Given n = (L - a)/d + 1 solve for L=? Critical Algebra Skills!
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- Опубликовано: 8 июл 2024
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Subtracting 1 from both sides gives n-1=L-a/multiply both sides by d creates d(n-1)=L-a subtracting a from both sides gives the answer L=a+d(n-1)
(n-1)=( (L-a)/d )
d(n-1)=(L-a)
a + d(n-1) = L
If I was a student and viewed the video solution, I would be very confused. There's too much extraneous manipulation going on. Just subtract 1 from both sides, multiply both sides by d and maybe factor out d. KISS
Why John would choose this method and NOT the following is beyond me, but here we go:
Step one is we need to isolate the variable so the equation becomes L = something.
a:) n - 1 = [(L-a) / d] + 1 -1. ( In this sub step we subtract 1 from both sides)
b:) d(n - 1) = d[(L - a) / d]. ( In this sub step we multiply both sides by d)
c:) d(n - 1) + a = L - a + a. ( In this sub step we add a to both sides )
Finally we get L = d(n - 1) + a. OR you can also have L = dn - d + a. Both are valid answers.
Bonus Round lets prove it. Let a = 1, d =3 and n = 71. Solve for L by plugging in known values.
L = 3(71 - 1) + 1. (First we start inside the parenthesis)
L = 3(70) + 1. (Next we do multiplication)
L = 210 + 1 or L = 211.
L = 3(71) -3 + 1. (No work inside parenthesis so we start with multiplication)
L = 213 - 3 + 1 Or -3 + 213 + 1. (Here we just do addition)
L = 210 + 1 so L = 211.
Both formula have been proven as valid.
Why John made this so much more complicated is beyond me. I would expect any of my passing students to be able to understand my above solution before the end of the school year.
(I teach "entry level" math, but I did intern as a remedial math teacher while earning my teacher's license and I also tutor older students that have decided to go back and finish school. All of these individuals are considered my students)
Three simple steps...
Subtract 1 from both sides, giving
n - 1 = (L - a) / d
Then multiply both sides by d, giving
d(n - 1) = L - a
Finally, add a to both sides, giving
a + d(n - 1) = L
He definitely went the long way round
Greetings.
L=d(n-1)+a.
If n=(L-a)/d +1,
dn=(L-a)+d, and
dn-d=L-a, d(n-1)=L-a,
d(n-1)+a=L. The answer is
L=d(n-1)+a. I think this expression reminds me of arithmetic series, number of terms, first term, last term, and common difference.
All you need is to make L the subject of the formula and you create the fraction both sides by dividing by one then you cross the multiplication to get rid of the fraction to form linear equation and finally make L the subject of the formula or in other ways you express L in terms of a,d and n. L= -a+b/dn
Seems like John just complicated it.
There ARE people who need to see a slow step by step process. A good math person does things in their head but he designs these videos for EVERY level of student.
There are at least two ways to solve that problem. Both are equally valid. Most of the comments use the other way.
It took me about 15 seconds to do this in my head but I realize that you are presenting this as a class and you have to justify each step in writing to teach the methodology. Thank You!
You said it better than i just wrote above.
A simpler way to solve is to begin by multiplying both sides by d:
dn = L - a + b
Then add a - b to both sides
dn +a - b = L
Swap around
L = dn - d + a
L = d(n-1) + a
This is what I attempted to do. I landed up with L=nd+a-d. Would this solution be acceptable?
@@InPursuitOfCuriosityIt is correct but incomplete. You should collect the two terms with d if possible.
L = nd- d + a
L = d( n- 1 ) + a
@@richardhole8429 Thanks for explaining.
Need to isolate L. So, start by subtracting 1 from both sides : n-1 = (L-a)/d
Next, multiply both sides by «d» : d(n-1)= d(L-a)/d → d(n-1) = L-a
Add «a» to both sides : a+d(n-1) = L. Place L on the left giving : L = a+d(n-1)
I'm going to have to watch this a few times, Mr RUclips Math Man...🤔 😅
Thank you
L = a + (n-1)d
The general antithetic sequence formula. ☺️
Do you have 7th grade curriculum? We are in the market for a good curriculum and would love to use yours.
Thanks
got it but w/o the last 'D' factor thanks for the fun
(L-a) / d = n-1 so L-a = (n-1).d and L = a + d.(n-1) What was this all about???
L = nd - d + a
Or
L = d(n - 1) + a
Why would you add the two terms on rhs side first? You wasted a lot of time explaining about adding fractions. I just multiplied by d all fractions then gone.
Same on a previous video. He goes around, and around, while simplicity wins.
What you call wasting is really educating
@@lamper2 well of course that is true but about a subject adding fractions that is unnecessary to this question
@@lamper2It is not helpful for an educator to make something seem way more complicated than it really is.
At the very least, they must be clear with their students that that is what they're doing. It needs to come with a "you don't need to do anything this complicated to solve this problem, but this problem provides a good way to illustrate a technique that you will find useful elsewhere" caveat.
However, that is still inferior to illustrating the manipulation of fractions with a scenario that actually requires the manipulation of fractions.
its about teaching on how to solve fractions, not solving with the most efficient method
D(n-1)+a
L = a + d(n - 1)
Pretty easy
I was close but failed teach I tried
Thank you, I’ll take the rest of time off from your videos, I did it in my head in about 10s.
(n-1)d+a=L
You repeat yourself over and over again. Why?
So hard to stay engaged.
I can see some students just losing interest and giving up
L= a+d(n-1)
L= a+d(n-1)