## Inopeki 3 years ago Can someone give me a only letter equation to solve?

1. across

2x+3=11

2. Inopeki

take 3 from both sides getting 2x=8 divide both sides by 2 gettin x=4

3. Inopeki

Btw, grats on Master rank

4. Inopeki

Can i have a harder one now?

5. jimmyrep

good job

6. FoolForMath

that's called only a letter equation? :D

7. Inopeki

Nope but i take whats coming lol

8. across

Let's do a slightly trickier one: (1/3)x+2=3

9. Inopeki

Ill take (1/3)x+2=3 1x+6=9, right sofar?

10. across

That's right thus far!

11. Inopeki

x+6=9 x=3

12. Inopeki

?

13. FoolForMath

That's correct.

14. across

Very good! Let's do this one now: (2/3)x+(5/2)=(5/4) (I love fractions.)

15. Inopeki

Please put that in equation form

16. across

$\frac{2}{3}x+\frac{5}{2}=\frac{5}{4}$

17. across

If you can do that without problems, then you would have mastered the solving of single-variable, linear equations. :P

18. Inopeki

Hmm

19. Inopeki

Im supposed to "equalize the fractions" right?

20. Inopeki

Wait!

21. across

We're trying to solve for $$x$$, so try and use the properties of addition, subtraction, multiplication and division to achieve that!

22. Inopeki

NONONONO

23. Inopeki

no clues :(

24. Inopeki

(2/3)x+(5/2)=(5/4)----->(4/6)x+(10/4)=(10/8)?

25. Inopeki

No that cant be it..

26. Inopeki

wait, just say yes or no, not why.

27. across

You rewrote the equation. :P

28. Inopeki

(2/3)x+(5/2)=(5/4) (2/3)x=(5/4)-(5/2) Right so far?

29. across

Yes

30. Inopeki

Now multiply both sides by 3 to get (3/3)x=(5/4)-(5/2)*3?

31. Inopeki

If so i have the answer

32. across

You have the right idea, but you're overlooking a few steps. :)

33. Inopeki

(3/3)x=(5/4)*3-(5/2)*3??

34. across

Let's first focus on trying to simplify this:$\frac{5}{4}-\frac{5}{2}.$Can you do that?

35. Inopeki

5-5 ---- 4-2?

36. across

There's where we're getting stuck. :P You need to learn how to simplify expressions of the form$\frac{a}{b}\pm\frac{c}{d}.$

37. Inopeki

:( i thought i knew this stuff...

38. across

It's not too bad, really, there's a simple rule which states that$\frac{a}{b}\pm\frac{c}{d}=\frac{ad\pm bc}{bd}.$

39. Inopeki

Whats the logic in that?

40. across

To put it simply, you cannot add nor subtract fractions with different bases. In other words,$\frac{5}{4}-\frac{5}{2}$cannot be simplified as it stands since they have different bases (4 and 2).

41. Inopeki

So i need to find the greates common (something i cant remember)?

42. across

However, if you have something like$\frac{3}{2}+\frac{5}{2},$they have similar bases and thus you can add them accordingly:$\frac{3+5}{2}=\frac{8}{2}=4.$

43. across

You can do that OR you could use the fact that$\frac{a}{b}\pm\frac{c}{d}=\frac{ad\pm bc}{bd}.$

44. Inopeki

How about if they have similar (the ones we have over the base)(whatever they call them)

45. across

The numerator? Do you mean something like$\frac{3}{5}+\frac{3}{6}?$

46. Inopeki

yeah

47. Inopeki

Oh wait, never mind i think i see why that wont work

48. across

It doesn't matter: they have different bases and you can't add them.

49. Inopeki

(2/3)x=(5/4)-(5/2) 5*2-4*5/2*5?

50. across

I saw something I liked: you know how to manipulate fractions. Could you manipulate$\frac{5}{4}-\frac{5}{2}$so that they have similar bases?

51. Inopeki

Im not sure i know what you mean by manipulate

52. Inopeki

You mean

53. Inopeki

make them higher so they match or something?

54. across

With "manipulate" I mean to say "play with" or "rearrange," like this:$\frac{5}{2}=\frac{10}{4}=\frac{15}{6}=etc.$

55. Inopeki

Oh you did mean like that

56. Inopeki

(5/2) (5/4)-------->2.5/2?

57. Inopeki

Aw man.. im really worthless at this..

58. Inopeki

At this rate ill never learn quantum physics..

59. across

That's good! But try and make the bases of the two fractions match and try to have only whole numbers in both the numerator and denominator.

60. Inopeki

Well how? i cant go upward.

