Question

solve the system the first equation is 3xy-4(x^2)=2 and the second equation is -5(x^2)+3y^2=7

Answer 1

\[{3xy-4x^2=2,- 5x^2+3y^2=7}\]

Answer 2

i need to know how to solve it for all its real solutions

Answer 3

Solve the first equation for y and substitute into the second equation.

Answer 4

yeah but there are 4 solutions

Answer 5

Yes. And when you do that you will get an equation containing x^4 which will have 4 solutions.

Answer 6

ok

Answer 7

\[y=\frac{2+4x^2}{3x}\]

Answer 8

Substitute that into the second equation.

Answer 9

so i just solve that

Answer 10

Do you know what substitute means?

Answer 11

ok yes got it

Answer 12

\[-5x^2+3(\frac{2+4x^2}{3x})^2=7\]

Answer 13

\[-5x^2+3(\frac{4+16x^2+16x^4}{9x^2})=7\]

Answer 14

ok

Answer 15

i already have two solutions i just need to know how to get the other two

Answer 16

\[-5x^2+\frac{12+48x^2+48x^4}{9x^2}=7\]

Answer 17

What are the two you have?

Answer 18

(-3,-3) and (2,3) and i know th other two are (-1,-2) and (1,20 i just don't know how to get them

Answer 19

You end up with:
\[(x^2-4)(x^2-1)=0\]

Answer 20

sorry the first one is (-2,-3)

Answer 21

So x = 2, -2, 1, -1

Answer 22

So go back to the equation:

Answer 23

\[y=\frac{2+4x^2}{3x}\]

Answer 24

And replace x with each one of the 4 x values and find the corresponding y.

Answer 25

k

Answer 26

so r (2,3) and (-2,-3) not solutions

Answer 27

(2,6) (-2,-6) (1,2) (-1,-2)

Answer 28

abc123 that is true

Answer 29

what is

Answer 30

That (2,3) and (-2,-3) are not solutions.

Answer 31

(2,3) and (-2,-3)

Answer 32

ok

Answer 33

What grade are you in bro?

Answer 34

11. don't judge

Answer 35

No wait. Those two are solutions. The 4 solutions are:(2,3, )(-2,-3, )(1,2), (-1,-2)

Answer 36

Sorry. I made a calculation error earlier.

Answer 37

ok