Equations

[AWOL]

How do I do them? I can do all that backwards figuring out stuff, hell, that's easy. What I don't get is stuff like this:

A certain number was divided by three, then quintupled to give 3 more than the number.

That confuses me. Halp please.

[Puchiko]

Mortalitas Wrote:How do I do them? I can do all that backwards figuring out stuff, hell, that's easy. What I don't get is stuff like this:

A certain number was divided by three, then quintupled to give 3 more than the number.

That confuses me. Halp please.

Let's name "the number" x.
A certain number was divided by three, then quintupled
That's one part of the equation.
(x/3).5

to give 3 more than the number
That's x+3

Now, because we have "to give", these should be equal.
(x/3).5=x+3
Can you figure this equation out?

Solution (will provide further explanation if needed)

Hidden stuff:

(x/3).5=x+3 We don't want brackets, they're complicated. Let's get rid of them

5x/3=x+3 We want to have only numbers on one side, and only unknowns on the other. Let's subtract x from both sides

4x/3=3 Let's multiply by 3

4x=9 Cool. But I want to know what one x is. Let's divide by 4

x=0.44 (rounded)

I can only hope I didn't make a mistake in these calculations.

[magikarp]

Hidden stuff:

5x/3 - x = 3
5x/3 - 3x/3 = 3 You change the x to a fraction with a denominator of 3 first.
2x/3 = 3 Then you subtract 3x from 5x and leave the bottom part the same.

I'm pretty sure the rest is right.

[AWOL]

*attempts equation* *obtains brain damage*

Still very confused...

[Milk2Go]

I don't get how anyone could not understand this. This is extremely basic math.

(y/3) x 5 = y+3
(y/3) = (y+3)/5
y = ((y+3)/5) x 3
y = (3(y+3)/5)
5y = 3(y+3)
5y = 3y + 9
2y = 9
y = 9/2
y=4.5

The number is 4.5. It works; plug it into the equation.
Puchiko, your answer is off by .1.