61. Inopeki

(5/2)---->(10/4)!!!! (5/4)

62. Inopeki

Now i can have (15)/(4)

63. across

I'll give you an example. I'm given$\frac{7}{3}+\frac{4}{5}.$I want their denominators match, so I multiply $$7/3$$ by $$5/5$$ to get $$35/15$$, and I multiply $$4/5$$ by $$3/3$$ to get $$12/15$$. They now have similar denominators and I can add them:$\frac{35}{15}+\frac{12}{15}=\frac{47}{15}.$

64. Inopeki

So i wasnt right?

65. across

Very good! You converted$\frac{5}{4}-\frac{5}{2}$into$\frac{5}{4}-\frac{10}{4}.$What's the next step?

66. Inopeki

(2/3)x=(5/4)-(10/4) (2/3)x=(10-5)/4?

67. across

That's very close! But you switched the signs:$\frac{5}{4}-\frac{10}{4}=\frac{5-10}{4}.$

68. Inopeki

Oh right! so its (2/3)x=(-5)/4

69. across

That's correct! Now, let's solve for x. :)

70. Inopeki

Can we multiply both sides by 3 now?

71. across

Yes we can. :)

72. Inopeki

(3/3)x=(-5)/4*3

73. Inopeki

x=(-5)/4*3

74. Inopeki

Is it right? :D

75. across

Well, let's look at the LHS only: when you multiplied$\frac{2}{3}x$by $$3$$, you somehow got$\frac{3}{3}x.$How did this happen?

76. Inopeki

Wait, should i divide both sides by 2/3?

77. Inopeki

x=(-5)/4*(2/3)

78. Inopeki

79. across

You could do that, yes, but let's do it one step at a time first so that you understand the process. When you have$\frac{2}{3}x,$to get rid of that $$3$$ in the denominator, we DO multiply the fraction by $$3$$ so that this happens:$\frac{2}{3}x\implies3\cdot\frac{2}{3}x\implies\not{3}\cdot\frac{2}{\not{3}}x\implies2x.$

80. Inopeki

Oh right, i knew it had something to do with multiplying with the denominator!

81. across

Now, let's do a cool little trick: what happens if I multiply$\frac{2}{3}x$by$\frac{3}{2}?$:)

82. Inopeki

5/5?

83. across

Close!$\frac{3}{2}\cdot\frac{2}{3}x=\frac{6}{6}x=x.$

84. Inopeki

so essentially, i was right?

85. across

Now, can you apply this to solve$\frac{2}{3}x=-\frac{5}{4}?$

86. across

Yes, you were right, but I don't think you understood the whole process, which now you do. :)

87. Inopeki

Both sides multiplied by 3/2?

88. across

That's correct.

89. Inopeki

x=11/4

90. Inopeki

If thats not right, im gonna go give up on physics and go cry in a corner..

91. across

I think you got it already, but you're trying to speed your way through it. Take your time! :) Let's check again:$-\frac{5}{4}\cdot\frac{3}{2}=?$

92. Inopeki

-5/4*6/4 so 1/4?

93. across

I see what you're trying to do, but remember that multiplication and division of fractions is totally different from addition and subtraction. :) They're easier, in fact!$\frac{a}{b}\cdot\frac{c}{d}=\frac{ac}{bd}.$

94. Inopeki

-30/4?

95. across

Look at our final expression:$-\frac{5}{4}\cdot\frac{3}{2}=-\frac{5\cdot3}{4\cdot2}=?$

96. Inopeki

Wait, do you know what this means?

97. across

What exactly?

98. Inopeki

That im a complete and utter failure and that i will never become a physicist.

99. across

Hey, it's your first time doing this. Don't beat yourself too hard. We're taking it one step at a time, and that's the best way there is to go!

100. across

Also, when I had your age, I knew much, much less than you do, and now look at me; I'm a math grad student! Your enthusiasm stretches beyond any of that I've seen in most students your age. Please keep at it; you will go really far. Take advantage of the fact that there are people out there that will gladly help you, and you can for sure include me in that group of people.

101. wasiqss

susanne its really sweet of u to help sumone like this with this devotion

102. wasiqss

n yehh that kid is damn focused on the job for sure!

103. LagrangeSon678

and you will never become a physicist

104. Inopeki

Ok, im done sulking. Wanna do some math?

105. Inopeki

Fumble you lagrangeson.

106. LagrangeSon678

those word came out of your mouth, fumble yourself

107. Inopeki

You suck. At least across tries to help me.

108. LagrangeSon678

your saying i havent helped you?

109. Inopeki

You just said i wasnt gonna become a physicist! Mark my words, I WILL!

110. LagrangeSon678

prove it then you little squirt

111. Inopeki

Screw you, ill get 2 phds

112. LagrangeSon678

we'll see ...

113. Inopeki

Damn right well see!

114. Inopeki

across, 15/6?