SOLUTION: Please solve algebraically: 2x^2 + 2y^2 + 4x = 6y x^2 + y^2 + 2x - 3y = 1 For the first equation, I was able to divide everything by 2 and complete the square. After, I comp

Algebra ->  Systems-of-equations -> SOLUTION: Please solve algebraically: 2x^2 + 2y^2 + 4x = 6y x^2 + y^2 + 2x - 3y = 1 For the first equation, I was able to divide everything by 2 and complete the square. After, I comp      Log On


   

Question 342986: Please solve algebraically:
2x^2 + 2y^2 + 4x = 6y
x^2 + y^2 + 2x - 3y = 1
For the first equation, I was able to divide everything by 2 and complete the square. After, I completed the square I factored it and got the answer:
( x+1)^2 + (y-3)^2 = 0
I did not know what to do after that though, it would be easy to solve the with a graph, but I cannot figure out how to solve both equations. Any help would be greatly appreciated, thanks so much!
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Here's the basic outline to solving this problem

1) Select an equation and complete the squares for both x and y. You've already done that. Call this equation 3.

2) Select a variable to solve for in equation 3. Let's say you choose to solve for 'y'.

3) Once you've solved for 'y', plug that solved expression into each 'y' of the other equation. Once you've done this, every 'y' will essentially be replaced by 'x' terms and you'll have an equation with only one variable.

4) Expand, combine like terms, and simplify to get the equation with one variable in the form ax%5E2%2Bbx%2Bc=0.

5) Now use the quadratic formula to solve for 'x'.

6) Use the solution(s) found in step 5) to find the corresponding solution(s) in terms of 'y'.


Let me know if this helps at all.