How can I solve this? Need Immediate Help!?

Question

4x^2 + y^2=16
x^2 - y^2= -4

I had a lesson on 2nd degree equations and I don't understand it.

Please help...it's URGENT!

Answer 1

let 4x^2+y^2=16 be equation 1
x^2-y^2=-4 be equation 2
solving equation 1 and 2 simultaneously
add equation1 to equation 2
(4x^2)+(x^2)+(y^2)+(-y^2)=16+(-4)
5x^2 +(y^2)-(y^2) =16-4
5x^2 +0 =12
5x^2=12
divide both sides of the equation by 5
(5x^2)/5=12/5
x^2=12/5
take the square root of both sides
x=square root of 12/5
x=sqrt(2.4)
x=+ or -1.549 or +or- 2.4^(1/2)

to find x substitute for x=2.4^(1/2) into equation 1
x^2-y^2=-4
2.4^(1/2)^2 -y^2=-4
2.4 -y^2=-4
y^2=2.4+4
y^2=6.4
take the square root of both sides
y=sqrt(6.4)
y=+2.5298 or -2.5298
y is approximately + 2.53 or -2.53

Answer 2

I suspect this is one system with 2 equations.

If you add them together, y^2 will disappear, leaving you with only 3 x^2 = 12.
Solve for x, then use that value of x to find y in an one of the original equations.

In Geometry, the top equation is a circle of radius 4, the bottom one is a hyperbola with two branches, the top one has two "arms" sticking up, symetrically on both sides of the y axis; the bottom branch has two "legs", also symetrical about the y axis.

Therefore, there will be 4 answers, all symmetrical in relation to the axes, in the form (x,y), (-x,y), (-x,-y),(x,-y)

Answer 3

Solve the second equation for y^2.
You get y^2 = x^2 + 4.

Now plug that into the first equation:
4x^2 + (x^2 + 4) = 16
5x^2 = 12
x^2 = 12/5

So there are two possible x-values: positive and negative square root of (12/5).

Plug each of the two x-values into one of the equations, and solve for y.

Hope that helps!

Answer 4

It would be easiest to replace x² with a and y² with b. Then you have:
4a + b = 16
a - b = -4

Now you can solve as you normally would with two equations and two unknowns:
5a = 12
a = 12/5

12/5 - b = -4
b = 4 + 12/5
b = 32/5

Now plug x and y back in:
x² = 12/5
x = sqrt(12/5)
x = sqrt(60/25)
x = sqrt(4) * sqrt(15) / sqrt(25)
x = ± 2 sqrt(15) / 5

y² = 32/5
y² = 160 / 25
y = sqrt(16) * sqrt(10) / sqrt(25)
y = ± 4 sqrt(10) / 5

Answer 5

4x^2 + y^2 = 16 eq'n 1

x^2 - y^2 = -4 eq'n 2

Rearrange equation 2 to give:
y^2 = x^2 +4
Then substitute this value for y^2 into equation 1:

4x^2 + (x^2 +4) = 16
5x^2 = 12
x^2 = 12/5 = 2.4
x= (+or-) sqrt2.4

Hope this helps

Answer 6

4x^2 + y^2=16
x^2 - y^2= -4
---
x^2 = -4 + y^2
---
4(x^2) + y^2=16
4(-4 + y^2) + y^2=16
---
-16 + 4y^2 + y^2 = 16
-16 + 5y^2 = 16
5y^2 = 32
y^2 = 32/5
y = +/-square root( 32/5 )

Answer 7

Add them togetherL

5x^2=12 --> x^2=12/5 --> x=(12/5)^-1

Insert in second eq.
12/5 -y^2=-4 --> y^2=12/5+4=22/5 - y=(22/5)^-1

Answer 8

Try adding the two equations together. Don'f forget the +/- when you take roots...

Answer 9

i have no idea