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MathEd-PII

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volunteer
Okay great to hear! So now we have two different denominators: x+3 and x-3
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volunteer
to get a common denominator we'd have to multiply them together to get (x+3)(x-3)
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volunteer
So doing one fraction at a time for the (x-1) / x+3 what would we have to multiply by to get that common denominator?
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student
oh jeez I'm sorry, I completly spaced that you asked me a question I'm so sorry-
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volunteer
It's all good!
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student
the only thing I can think of is 1x3
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volunteer
Really close...I like to look at it by what is missing from each fraction if that helps. For example, in the first one, you have x+3 in the denominator so you'd be missing the x-3 and have to multiply the numerator and denominator by that if that makes sense.
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volunteer
So for the next one, which term are you missing and what would you then multiply by?
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student
would it still be x-3?..
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volunteer
https://example.comis time you already have an x-3
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[ { "pii_type": "URL", "surrogate": "https://example.com", "start": 0, "end": 19 } ]
student
oh so x+3
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student
?
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volunteer
Yeah you got it!
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volunteer
Then for the last one, we already have the x+3 so it would be....?
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student
x-3
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volunteer
Yes!!
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volunteer
Now what ideas do you have for possible next steps?
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student
uh simplify?
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volunteer
Yeah! :)
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volunteer
And how would you go about doing that?
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student
https://example.com not sure actually. normally, i'd just cancel out the number that's the same or opposite of it but there's both of those..
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[ { "pii_type": "URL", "surrogate": "https://example.com", "start": 0, "end": 19 } ]
volunteer
Okay so try thinking about it like this.... if you had something simple like 1/5 + 2/5 how would you simplify that?
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volunteer
If the denominators are the same we can just add right?
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student
yeah
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volunteer
so then what happens is if you have one fraction equal to another fraction with the same denominator then the numerators must be equal right?
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volunteer
like if I had something simple like x/5 = y/5 then the x and y would be equal right?
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volunteer
So I know there's a lot more going on here, but we could apply the same idea
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student
I'm so sorry, i'm still a little lost...:(
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volunteer
Okay I'll try breaking it down a bit simpler, where did you first start feeling a little lost and I'll go back to there
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student
probably the message after my last response
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volunteer
Okay cool
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volunteer
okay let's start here, if I have x/5 and y/5 in what I just drew out we can agree that x and y will be equal right?
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student
yeah
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volunteer
Great, so now if I take it a step further and add more stuff to the numerator the two expressions would still be equal right?
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volunteer
like the x+1 and the y-3
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student
?
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volunteer
So they have the same denominator right? So if those expressions are equal what also must be equal?
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student
the numerator?
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volunteer
Exactly!
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volunteer
So in your original question, if all of your terms have the same denominator and you simplified the one on the left by adding it....what must be the same?
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student
the numerator...would it be x-3
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volunteer
Yep it would be the numerator! So you'd have the (x-1)(x-3) + (x-2)(x+3) = -(1-2x)(x-3) expression
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volunteer
does it make sense how we got there?
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student
uh partially...
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volunteer
what can I clarify that doesn't quite make sense?
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student
Sorry, I'm just trying to think of how x-3 would be the answer still
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volunteer
like the denominator
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volunteer
?
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student
for the numerator part
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volunteer
Like why it is still included in the numerator? I'm not sure what you are asking sorry
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student
wait wait wait is this the numerator answer?
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volunteer
Yes that's right!
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volunteer
since the denominators are the same we ONLY have to solve for the terms in the numerator! :)
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student
so does it just stay like that or does it factor out into something from there?
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volunteer
So you would have to factor out the (x-1)(x-3) + (x-2)(x+3) = -(1-2x)(x-3)
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volunteer
and then solve for x by simplifying both sides
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volunteer
Like the top part I just circled in red is equal to the part you circled
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volunteer
Since I believe this problem is looking for the value of x not just an expression for it
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student
oh ok
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student
how would we factor it out
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volunteer
You can use the distributive property on each set of terms and then add the like ones if that makes sense?
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student
huh?..
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volunteer
Have you talked about the distributive property? I can do a quick example if that would help?
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student
I have but it didn't make a ton of sense
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volunteer
Okay I'll do a quick example down at the bottom
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volunteer
Did the example I just did make sennse?
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volunteer
I tried to use different colors, but I can type out an explanation if that helps too
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student
yeah it makes sense
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volunteer
Yayy! I'm glad
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volunteer
Do you want to try the distributive property on your problem and then I can check your work?
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student
yeah. is it ok if I do it on my paper irl and then say my answer before I type out what I did if something is wrong
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volunteer
Yeah that's totally fine!
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volunteer
You can also upload a picture of your work if you want me to check it or if you feel confident and don't need me to check it that's fine too
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volunteer
I'm here to help you so whatever is best for you!
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student
I'll probably upload it
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volunteer
What you have there looks right! Now you just need to do the other side of the equation:(x-1)(x-3) + (x-2)(x+3)
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student
wait what??
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volunteer
2x^2 -7x + 3 = (x-1)(x-3) + (x-2)(x+3)
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volunteer
I didn't see the simplification of the other side, but maybe I missed it?
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student
I didn't know there was another thing I needing to do-
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volunteer
Oh okay I thought we clarified that earlier, my bad
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volunteer
Because if you have the 2x^2 -7x +3 it has to be equal to something to solve for x right?
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student
Oh!
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student
wait how do I start that...I wrote down the equation but don't know where to start
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volunteer
It would be the same as before, you start with distributive property for the terms on the other side. Or did you already do that?
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student
uh nope I just have this written down- 2x^2 -7x + 3 = (x-1)(x-3) + (x-2)(x+3)
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volunteer
Okay so follow the same process you did before for the (x-1)(x-3) + (x-2)(x+3) stuff
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volunteer
where you use the distributive property and then simplify it out
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student
ok I'll do that on paper
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volunteer
Sounds good!
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student
should i simplify my answers down so they're in a shorter solution?
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volunteer
What do you mean by shorter solution?
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volunteer
Like when you are doing the distributive property?
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volunteer
Okay I see it!
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student
The bottom circled ones
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volunteer
Now you have to move all the terms to one side of the equation and then solve for x using the quadratic formula!
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volunteer
We are almost there!
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student
uh how do i do that
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volunteer
How to move the terms to one side or the quadratic formula part?
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student
both
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