PDA

View Full Version : Need math help

Valkyrie144
9th April 2011, 03:17 PM
I have done this problem over and over and keep getting the same wrong answer. Any help would be much appreciated.

5-1/3x= 1/12x

The first step I find the common denominator to be 12. So, I take 12x5=60. I then take 12x1/3, which would be 4x.

On the right side, my 12 would cancel out and just give me x

Now, I have 60-4x=x

I then add 4x to both sides and get 60=4x. Now, my final answer is x=15. But, my math book says my answer should be 12.

I have a couple more math problems if I can get this one figured out to ask. Thank you

Jorghnassen
9th April 2011, 03:18 PM
x+4x = 5x, thus 60 = 5x => x=12.

9th April 2011, 03:28 PM
I have done this problem over and over and keep getting the same wrong answer. Any help would be much appreciated.

5-1/3x= 1/12x

The first step I find the common denominator to be 12. So, I take 12x5=60. I then take 12x1/3, which would be 4x.

On the right side, my 12 would cancel out and just give me x

Now, I have 60-4x=x

I then add 4x to both sides and get 60=4x. Now, my final answer is x=15. But, my math book says my answer should be 12.

I have a couple more math problems if I can get this one figured out to ask. Thank you

When you added 4x to both sides you get 60 -4x + 4x = x + 4x
You got 60 = 4x
but it should have been 60 = 5x where 60 + x*(-4 +4) = x*(4+1)
(-4 +4) = 0 & (4+1) = 5

As a general rule, expand then simplify and there always exists that 1 multiplied by everything.

Valkyrie144
9th April 2011, 04:37 PM
oh my gosh. These little mistakes are killing me.

Vorpal
9th April 2011, 04:42 PM
I hate that notation in ASCII. Does 1/3x mean (1/3)x or 1/(3x)? Based on the answer, I guess the former, but this would not be at all obvious if the answer was not provided.

Valkyrie144
9th April 2011, 04:44 PM
Here is another if you guys don't mind.

1/5 (x + 6) = 1/7 (x + 8)

my first step is to use distributive property and multiply 1/5 x x and 1/5 x 6. I would get x/5 + 6/5 = x/7 + 8/7

Now I have to find the common denominator which is 35. So, I multiply x/5 by 35 and get 7x. Do I do the same for 6/5? Would I take 6/5 and multiply by 35 to get 42?

Thank you:)

Vorpal
9th April 2011, 04:47 PM
Now I have to find the common denominator which is 35. So, I multiply x/5 by 35 and get 7x. Do I do the same for 6/5? Would I take 6/5 and multiply by 35 to get 42?
Yes, you should.

But an easier way would be to multiply through by 35 as the first step, to get 7(x+6) = 5(x+8). Then distributive property works without any fractions.

rjh01
9th April 2011, 08:16 PM
The answer to the first question is here http://www.wolframalpha.com/input/?i=5-1%2F3x%3D+1%2F12x

The answer to the second question is here http://www.wolframalpha.com/input/?i=1%2F5+%28x+%2B+6%29+%3D+1%2F7+%28x+%2B+8%29
x=-1

The home page is this http://www.wolframalpha.com/ Just plug in the maths question and the answer will appear as above.

GodMark2
9th April 2011, 08:57 PM
double

GodMark2
9th April 2011, 09:01 PM
I hate that notation in ASCII. Does 1/3x mean (1/3)x or 1/(3x)? Based on the answer, I guess the former, but this would not be at all obvious if the answer was not provided.
Everyone should learn
\LaTeX \begin {align} 5 - \frac {1} {3} x &= \frac{1}{12}x \\ 60-\frac{12}{3}x &= \frac{12}{12}x\\ 60-4x &= x \\ 60-4x+4x &= x+4x \\ 60 &= 5x \\ 12 &= x \\ \end {align}

69dodge
9th April 2011, 10:38 PM
The answer to the first question is here http://www.wolframalpha.com/input/?i=5-1%2F3x%3D+1%2F12x

Oops. See Vorpal's post #5.

The Norseman
9th April 2011, 10:40 PM
Everyone should learn
\LaTeX \begin {align} 5 - \frac {1} {3} x &= \frac{1}{12}x \\ 60-\frac{12}{3}x &= \frac{12}{12}x\\ 60-4x &= x \\ 60-4x+4x &= x+4x \\ 60 &= 5x \\ 12 &= x \\ \end {align}

So where were you in my Latex question thread? :D

Beerina
10th April 2011, 12:18 AM
So where were you in my Latex question thread? :D

He...thought it was a different Latex.

rjh01
10th April 2011, 01:08 AM
Oops. See Vorpal's post #5.
Fine. Then the answer is http://www.wolframalpha.com/input/?i=5-%281%2F3%29*x%3D+%281%2F12%29*x

Edit. Latex on the forum. All you need to know http://forums.randi.org/tags.php?tag=Latex

Valkyrie144
10th April 2011, 03:44 AM
wolfram is really nice, but I can't see where it shows the steps.

here is another I'm having problems with.

.6x + 5.9 = 3.8

First, I take 5.9 from both sides to get .6x = -2.1
Next, I take -2.1 and divide it by .6x

So, my answer should be x= -2.1/6

Dancing David
10th April 2011, 05:01 AM
.6x + 5.9 = 3.8

(.6x + 5.9)-5.9 = (3.8)-5.9

.6x = -2.1

x= (-2.1)/6

Yup :)

rjh01
10th April 2011, 05:09 AM
wolfram is really nice, but I can't see where it shows the steps.

here is another I'm having problems with.

.6x + 5.9 = 3.8

First, I take 5.9 from both sides to get .6x = -2.1
Next, I take -2.1 and divide it by .6x

So, my answer should be x= -2.1/6

At the very least it will tell you if you have the right answer. For example there would be no need to ask the above question. It would tell you that you have the right answer.

10th April 2011, 07:28 AM
.6x + 5.9 = 3.8

(.6x + 5.9)-5.9 = (3.8)-5.9

.6x = -2.1

x= (-2.1)/6

Yup :)

Nope.

x = (-2.1)/.6 = -3.5

Don't lose the decimal point. Dividing by .6 is not the same as dividing by 6.

Respectfully,

69dodge
10th April 2011, 09:20 AM
.6x + 5.9 = 3.8

You could multiply everything by 10 first, to get rid of all the decimal points. Then, you don't have to worry about them anymore. That is, as your first step, turn the equation into 6x + 59 = 38.