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Have you learned the keep, change, flip method?
Never heard of it!
Ok, I'll teach you. First, we keep the top fraction as it is.
Next, we change the division sign to a multiplication sign. So this will look like "fraction 1" x "fraction 2"
step 3 is to Flip the bottom fraction. In this case you have bottom fraction 1+ bottom fraction 2, so you would flip both of them.
so 10/x+1 x 2/1 x+1/3?
close. \[(10/x+1)(2/1+ x+1/3)\]
you aren't multiplying the last two fractions; maybe you just missed the + sign in between?
Oh I get it, so what do you do next? Now that I have (10/x+1)(2/1 + x+1/3)?
now we need to get a common denominator for the second term. so in other words, how can we change the expression \[(2/1 + (x+1)/3) \] to have a common denominator?
Hmm, you have to find the GCF for the denominators right?
Well 2/1 has a denominator of 1, so 2/1=2. (x+1)/3 has a denominator of 3. So if we want 2/1 to have a denominator of 3, what we do to the top, we must do to the bottom, so what would the top become?
6 right? Because you have to multiply 3 to the bottom and top?
Ok, so now we have (10/x+1)(6/3+(x+1)/3). Now we can combine the numerators in the right hand term since they both have a denominator of 3. What do you get?
3+x/3? and you would just cancel out the 3 and be left with (10/x+1)(x)?
not exactly. Let's look at the second term only. we have (6/3 + (x+1)/3). So if 6 and x+1 are our numerators, we can add them together because they have the same denominator. What do you get?
Woops, I looked at my work wrong. You would just have 7+x/3
Actually. I have a much easier way. Don't hate me, but it's simpler to do it the second way in this case.
ok, so starting from the beginning, we have two different denominators. one of them is 2 and one of them is x+1. go through each fraction individually and make it so that the denominator is 2(x+1)...don't multiply anything out yet, just leave it as multiplication and let me know what you get. remember that what you do to the top, you must do to the bottom.
Would it be |dw:1377730424737:dw|
are you starting from the beginning?
Yeah, I multiplied the 1 and the 2 in 1/2 by (x+1) and I multiplied the 3/x+! by 2
Yeah that is what I did
Ok, so now you see from the attached example that all of the smaller fractions have the same denominator. When all fractions have a common denominator, we can eliminate the denominators from the problem and work with the numerators. So if we cross out all of those denominators, what are we left with?
sorry for the confusion on the first part!
It's alright! It was worth it in the end, I actually kinda understand now. Thanks!
okay good! let me know if you have any questions!!