Question

solve system of equations:

3x-4y=25
x^2+y^2=25

Answer 1

square equation (1) : 9x^2 - 16y^2 = 625
(Multiply equation (2) by 9 ) : 9x^2 + 9 y^2 = 225
solve : - 16y^2 - +9y^2 = 625 - 225
-25y^2 = 400
- y =\[\sqrt{400/25}\]
- y = \[\sqrt{16}\]
y = -4
if y = -4 then 3x = 25+4y
= 25 + 4 ( -4)
= 25-16
= 9
therefore x = 9 and y = -4
and it should be the right answer,,,, :)

Answer 2

i didn't know you could square the first one that makes it simple then ! thanks

Answer 3

you just need to cancel out either x or y... to obtain the answer..
i prefer to square the equation and multiply by 9 to the other equation rather than :
\[x = (25 + 4y) \div 3\]
and then plug in value of x into the equation : x^2 + y^2 = 25

Answer 4

yea i did it like that and it was a mess .. squaring seems better in this case thank you!

Answer 5

btw i made a mistake...
when replacing y value into the equation:
3x = 25 + 4y
3x = 25 + 4 (-4)
3x = 9
x = 3
sorry..

Answer 6

i get sqrt of -16 which will give me an imaginary how did you get a whole number

Answer 7

i did it like : -y = sqrt of 16... just to avoid the imaginary
so , y= -4

Answer 8

is that 'legal' ?

Answer 9

guess not :s i just did that to avoid - sqrt of 16
but let me try again another way ..

Answer 10

thank you so much!

Answer 11

ok

Answer 12

3x - 4y = 25 , x^2 + y^2 = 25
so : 3x -4y = x^2 + y^2
x^2 - 3x = - y^2 -4y
x (x -3) = -y ( y + 4)
( x-3 ) = 0 or (y + 4 ) =0
x= 3 and y = -4

Answer 13

that would mean that x is also 0 and y is also 0

Answer 14

yep but you cant use this values because if x and y would be 0 it would mean that the curve does exist so it is rejected x = 0 y=0 ( Rejected )

Answer 15

this is the graph of x^2 + y^2 = 25