Systems of Equations...... how do i solve this????

Question

I understand how to solve simpler ones. But i dont get what top do when i have squared terms.

3x²-y²=9
x²+2y²=10

x²+y²=25
9y=4x²

how do i solve these?

(to make a square, press ALT + 0178)

Answer 1

For example:

x² + y² = 25
x² = 25 - y²

Then from the second equation,

9y = 4x²

so 9y = 4( 25 - y²)

9y = 100 - 4y²

4y² +9y -100 = 0

Answer 2

If you have a term that is squared in both and can eliminate, then do so. Otherwise you may need to use the substitution method
3x^2 - y^2 = 9
x^2 + 2y^2 = 10 multiply top equation by 2 to get the y's elim.

6x^2 - 2y^2 = 18
x^2 + 2y^2 = 10 Now add

7x^2 = 28
x^2 = 4
x = 2 or -2 put these back into one of the equations and find y

I used x^2 + 2y^2 = 10
x = 2: 4 + 2y^2 = 10
2y^2 = 6
y^2 = 3
y = sqrt3 or -sqrt 3 (2,sqrt3), (2, -sqrt 3)

x = -2: get the same y answers: (-2, sqrt3) and (-2, -sqrt3)
Four points of intersection.

You can do the second one by rewriting the second equation as -4x^2 + 9y = 0

Multiply the top eqution by 4: 4x^2 + 4y^2 = 100
Now adding the two cancels out the x^2 terms
4y^2 + 9 y = 100
4y^2 + 9y - 100 = 0
(4y + 25)(y - 4) =0
y = -25/4 and y = 4

Put these back into 9y = 4x^2 and solve for the x's that go with each y

Answer 3

These can usually be done with substitutions or addition and subtractions.

The first one, I would do by adding so multiply the top equation by 2 and then add it to the second one and get
6x²-2y²=18
x²+2y²=10

The y's cancel and you get 7x²=28 which gives us that x=-2,2. Go back to the second equation and then plug in both of the values and keep track of what you plugged in.

x=2 => 4+2y²=10 => y=Sqrt(3),-sqrt(3)
x=-2 gives us the same two answers so there are four solutions

(2,sqrt(3)), (2,-sqrt(3)), (-2,sqrt(3)), (-2,-sqrt(3))

and the answer makes sense because the first equation is a hyperbola and the second equation is an ellipse and it is possible for them to intersect at four distinct point.

The second system, I would do by substitution.

9y=4x² => (9/4)y=x² so plug it in the first equation
(9/4)y+y²=25 which is a quadratic and gives us
y=4, -14.5

So plugging them into the second equation gives us
y=4 => 36=4x² => x=3, -3

y=-14.5 => -32.625=x² which has no solution in the real numbers because the square of a real number cannot be a negative number so you have only two solutions.

The circle and the parabola intersect at two points
(3,4) and (-3,4)

Answer 4

3x²-y²=9 [a]
x²+2y²=10 [b]

Multiply equation [a] by 2.
(3x²-y²=9)2
6x²-2y²=18 [c]

Add equations [c] and [b] together.
6x²-2y²=18
x²+2y²=10

The -2y² and 2y² cancel each other out, and you're left with
7x²=28

Divide both sides by 7
x²=4

Take the square root of both sides
x=±2

Plug x=±2 into either equation (I'll choose equation [a])
3x²-y²=9
3(±2)²-y²=9
3(4)-y²=9
12-y²=9
-y²=-3
y²=3
y=√3 (square root of 3)

So your answers are x=±2 and y=√3

Answer 5

x²+y²=25 is a version of the pathagorem theory
3 & 4 = 5

9y=4x²
y-can be 4, 16, 36, ....
x-can be 3, 6, 9, ...

x²+2y²=10
is going to involve negative number

Answer 6

multiply the first eq by 2
6x^2-2y^2=18
x^2+2y^2=10
------------------- >simplify
7x^2=28
x^2=4
x=+ or - 2

now plug either them into any eq
1st one-
12-y^2=9
y^2=3
y=square root of 3
so ur solutions are
(2,sqr(3)) and (-2,sqr(3))

for the 2nd one
make the second eq = 0