Question

In the system shown below, what is the sum of the x-coordinates of all solutions?

3y^2-5x^2=7
3y^2-x^2=23

Answer 1

3y^2 - 5x^2 = 7
3y^2 = 7 + 5x^2
y = √[(7 + 5x^2)/3]

3y^2 - x^2 = 23
3√[(7 + 5x^2)/3]^2 - x^2 = 23
(7 + 5x^2) - x^2 = 23
4x^2 = 16
x^2 = 4
x = -2 [Not +2 because in the second quadrant, x is negative]

y = √[(7 + 5x^2)/3]
y = √[(7 + 5(-2)^2)/3]
y = √[(7 + 20)/3]
y = √9
y = +3 [Not -3 because in the 2nd quadrant y is positive]
(-2,3)

you asked this ques... once again!!
but hope this helps :-)

Answer 2

Well, neglect what the above person said.. I don't really know how come the term '' second quadrant'' appears :S

To do this, just solve for x and add the x value together
3y^2-5x^2=7 -(1)
3y^2-x^2=23 => 3y^2 = 23 +x^2 -(2)
Put (2) into (1)

23 + x^2 -5x^2=7
-4x^2 = -16
x^2 = 4

Can you solve it from here and get your answer?

Answer 3

can you help me out a little more please

Answer 4

x = 2 or x=-2
Can you do it?

Answer 5

-2?

Answer 6

sum of the x-coordinates of all solutions
For that i think you should add all the x values together?

Answer 7

ah I'm sorry but can you just be straight forward with me I'm tired and haven't gone to bed yet :/

Answer 8

First, I'm not sure if it is like that
x-coordinates of all solutions = sum of x values = 2+ (-2) =0
I don't really know it :S

Answer 9

sum of the x - coordinate is zero

Answer 10

may you please rewrite again the system of eqations