Help! How would you explain how to solve this equation?

SplixBeast

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Jun 24, 2020
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Find the value of z:

(-12+z/3)=-6

*By explain I'm referring to "move the numbers so all like terms are on one side" sort of answer

Thanks!
 
How about: Drop the parentheses. Add 12 to both sides. Multiply both sides by 3.
 
We don't give sort of an answer or an answer on this site. After all this is a help site.

Can you tell us where you are stuck so we know what typwe of help you need? Can you follow LCKurtz's instructions? Please post back.
 
I have always hated the "move to the other side of the equation" wording. I have known students who solve "3x= 6" as "x= -3+ 6= 3" saying "my teacher told me that when you move a number to the other side of the equation you have to change the sign"!

Instead I would explain that to "solve" -12+ 6z= -6 (as pointed out by LCKurtz the parentheses are irrelevant) we have to "undo" what is done to z by doing the opposite of each thing done to
z in the opposite order. If we were to start with "z" to get "-12+ 6z" we would
1) multiply z by 6
2) subtract 12

To "undo" that we
1) add 12 (the opposite of "subtract 12")
2) divide by 6 (the opposite of "multiply by 6")
and of course anything we do to one side we have to do to the other

So from -12+ 6z= -6
Add 12 to both sides: -12+ 6z+ 12= -6+ 12. 6z= 6
Divide both sides by 6: 6x/6= 6/6. z= 1.

Check: If z= 1, -12+ 6z= -12+ 6(1)= -12+ 6= -6.
 
… "move the numbers so all like terms are on one side" sort of answer
Hello SB. I'd start the explanation by distinguishing between constant and variable terms. There are three terms, in the given equation.

-12 + z/3 = -6

z/3 is a variable term, and a pair of like-terms are the constants -12 and -6.

In general, the first steps in solving such equations are to separate variable terms from constant terms; that is, we collect all variable terms on one side of the equation and all constant terms on the other side. In this particular equation, it's easier to leave the variable term on the left-hand side and move that constant term to the right-hand side. (Otherwise, we'd need to move two terms, instead of one.)

Next, I'd remind why we move like-terms to the same side: so we can combine them into a single term. (That's called 'simplifying'.) We move terms by adding their opposite -- we add that to each side. The opposite of -12 is 12. Therefore, we add 12 to each side. On the left-hand side, the opposites 12 and -12 combine to make zero, so there's no longer a constant term on the left-hand side. On the right-hand side, we simplify, by adding the constants -6 and 12.

z/3 = 6

Lastly, I'd remind that 'solving for z' means finding the value of z that makes the original equation a true statement. I'd explain that our result above shows one-third of the solution -- it's 6. But, we're not interested in a third of the solution; we want the whole z. In other words, we want three times as much as what we see above. Therefore, we multiply each side by 3, and we have solved for z.

z = 18

?
 
I have always hated the "move to the other side of the equation" wording. I have known students who solve "3x= 6" as "x= -3+ 6= 3" saying "my teacher told me that when you move a number to the other side of the equation you have to change the sign"!
I remember walking pass a classroom and hearing one of our adjuncts having her students recite "a negative and a negative is a positive".
 
… … having her students recite "a negative and a negative is a positive".
You stepped in straight away, to improve upon her lesson, yes? ;)

Seriously, I'd hope that her students understood the proper context, but it was a sloppy thing to teach, regardless.

\(\;\)
 
No, I did not step in. I was in the department at that time for about a week. I just felt too new to say anything at that time. Would I say anything now? No, as I have given up on most adjuncts.
 
I have given up on most adjuncts.

I work with some very capable adjuncts, who rightly win teaching awards.

I've also seen students being taught things like "keep-change-flip" (for dividing fractions) by full-time faculty (in remedial arithmetic courses, which are not taught by the math department). In that case, that may well be the best way to communicate with these students.
 
Dr Peterson, you are I think an adjunct professor. Of course you are quite competent. Students should be honored to have a professor of your ability. I know for a fact that in South Florida the adjuncts are usual very qualified as they are retired university math faculty. Most adjuncts that taught where I have worked are current high school teachers. I am sorry but most of them would fail a pre calculus class. Of course there are exceptions but most are not very good.
 
We do also have some high-school teacher adjuncts (some of whom I am definitely not as confident of), in addition to the retired professors, retired bankers or engineers, hopeful future full-timers, and so on. A certain number of former teachers probably have no desire (or ability) to teach beyond elementary algebra. But I will admit that I have tutored students in pre-calc with notes that suggested their teacher (I didn't ask who) didn't understand vectors, so it may be that some are teaching beyond their knowledge despite class observations and other checks of their competence.

My only concern was your word "most". I have to agree on "some".
 
In following this conversation, I'm thinking when you refer to "adjuncts" you are talking about non-tenure track or perhaps non-phd people staffing remedial or freshman mathematics courses in college or university settings. Is that correct? Or are you talking about perhaps non-teaching certificated people staffing high school mathematics because of a shortage of fully qualified teachers?
 
My experience with "adjunct" professors is that it is a way for the college or university to avoid calling them "real" employees and so not pay for health insurance or retirement. Yes, "adjunct" professors can be excellent (often better teachers than the full professors who are mainly interested in research) but I wouldn't want to be one or work for a college that had a large number of adjunct faculty.

(But that's just me blowing off steam.)
 
I have always hated the "move to the other side of the equation" wording. I have known students who solve "3x= 6" as "x= -3+ 6= 3" saying "my teacher told me that when you move a number to the other side of the equation you have to change the sign"!

Instead I would explain that to "solve" -12+ 6z= -6 (as pointed out by LCKurtz the parentheses are irrelevant) we have to "undo" what is done to z by doing the opposite of each thing done to
z in the opposite order. If we were to start with "z" to get "-12+ 6z" we would
1) multiply z by 6
2) subtract 12

To "undo" that we
1) add 12 (the opposite of "subtract 12")
2) divide by 6 (the opposite of "multiply by 6")
and of course anything we do to one side we have to do to the other

So from -12+ 6z= -6
Add 12 to both sides: -12+ 6z+ 12= -6+ 12. 6z= 6
Divide both sides by 6: 6x/6= 6/6. z= 1.

Check: If z= 1, -12+ 6z= -12+ 6(1)= -12+ 6= -6.
My teacher from old times used exactly your explanation.
 
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