Question

Solve the system using any method (provide steps please)
\sqrt(3x)-\sqrt(2y)=2\sqrt(3)
2\sqrt(2x)+\sqrt(3y)=\sqrt(2)

Answer 1

I don't understand the question, can u upload the picture?

Answer 2

@kevin no i can't screenshot from my computer. I have to solve the system using substitution or elimination. Its square root of 3x minus square root of 2y is equal to 2 square root 3. The second equation is 2 square root of 2x plus square root of 3y is equal to sqaure root too.

Answer 3

*square root 2 not too sorry

Answer 4

Like this?
\[\sqrt(3x)-\sqrt(2y)=2\sqrt(3)\]
\[\ 2\sqrt(2x)+\sqrt(3y)=\sqrt(2)\]

Answer 5

@ AloneS yes

Answer 6

Alright lol, so yo need help with them..or something?

adn @alones like this LOL

Answer 7

@AloneS yes. I have to use substitution or elimination.

Answer 8

Ahha aright.

Answer 9

waitt so those two problems come together, or they're separate?

Answer 10

together. I am solving a system.

Answer 11

\[\large\rm \color{royalblue}{\sqrt{3x}-\sqrt{2y}=2\sqrt3}\]Let's try to line up the x's.
We'll multiply this first equation by 2sqrt2,\[\large\rm \color{royalblue}{2\sqrt{6x}-4\sqrt{y}=4\sqrt6}\]Then,\[\large\rm \color{orangered}{2\sqrt{2x}+\sqrt{3y}=\sqrt2}\]Multiplying our second equation by sqrt3 gives us,\[\large\rm \color{orangered}{2\sqrt{6x}+3\sqrt{y}=\sqrt6}\]Now the x's match, yay!
So you can subtract one from the other, and solve for y.