Basic Algebra problem giving me problems

I have the basic algebra problem: $3 \dfrac{1}{4} × x - 2 = 14$

I've gotten this far:

$3 \dfrac{1}{4} × x - 2 = 14$

$3 \dfrac{1}{4} × x = 16$

$3 \dfrac{1}{4} × 4 = 13$

$3 \dfrac{1}{4} × 4 \dfrac{37}{40} = 16.00625$

The problem is, I have the answer, which is 64/13, but I have absolutely no idea how to come about that number. Even odder is 64 / 13 = 4 23076448177/24999485525, so you could say I'm "stuck". Can somebody please explain to me how to get to the answer? Sorry if this is overly-basic, usually I can somehow figure this out, but it's giving me trouble.

EDIT: The reasoning for $3 \dfrac{1}{4} × 4 = 13$ is because I picked a random number for $x$, and if $x = 5$, the output is $16.25$, which is too large, and if I pick $4$ than the output is too small, so I know the answer lies somewhere between 4 and 5.

EDIT 2: Thank you all for your answers, all very helpful. I feel thoroughly stupid for not using the fundamental properties of algebra. My mistake. Again thanks for all your answers!

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What did you do between $3\frac{1}{4} \cdot x = 16$ and $3\frac{1}{4} \cdot 4 = 13$? – JiK Mar 24 '14 at 21:43
@JiK I picked a random number for $x$ and continued to do that until I got $13$. – Jacedc Mar 24 '14 at 21:46
That's not the correct way to solve an equation @Jacedc, the point is to find $x$ through algebra, not trial and error. Check my answer below for the algebraic method. – Disousa Mar 24 '14 at 21:48

Once you have $3\tfrac{1}{4}\cdot x=16$, multiplying both sides by $4$ removes all the fractions and leaves an equation with only integers. Then dividing both sides by the appropriate integer yields the result.

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Ahh, okay. That's right. I have no idea why I tried solving it through trial and error, my mistake. Thanks for your help! – Jacedc Mar 24 '14 at 21:51

$$3 \dfrac{1}{4} × x - 2 = 14$$

Remember in algebra - what you do to one side, you must to do the other!

First, let's make the fraction into a more usable fraction. What's $3 \dfrac{1}{4}$? Well, we know that to have $1$ we would need $4 \times \frac{1}{4}$. Hence, to have 3, you would need 3 times that! That's $\frac{12}{4}$ for 3. Keep in mind that we have 3 AND $\frac{1}{4}$, meaning that you'll have to add the $\frac{12}{4}$ to $\frac{1}{4}$. So the fraction ends up being: $\frac{13}{4}$.

$$\frac{13}{4} × x - 2 = 14$$

Now let's start using the fundamental rule of algebra that I mentioned earlier. Whatever you do to one side, you have to do to the other.

We first see the $-2$ next to the $x$ term. We don't want that. How do we eliminate? Add 2 to that side to make it zero:

$$\frac{13}{4} × x - 2 \color{red}{+2} = \frac{13}{4} × x \color{red}{+0}=\frac{13}{4} × x$$

Yay! Got rid of it! BUT WAIT! That's just the left side. What about the right side?

Must apply the same deal. Add 2 as well:

$$14 \color{red}{+2} =16$$

Therefore, we now have:

$$\frac{13}{4} × x=16$$

We want $x$ to be alone so we can get it's value. What else can we do to cancel? Hmm... OH! I GOT IT! Multiply by the RECIPROCAL.

The reciprocal of $\frac{13}{4}$ is, very simply, $\frac{4}{13}$. A reciprocal is basically taking a fraction and flipping it. The reciprocal times the original number will equal 1. This is because the denominator will cancel the other numerator, and the numerator will cancel the other denominator.

Thus, we'll have to multiply the left side by the reciprocal:

$$\frac{13}{4} \color{red}{\times \frac{4}{13}} \times x=\color{red}{1}x=x$$

YES! We've got $x$ alone. Woohoo! Now just apply it to the right side:

$$16 \color{red}{\times \frac{4}{13}}=\frac{64}{13}$$

Combining it all~

$$x=\frac{64}{13}$$

YAY! DONE.

Cheers! -Shahar

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We start with the initial equation: $$(3+\frac 14)x-2=14$$

We simplify: $$\frac {13}4x-2 = 14$$

We add 2 to both sides: $$\frac {13}4x = 16$$

Then we multiply both sides by 4: $$13x = 64$$

And then divide both sides by 13: $$x = \frac{64}{13}$$

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I think OP uses the notation $3\tfrac{1}{4}$ to mean $3+\tfrac{1}{4}$, or $\tfrac{13}{4}$. – Servaes Mar 24 '14 at 21:47
I did, @Servaes is correct. – Jacedc Mar 24 '14 at 21:50
EDIT: Corrected the answer to reflect this. – Disousa Mar 24 '14 at 21:58
It should be $(3+\frac{1}{4})\times x$, not $3+\frac{1}{4}\times x$. – Eff Mar 24 '14 at 22:01
You are correct. Editing again xD – Disousa Mar 24 '14 at 22:02

$$3 \dfrac{1}{4} × x - 2 = 14$$ $$\frac{13}{4} × x - 2 = 14$$ $$4\cdot\frac{13}{4} × x - 4\cdot2 =4\cdot 14$$ $$13 x - 8 =56$$ $$13x=56+8$$ $$13x=64$$ $$x=\frac{64}{13}$$ because $$3 \dfrac{1}{4}=3+\frac{1}{4}=\frac{13}{4}$$

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Correction, $14*4\neq64$ :) – Disousa Mar 24 '14 at 21:57
You did $4⋅\dfrac{13}{4}x - 4⋅2 = 4⋅14$ thus multiplying both sides by 4 (which is correct) but you did it twice on the left side? ($4⋅2$) – Jacedc Mar 24 '14 at 22:00
it is my mistake – Adi Dani Mar 24 '14 at 22:00