## Inopeki Group Title Can someone give me a only letter equation to solve? 2 years ago 2 years ago

1. across Group Title

2x+3=11

2. Inopeki Group Title

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

3. Inopeki Group Title

Btw, grats on Master rank

4. Inopeki Group Title

Can i have a harder one now?

5. jimmyrep Group Title

good job

6. FoolForMath Group Title

that's called only a letter equation? :D

7. Inopeki Group Title

Nope but i take whats coming lol

8. across Group Title

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

9. Inopeki Group Title

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

10. across Group Title

That's right thus far!

11. Inopeki Group Title

x+6=9 x=3

12. Inopeki Group Title

?

13. FoolForMath Group Title

That's correct.

14. across Group Title

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

15. Inopeki Group Title

Please put that in equation form

16. across Group Title

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

17. across Group Title

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

18. Inopeki Group Title

Hmm

19. Inopeki Group Title

Im supposed to "equalize the fractions" right?

20. Inopeki Group Title

Wait!

21. across Group Title

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

22. Inopeki Group Title

NONONONO

23. Inopeki Group Title

no clues :(

24. Inopeki Group Title

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

25. Inopeki Group Title

No that cant be it..

26. Inopeki Group Title

wait, just say yes or no, not why.

27. across Group Title

You rewrote the equation. :P

28. Inopeki Group Title

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

29. across Group Title

Yes

30. Inopeki Group Title

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

31. Inopeki Group Title

If so i have the answer

32. across Group Title

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

33. Inopeki Group Title

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

34. across Group Title

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

35. Inopeki Group Title

5-5 ---- 4-2?

36. across Group Title

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 Group Title

:( i thought i knew this stuff...

38. across Group Title

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 Group Title

Whats the logic in that?

40. across Group Title

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 Group Title

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

42. across Group Title

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 Group Title

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 Group Title

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

45. across Group Title

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

46. Inopeki Group Title

yeah

47. Inopeki Group Title

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

48. across Group Title

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

49. Inopeki Group Title

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

50. across Group Title

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 Group Title

Im not sure i know what you mean by manipulate

52. Inopeki Group Title

You mean

53. Inopeki Group Title

make them higher so they match or something?

54. across Group Title

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

55. Inopeki Group Title

Oh you did mean like that

56. Inopeki Group Title

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

57. Inopeki Group Title

Aw man.. im really worthless at this..

58. Inopeki Group Title

At this rate ill never learn quantum physics..

59. across Group Title

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 Group Title

Well how? i cant go upward.

61. Inopeki Group Title

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

62. Inopeki Group Title

Now i can have (15)/(4)

63. across Group Title

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 Group Title

So i wasnt right?

65. across Group Title

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

66. Inopeki Group Title

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

67. across Group Title

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

68. Inopeki Group Title

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

69. across Group Title

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

70. Inopeki Group Title

Can we multiply both sides by 3 now?

71. across Group Title

Yes we can. :)

72. Inopeki Group Title

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

73. Inopeki Group Title

x=(-5)/4*3

74. Inopeki Group Title

Is it right? :D

75. across Group Title

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 Group Title

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

77. Inopeki Group Title

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

78. Inopeki Group Title

79. across Group Title

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 Group Title

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

81. across Group Title

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

82. Inopeki Group Title

5/5?

83. across Group Title

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

84. Inopeki Group Title

so essentially, i was right?

85. across Group Title

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

86. across Group Title

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

87. Inopeki Group Title

Both sides multiplied by 3/2?

88. across Group Title

That's correct.

89. Inopeki Group Title

x=11/4

90. Inopeki Group Title

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

91. across Group Title

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 Group Title

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

93. across Group Title

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 Group Title

-30/4?

95. across Group Title

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

96. Inopeki Group Title

Wait, do you know what this means?

97. across Group Title

What exactly?

98. Inopeki Group Title

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

99. across Group Title

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 Group Title

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 Group Title

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

102. wasiqss Group Title

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

103. LagrangeSon678 Group Title

and you will never become a physicist

104. Inopeki Group Title

Ok, im done sulking. Wanna do some math?

105. Inopeki Group Title

Fumble you lagrangeson.

106. LagrangeSon678 Group Title

those word came out of your mouth, fumble yourself

107. Inopeki Group Title

You suck. At least across tries to help me.

108. LagrangeSon678 Group Title

your saying i havent helped you?

109. Inopeki Group Title

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

110. LagrangeSon678 Group Title

prove it then you little squirt

111. Inopeki Group Title

Screw you, ill get 2 phds

112. LagrangeSon678 Group Title

we'll see ...

113. Inopeki Group Title

Damn right well see!

114. Inopeki Group Title

across, 15/6?