True & False Problem

[maggie0160]

I think the answer is True, but I maybe be wrong when ti comes to ordered pairs.

Solve the system:
2x^2 - 3y^2 = 16
x^2 + 2y^2 = 15.
There are exactly three different ordered pairs which satisfy this system.

True or False

 

[pka]

Can you graph the system?
I find four pair.

 

[soroban]

Hello, maggie0160!

I think the answer is True, but I maybe be wrong when it comes to ordered pairs. . . . are you guessing?

Solve the system:
. . . \(\displaystyle 2x^2\,-\,3y^2\:=\:16\)
. . . .\(\displaystyle x^2\,+\,2y^2\:=\:15.\)

There are exactly three different ordered pairs which satisfy this system. .True or False?

Multiply the second equation by -2: .-\(\displaystyle 2x^2\,-\,4y^2\:=\) -\(\displaystyle 30\)
. . . . . . . . . . Add the first equation: . .\(\displaystyle 2x^2\,-\,2y^2\:=\:16\)

. . . . . and we get: .-\(\displaystyle 7y^2\,=\,-14\;\;\Rightarrow\;\;y^2\,=\,2\;\;\Rightarrow\;\;y\,=\,\pm\sqrt{2}\)

. . . . . then: .\(\displaystyle x\,=\,\pm\sqrt{11}\)


There are four intersections: .\(\displaystyle (\sqrt{11},\,\sqrt{2}),\\sqrt{11},\,\)-\(\displaystyle \sqrt{2}),\\)-\(\displaystyle \sqrt{11},\,\sqrt{2}),\\)-\(\displaystyle \sqrt{11},\,\)-\(\displaystyle \sqrt{2})\)