Question

Solve the equations:
5x^2+4y^2=16 -----eqn 1
13x^2-5y^2=57-----eqn 2

Answer 1

anyone there

Answer 2

prolly try elimination, or a cramer rule

Answer 3

the y parts are prolly easiest to get rid of visually; multiply the top by 5 and the bottom by 4, than add em up

Answer 4

first i multiplied eqn 1 by 5 and 2 by 4 to get

Answer 5

and then added both equations

Answer 6

and the result is 77x=308

Answer 7

then the value of x will be coming+2,-2

Answer 8

that seems reasonable :)

5(5)x^2+4(5)y^2=16(5)
13(4)x^2-5(4)y^2=57(4)
------------------------
x^2(25+52) = 80+228

it looks like youve already solved it for the most part

Answer 9

then i substituted the value of x in eqn 1

Answer 10

i dont know the last part so that is why i am giving you all hints to solve last part of the question

Answer 11

are you allowed complex solutions?

Answer 12

5x^2 + 4y^2 = 16

the last part is to finish it up ....

Answer 13

on substituting in eqn 1 we will get the value of y i.e, what??????

Answer 14

if you did all that work by yourself above the rest should be a walk in the park for you

5(2)^2 + 4y^2 = 16

5(4) + 4y^2 = 16

20 + 4y^2 = 16

4y^2 = -4

y^2 = -1

Answer 15

first equation is an ellipse
second is a hyperbola
they do not intersect

Answer 16

since as amistre wrote you get \(y^2=-1\) which is certainly not possible with real numbers

Answer 17

it will be -1=i am i right

Answer 18

iota

Answer 19

if you are allowed numbers that fall outside of the set of "Real", then yes, it would be "i"

otherwise there are no "Real" solutions

Answer 20

it is allowed

Answer 21

it will be +i and -i on squaring on both sides

Answer 22

its not done yet

Answer 23

now for x=-2
again the answer will be coming +i and -i

Answer 24

of course it is sine (+-2)^2 = 4

Answer 25

since

Answer 26

thank u all i was making a silly mistake

Answer 27

we all make the silly mistakes :)

Answer 28

thanks its done the solution set is {(2,+i,-i),(-2,+i,-i)} thanks the book shows the answer is absolutely right