. The second one, I've tried substitution but I couldn't get the answer. Please help. I should probably add that I'm suppose to find the solution set for each.

I attempted the first one and I got:

x = 2 (+ or -) 2sqrt(-2)

and then I got y^2 to equal either -4 or 12, plugging in both forms of x. The answers aren't very pretty, so I'm guessing I did something terribly wrong.

 

[stapel]

Substitution is the customary method for solving systems of equations.

1) I would solve each for "y² =", set the right-hand sides equal, and solve the resulting quadratic for x. Then plug each x-value into one of the original equations, and solve for the corresponding y-values.

2) Solve the second equation for one of the variables (your choice). Plug that into the first equation. Solve the resulting quadratic. Plug the solutions into the first equation (in its original form), and solve for the corresponding values of the other variable.

Note: You can check your solutions by graphing.

Eliz.

 

[tkhunny]

Re: I forget how to work these problems

1.

x^2 + y^2 = 8
16x + 4y^2 = 80

2.

x^2 + y^2 = 25
x - y = 1

These are systems of equations (obviously). The first one, I can't remember what to do. The second one, I've tried substitution but I couldn't get the answer. Please help. I should probably add that I'm suppose to find the solution set for each.

I attempted the first one and I got:

x = 2 (+ or -) 2sqrt(-2)

and then I got y^2 to equal either -4 or 12, plugging in both forms of x. The answers aren't very pretty, so I'm guessing I did something terribly wrong.

Remember, when you have circles, you have limited Domain. It is not so unusual to have no solution.

On the first, I'll assume you have posted it correctly.

x^2 + y^2 = 8
16x + 4y^2 = 80

#1 - Remove common factors

x^2 + y^2 = 8
4x + y^2 = 20

#2 - It looks quite convenient to solve the second for y^2 and substitute into the first.

4x + y^2 = 20 ==> y^2 = 20-4x
x^2 + y^2 = 8 ==> x^2 + (20-4x) = 8

You should have no trouble solving that for 'x', using the Quadratic Formula, if nothing else.

On the second:
x^2 + y^2 = 25
x - y = 1

Solve and substitute.

x - y = 1 ==> x = 1+y
x^2 + y^2 = 25 ==> (1+y)^2 + y^2 = 25

1 + 2*y + y^2 + y^2 = 25
2*y^2 + 2*y - 24 = 0

It looks to me like you should be able to solve that, again, using the Quadratic Formula if nothing else leads anywhere useful.

 

[Paradox]

*crys* But what do I do about the insane answers that I get when I use the quadratic formula!? I undertand now how to DO these problems.... but why am I not able to get resonable answers? Are these two problems SUPPOSE to be that way?

 

[Denis]

"But what do I do about the insane answers that I get when I use the quadratic formula!?"

Nothing!

"I undertand now how to DO these problems.... but why am I not able to get resonable answers?"

Because the equation itself is not reasonable...

"Are these two problems SUPPOSE to be that way?"

Yes, the ANSWER is SUPPOSED to be that way.
By the way, the 2nd problem has VERY REASONABLE answers; what did you get?


AND I'll bet you if you check this equation you posted: 16x + 4y^2 = 80 ,
you'll find you made a typo and should be 16x^2 + 4y^2 = 80 ;
then you'd get a NICE answer of x=+-2 and y=+-2