# Please help me figure out the steps to solving this problem?

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#4/sqrt2+2/sqrt3#

##### 1 Answer

#### Answer:

#### Explanation:

The first thing that you need to do here is to get rid of the two radical terms from the denominators.

To do that, you must **rationalize** the denominator by multiplying each radical term by itself.

So what you do is you take the first fraction and multiply it by *value* the same. This will get you

#4/sqrt(2) * sqrt(2)/sqrt(2) = (4 * sqrt(2))/(sqrt(2) * sqrt(2))#

Since you know that

#sqrt(2) * sqrt(2) = sqrt(2 * 2) = sqrt(4) = sqrt(2^2) = 2#

you can rewrite the fraction like this

#(4 * sqrt(2))/(sqrt(2) * sqrt(2)) = (4 * sqrt(2))/2 = 2sqrt(2)#

Now do the same for the second fraction, only this time, multiply it by

#2/sqrt(3) * sqrt(3)/sqrt(3) = (2 * sqrt(3))/(sqrt(3) * sqrt(3))#

Since

#sqrt(3) * sqrt(3) = sqrt(3^2) = 3#

you will have

#(2 * sqrt(3))/(sqrt(3) * sqrt(3)) = (2 * sqrt(3))/3#

This means that the original expression is now equivalent to

#4/sqrt(2) + 2/sqrt(3) = 2sqrt(2) + (2sqrt(3))/3#

Next, multiply the first term by

#2sqrt(2) * 3/3 + (2sqrt(3))/3 = (6sqrt(2))/3 + (2sqrt(3))/3#

The two fractions have the same denominator, so you can add their numerators to get

#(6sqrt(2))/3 + (2sqrt(3))/3 = (6sqrt(2) + 2sqrt(3))/3#

Finally, you can use

#(6sqrt(2) + 2sqrt(3))/3 = (2(3sqrt(2) + sqrt(3)))/3#

And there you have it

#4/sqrt(2) + 2/sqrt(3) = (2(3sqrt(2) + sqrt(3)))/3#