How do u solve the system x^2+y^2=25 and 2x^2-3y=6 without a calculator?

Answer 1

You should make sure you have one variable to solve:

x^2=25-y^2

Place x^2 in the second equation

2 (25-y^2)-3y=6
50-2y^2-3y =6
44-2y^2-3y=0
(2y+11) (-y+4) =0

2y+11 = 0 or -y+4 =0

y= -11/2 or y= 4

Answer 2

You have to set one of the parts equal to each other in both equations. Meaning you can multiply the first equation by 2 to make the X's equal. that leaves you with 2y^2+3y=44. Move the 44 to the left hand side leaving 2y^2+3y-44. Now you have a quadratic equation. The equation leaves y= 4. Then you subsitute the Y's for the number 4, which leads to x^2+4^2=25. Subtract 4^2 from 25 and you get x^2=9. So x = 3. If you plug x=3 and y=4 for both equations, you'll see that they "check." So that is your answer and explanation.

Answer 3

hi,

a: x^2+y^2=25 => x^2 = 25-y^2

b: 2*(25-y^2) - 3y = 6

c: -2y^2-3y+44 = 0

d: -2y^2-11y+8y = 0

e:(2y+11)(4-y) = 0 ==> y1 = -5.5, y2 = 4

and you can take it from here.

Talk to me if there is anything you don't understand in what I did.

Answer 4

x^2+y^2=25
2x^2-3y=6

solve for y in the second equation
2x^2 - 3y = 6
-3y = -2x^2 + 6
y = 2/3x^2 - 2

plug 2/3x^2 - 2 for y in the first equation
x^2 + (2/3x^2 - 2)^2 = 25
x^2 + 4/9x^4 - 8/3x^2 + 4 = 25

multiply 9 for both sides
9x^2 + 4x^4 - 24x^2 + 36 = 225

4x^4 -15x^2 - 189 = 0

factor

(x - 3) (x + 3) (4x^2 + 21) = 0

x = 3 or -3

Answer 5

-2x^2 -2y^2 = -50
2x^2 -3y = 6
-2y^2-3y = -44
(2y-11)(y+4)=0
y = -4 and y = 11/2 Reject y= 11/2 because its abs is > 5
So x = 3 or -3
The curves intersect at (3,-4) and (-3,-4)