can u solve x and y simalteneously for me thanks in advance x-3+2y=0 2y^2+3x^2+4xy=9
from the first linear equation:
x = 3 - 2y
plug that into
2y^2 + 3x^2 + 4xy = 9
2y^2 + 3(3-2y)^2 + 4y(3-2y) = 9
2y^2 + 27 - 36y + 12y^2 + 12y - 8y^2 -9 = 0
6y^2 - 24y + 18 = 0
y^2 - 4y + 3 = 0
(y - 1)(y - 3) = 0
take it from there
from this step how did u come up with 2y^2+27-36y+12y^2+12y-8y^2-9=0
I expanded the 3(3-2y)^2 part
= 3(9 - 12y + 4y^2
= 27-36y+12y^2
straightforward algebra
Sure, I can help you solve the system of equations simultaneously. To solve for x and y simultaneously, we need to find the values of x and y that satisfy both equations.
Let's start by rearranging the first equation:
x - 3 + 2y = 0
Move the constant term (-3) to the other side of the equation:
x + 2y = 3 ---> Equation 1
Now, let's take a look at the second equation:
2y^2 + 3x^2 + 4xy = 9
To simplify this equation, we'll need to apply some algebraic techniques. One approach is to try to factorize or manipulate the equation into a more manageable form.
Since the equation contains terms with the variables squared, we can try factoring or rearranging the equation. However, in this case, it isn't that straightforward.
Alternatively, we can try solving the equation using numerical methods or graphically. Graphing the equations will give us a visual representation of the points where they intersect, which are the solutions.
Do you have any specific values you would like to substitute for x or y to obtain a numerical solution, or would you like to proceed with graphing the equations?