## cutiecomittee123 one year ago Solve this system of conic section equations 7y^2+x^2=64 x+y=4

1. jim_thompson5910

First solve the equation x+y = 4 for y \[\Large x+y=4\] \[\Large y=4-x\] Now plug this into the other equation so you boil things down to one variable only \[\Large 7y^2+x^2=64\] \[\Large 7(4-x)^2+x^2=64\] \[\Large 7(16-8x+x^2)+x^2=64\] \[\Large 112-56x+7x^2+x^2=64\] \[\Large 112-56x+8x^2=64\] \[\Large 112-56x+8x^2-64=0\] \[\Large 8x^2-56x+48=0\] From this point, you need to solve for x. To do so, I recommend using the quadratic formula. I'll let you do this part. Tell me what x values you get as the solutions.

2. cutiecomittee123

I get x=1 and x=6

3. jim_thompson5910

me too

4. jim_thompson5910

if x = 1, then y = ???

5. cutiecomittee123

sweet and then just plug that into on of the original equations to solve for y

6. jim_thompson5910

yes or plug into y = 4-x since that already has y isolated

7. cutiecomittee123

so like y=4-1 and y=4-6

8. cutiecomittee123

y=3 and y=-2

9. jim_thompson5910

correct, so the solutions are these two ordered pairs (1,3) and (6,-2) and this confirms it http://www.wolframalpha.com/input/?i=7y^2%2Bx^2%3D64%2Cx%2By%3D4

10. cutiecomittee123

wanna help me with another one?

11. jim_thompson5910

sure

12. jim_thompson5910

13. cutiecomittee123

x^2+y^2+2x+2y=0 x^2+y^2+4x+6y+12=0 solve this system of equations

14. jim_thompson5910

ok at first this looks really complicated, but we can eliminate quite a bit here notice how each equation has x^2+y^2 in it so we can subtract the equations (either equation1 - equation2 or equation2-equation1) to eliminate the x^2+y^2 terms what do you get when you subtract?

15. cutiecomittee123

2x+4y+12=0

16. jim_thompson5910

Now let's solve 2x+4y+12=0 for x 2x+4y+12=0 2x+4y+12-12=0-12 2x+4y = -12 2x+4y-4y = -12 - 4y 2x = -4y - 12 2x/2 = (-4y-12)/2 x = -2y - 6

17. cutiecomittee123

now plug that into the other equation

18. jim_thompson5910

Next, plug x = -2y - 6 into either original equation. I'll pick the first equation x^2+y^2+2x+2y=0 (-2y-6)^2+y^2+2(-2y-6)+2y=0 ... replace x with -2y-6 (4y^2+24y+36)+y^2+2(-2y-6)+2y=0 4y^2+24y+36+y^2-4y+12+2y=0 5y^2+22y+48=0 solve that for y (use the quadratic formula). Tell me what you get

19. jim_thompson5910

yes correct

20. jim_thompson5910

hmm I messed up, let me fix

21. jim_thompson5910

x^2+y^2+2x+2y=0 (-2y-6)^2+y^2+2(-2y-6)+2y=0 ... replace x with -2y-6 (4y^2+24y+36)+y^2+2(-2y-6)+2y=0 4y^2+24y+36+y^2-4y-12+2y=0 ... it should be -12, not +12 5y^2+22y+24=0

22. cutiecomittee123

well I tried and I got the b^2-4(a)(c) = -476 you cant take a square root of a negative number

23. jim_thompson5910

I fixed my mistake and got 5y^2+22y+24=0

24. cutiecomittee123

oh gotcha let me try that

25. cutiecomittee123

y=-2 and y=-2.4

26. jim_thompson5910

-2.4 or -12/5

27. jim_thompson5910

now use each y value to find the corresponding x value

28. cutiecomittee123

-12/5? how did you get that?

29. cutiecomittee123

-22+ or =- the sqrt of 4

30. cutiecomittee123

over 10

31. jim_thompson5910

I used the quadratic formula or you can convert -2.4 to fraction form -2.4 = -2.4*(10/10) = -24/10 = -12/5

32. cutiecomittee123

makes sense

33. cutiecomittee123

so now we plug in the y values to get x

34. jim_thompson5910

correct

35. jim_thompson5910

you can use x = -2y - 6

36. cutiecomittee123

x=-10.8 and x=-10

37. jim_thompson5910

incorrect on both

38. jim_thompson5910

if y = -2, then x = -2y - 6 x = -2(-2) - 6 x = 4 - 6 x = -2

39. jim_thompson5910

if y = -2.4, then x = -2y - 6 x = -2(-2.4) - 6 x = 4.8 - 6 x = -1.2

40. cutiecomittee123

so the solutions are (-2,-2) and (-1.2, -12/

41. jim_thompson5910

correct and the graph confirms it

42. jim_thompson5910

I think you meant to say -12/5

43. jim_thompson5910

if you use -12/5 then convert -1.2 to fraction form as well -1.2 = -1.2*(10/10) = -12/10 = -6/5

44. cutiecomittee123

Yeah I did mean -12/5 thanks:))

45. jim_thompson5910

np