## jagr2713 one year ago x-3y=-9 6x+y=3

1. jagr2713

i dont understand this @freckles

2. steve816

How does a 99 rated mathlete not know how to do system of equations O.o

3. jagr2713

Well when you haven't done math in a while u forget things.....

4. IrishBoy123

|dw:1442268783821:dw|

5. jagr2713

ok i know the part where u add them and stuff but i just need help in the beginning. How do u find what to multiply them with?

6. IrishBoy123

when they are set out like that, just imagine adding them and knocking out an x or a y so, let's do it again

7. IrishBoy123

|dw:1442269218613:dw|

8. jagr2713

wait why did u use 1 and 2

9. IrishBoy123

if you meant, the (1) and (2) in brackets to identify the equations, and just a habit is that what you were referring to?

10. jagr2713

Because my teacher taught me to multiply them by a number and then add then and cancel the x or y u used etc

11. IrishBoy123

loads of ways to do these eg |dw:1442269699628:dw|

12. IrishBoy123

can i ask , do you work on the OS website??

13. jagr2713

whats the easiest way? can u do in equation form cause the drawing is losing me

14. jagr2713

do i work on the website?

15. IrishBoy123

yes

16. jagr2713

what does that mean?

17. IrishBoy123

are you part of the OS team?

18. jagr2713

Oh yes...

19. IrishBoy123

easiest way is substitution, IMHO. as in, it's pure method. so, from the first equation, which is $$x - 3y = -9$$, you can say $$x = 3y - 9$$ then in the second, which is $$6x + y = 3$$, you can say $$6(3y - 9) +y = 3$$ so $$18y - 54 + y = 3$$ or $$19y = 57$$, $$y = 3$$

20. IrishBoy123

so you code this site?!

21. jagr2713

OOOHOHO you have to change the x-3y=*9 in x=3y-9 and lets stay relevant :D u can message me the other questions

22. IrishBoy123

OK!!

23. jagr2713

But what if its x+3y=-9 does 3y become positive cause u move it?

24. IrishBoy123

if x + 3y = -9 then x = -3y - 9

25. IrishBoy123

because x+3y-3y = -9-3y

26. jagr2713

so i got 6(3y-9)+y=3 18y-54+y=3 +y +y 19y-54=3 +54 +54 19y=57 y=3

27. jagr2713

so now u plug it into the x too get x right? x-3(3)=-9 x+9=-9 -9 -9 x=-18 so we got (-18,3)

28. jagr2713

??

29. IrishBoy123

if you know that y = 3, plug that into one of the original equations to find x

30. IrishBoy123

eg, if x - 3y = -9 what is x if y = 3

31. IrishBoy123

or if 6x + y = 3 what is x if y = 3??

32. jagr2713

you plug it into the x-3y=-9 right?

33. IrishBoy123

into either, you want to know when they meet

34. jagr2713

it says -18,3 is wrong

35. jagr2713

wait don't i need another point?

36. IrishBoy123

i thought we had agreed 0, 3 ?!?!

37. IrishBoy123

as in x = 0

38. IrishBoy123

and y = 3

39. jagr2713

OOHOHOHO i did it wrong ok so i need another set of points to connect it.

40. IrishBoy123

indeed you might! gtg

41. jagr2713

So how do i get it?

42. jagr2713

@freckles

43. IrishBoy123

really sorry @jagr2713 i gotta go sleep i am in europe, i guess you are somewhere else my condolensces!!!!

44. jagr2713

Its ok

45. jdoe0001

are you meant to solve by either substitution or elimination?

46. jagr2713

Graphing them

47. jagr2713

i got the answer 0,3 just need another point

48. jdoe0001

well... where the graph of two equations meet, is where the answer is :) hm if 0,3 is the answer.... isn't that what you needed? :)

49. jagr2713

Yea but thats only one point need 2 for a line

50. jdoe0001

hmmm one sec

51. jagr2713

ok :D

52. jagr2713

here use this i did another question x-3y=-3 3x+y=1 i got (0,1)

53. jdoe0001

..is that another one?

54. jagr2713

Yea i just change questions i got 0,1 but need another point

55. jdoe0001

hmmm thought you were supposed to just graph them?

56. jagr2713

Yea it say solve the following system of equation by graphing them

57. jdoe0001

well... I'd just graph them then :) lemme do a quick table

58. jagr2713

ok thanks

59. jagr2713

wait do we just do the x|y thing and put 0 for x and ory?

60. jagr2713

|dw:1442273690049:dw| like that?

61. jdoe0001

yeap to make a line, all you need is 2 points so just pick two random "x" to get two "y"s

62. jagr2713

OHH so |dw:1442273743049:dw|

63. jagr2713

other is 0,3 2,0

64. jdoe0001

hmm yes $$\begin{array}{llll} x-3y=-3\implies &\cfrac{x+3}{3}={\color{brown}{ y}} \\ \quad \\ 3x+y=1\implies &{\color{brown}{ y}}=1-3x \end{array} \\ \quad \\ \qquad \qquad \begin{array}{ccllll} x&y \\\hline\\ &\frac{x+3}{3}\\ \\\hline\\ &\frac{x+3}{3}\\ 0&1\\ -3&0 \end{array}\qquad \begin{array}{ccllll} x&y \\\hline\\ &1-3x \\\hline\\ 0&1\\ 4&-11 \end{array}$$

65. jagr2713

where did u get they -3 from?

