(Will give medal to the person who helps)
Find the ratio of x to y.
~ x/5 = 2/3 = 5/y ~
1.) 2/3
2.) 4/9
3.) 1

- anonymous

- jamiebookeater

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- mathstudent55

Use x/5 = 2/3 to solve for x.
It's a proportion, so cross multiply.

- anonymous

Alright I tried it , but I came out with decimal. Is that ok or ?

- mathstudent55

|dw:1439168515677:dw|

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## More answers

- mathstudent55

It's better to leave it as a fraction like I did above.

- anonymous

wait nvm I looked at it wrong

- mathstudent55

Do you see what I did to solve for x?
I used the first two fractions as a proportion, and solved the proportion for x.
Are you ok so far?

- anonymous

yea I am ~

- mathstudent55

Great.
Now take the second and third fractions, and use them as a proportion and solve for y.

- mathstudent55

Now cross multiply and solve the proportion below for y.
|dw:1439168686577:dw|

- anonymous

2y = 15

- anonymous

right ??

- mathstudent55

good so far
Now solve for y

- mathstudent55

Leave the result as a fraction.

- anonymous

\[15 \div 2 ?? \]

- mathstudent55

|dw:1439168922451:dw|

- mathstudent55

Yes, 15 divided by 2 is simply the fraction 15/2.

- mathstudent55

Now we want the ratio of x to y.
The ratio of x to y can be written as x:y, and also as the fraction
\(\dfrac{x}{y} \)

- anonymous

do we cross multiply them now ? the x y ?

- mathstudent55

Since we know what x and y are equal to, we write a ratio using those two numbers:
|dw:1439169044897:dw|

- mathstudent55

|dw:1439169083196:dw|

- anonymous

thats none of my choices ~

- mathstudent55

We haven't finished yet. We now have a complex fraction, a fraction of fractions.
We need to simplify it.
Remember that a fraction means a division.
|dw:1439169173036:dw|

- mathstudent55

To divide fractions, we multiply the first fraction by the reciprocal of the second one.
Reciprocal is the math term that means to flip the fraction.

- mathstudent55

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- mathstudent55

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- anonymous

ohhh I see ! ~ :)

- mathstudent55

Great!

- anonymous

i understand now ! thanks !

- mathstudent55

You're welcome.

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