## Matt71 2 years ago <Simultaneous equation> y=2-x, x(x+y)=5-3y^2 a solution would be helpful please, no idea how these things are solved

1. Helper93

You know that y=2-x, so plug in 2-x instead of y in the second equation and solve.

2. Matt71

Hold on, lemme try

3. Matt71

What's (2-x)^2? Is it the long answer or like, the one with 2 numbers?

4. Matt71

can't do it man, help me out here

5. Mertsj

If y = 2-x, then x = 2-y. Replace each x with 2-y

6. Matt71

where do I go from 2-2y=5-3y^2?

7. Mertsj

Put it in the standard form for a quadratic equation and then solve by factoring (if it will factor) or by the quadratic formula if it will not.

8. Matt71

English man, English

9. Mertsj

Rearrange the equation so that it is equal to 0. Make the y^2 term positive.

10. Mertsj

Also, it should be 4-2y not 2-2y

11. Matt71

Oh

12. Helper93

Actually, Mertsj is right, it's easier to replace x other than y.

13. Matt71

Eh?

14. Matt71

could one of you dudes just give me the answer +solution? It's REALLY late

15. Mertsj

\[(2-y)(2-y+y)=5-3y^2\] \[(2-y)(2)=5-3y^2\] \[4-2y=5-3y^2\] \[3y^2-2y-1=0\] \[(3y+1)(y-1)=0\]

16. Mertsj

Can you solve that for y?

17. Matt71

waaaait

18. Matt71

how did 3y^2-2y-1=0 become (3y+1)(y-1)=0 ?

19. Mertsj

By factoring.

20. Matt71

what method?

21. Helper93

You can also use discriminant if you want.

22. Mertsj

Too many cooks.

23. Matt71

mmm, well

24. Matt71

thanks dudes, I'll try to figure this dang thing out

25. Matt71

Sorry Helper Can't give more than one medal