66. jdoe0001

x-3y=-3 <--- 3x+y=1

67. jagr2713

is this correct (0,3) (-9,0) and (0,3) (2,0) ?

68. jdoe0001

hmmm 0, 3 is not, -9,0 isn't either

69. jdoe0001

x =0 x -3y=-3 0-3y=-3 y = (-3)/(-3) y =1

70. jagr2713

x-3y=-9 -3y=-9 y=1/3 x-3y=-9 x=-9

71. jdoe0001

x=-9 (-9)-3y=-3 -3y=-3+9 -3y=+6 y=(6)/(-3) y=-2

72. jagr2713

wait this is confusing me dont u put 0 for them?

73. jdoe0001

wel... you said it was x-3y=-3 3x+y=1 but now you seem to have changed it some

74. jagr2713

OHOHOHO I've been doing the wrong problem... :D KML

75. jagr2713

I WAS LOOKING AT THE QUESTION LOL

76. jagr2713

i got (0,1) (-3,0) and (0,1) (3,0)

77. jdoe0001

but all you need to graph a line is two points $$\begin{array}{llll} x-3y=-3\implies \cfrac{x+9}{3}={\color{brown}{ y}}\\ 6x+y=3\implies {\color{brown}{ y}}=3-6x \end{array}$$ http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiIoeCs5KS8zIiwiY29sb3IiOiIjM0IxRENGIn0seyJ0eXBlIjowLCJlcSI6IjMtNngiLCJjb2xvciI6IiNENDExMTEifSx7InR5cGUiOjEwMDAsIndpbmRvdyI6WyItNi41NCIsIjYuNDYiLCItMi40OCIsIjUuNTIiXX1d

78. jagr2713

D: it says my answer is wrong... i did the x|y thing

79. jdoe0001

check yours PEMDAS I'd think

80. freckles

I'm confused... Are you saying you got 3 different solutions for x-3y=-3 3x+y=1 ? You should only get one solution, infinitely many solutions, or no solutions for system of linear equations.

81. jagr2713

OHH my friend told me to use y=mx+b

82. freckles

why? aren't you trying to solve the system x-3y=-3 3x+y=1 ? or is that not the question?

83. jagr2713

yes by graphing it thou but i did it and getting it wrong D:

84. freckles

you should just get a point for the solution of this system: you will see that when graphing these two lines that they will share a common point that point is the solution x-3y=-3 -3y=-x-3 y=1/3*x+1 --- 3x+y=1 y=-3x+1 so graphing both y=1/3*x+1 and -3x+1 we have something that looks like this: |dw:1442275696567:dw| the solution is given at the point where both lines touch each other (cross each other; intersect each other)

85. freckles

|dw:1442275792808:dw| so this point I darkened is the solution because is the point that two lines have in common

86. jagr2713

wait can u try thing one x-2y=-2 4x+y=1

87. freckles

were you trying to write in y=mx+b form to graph the lines to find the point ?

88. jagr2713

yea i was using that

89. freckles

ok well try writing x-2y=-2 in that form

90. jagr2713

2y=x+2 right?

91. freckles

so you subtracted x on both sides and multiplied -1 on both sides? x-2y=-2 subtracted x on both sides -2y=-x-2 multiplied both sides by -1 2y=x+2 this is correct so far then just divide both sides by 2

92. jagr2713

yea i got y=1/2x+1

93. jagr2713

then u graph it by going to (0,1) then rise over run

94. freckles

that sounds great so y=1/2x+1 says the y-intercept is 1 and the slope is 1/2 |dw:1442276083294:dw| right so you get something like this now we also need to graph 4x+y=1 on that same graph

95. freckles

4x+y=1 is equivalent to y=-4x+1 can you try graph that we should actually already see the solution even before graphing

96. jagr2713

yea we got 0,1 then rise over run -4/1

97. freckles

so before graphing you should see that (0,1) is a point on both graphs where both graphs differ in slope therefore (0,1) is the solution

98. jagr2713

yea 0,1 is the slope so we go down 4 and move 1 to the left right?

99. freckles

(0,1) is the y-intercept the slope is -4/1 so we move down from the y-intercept 4 units and to the right once

100. jagr2713

Oh right 1 sorry my friend told me left....

101. jagr2713

102. jagr2713

can u help with one more but this one is substitution

103. freckles

sure

104. freckles

one more then I have to roll my croissant dough

105. jagr2713

ok :D 4x+y=24 y=2x

106. freckles

replace first y with (2x) solve 4x+2x=24 for x

107. jagr2713

don't you reverse it like y=4x+24 and plug into x y=2(4x+24)

108. freckles

the second equation says y=2x so you insert that y here: 4x+y=24 to get 4x+2x=24 anyways you can't replace x with 4x+24 because we aren't given x=4x+24 unless you already know that x=-8 which x doesn't -8 and also 4x+y=24 is not equivalent to y=4x+24

109. jagr2713

wait how did u get 4x+2x=24

110. freckles

$\color{red}{y=2x} \\ 4x+\color{red}{y}=24 \\ \\ 4x+\color{red}{2x}=24$

111. freckles

I replaced y with 2x because one of the equations said y=2x

112. jagr2713

Wait i understand igtg thanks for the help

113. freckles

k peace