multi variable equation help
Dale
following set of problems. What I actually need is the method to help show
her how to do it. I know it's fairly simple but I guess I'm getting old.
Thanks for the help.
a + b -3c = 8
3a + 4b - 2c = 20
2a - 3b + c = -6
Darrell
news:vfIF7.24353$Tb.13...@news1.sttln1.wa.home.com...
For easy reference let's number the equations:
1. a + b - 3c = 8
2. 3a + 4b - 2c = 20
3. 2a - 3b + c = -6
There are several ways to solve this depending on where she is in her
studies. One way is by repeated use of the addition method (a.k.a.
elimination method.) This method is often discussed when first
exposed to three eq. with 3 variables (after already learning to solve
2 eq. with 2 variables.)
Eliminate a variable from two of the three equations. Let's take 1
and 2...
1. a + b - 3c = 8
2. 3a + 4b - 2c = 20
We can "eliminate" a from these by multiplying eq. 1 by -3 and adding
them.
-3a - 3b + 9c = -24
3a + 4b - 2c = 20
---------------------
4. b + 7c = -4
Next, eliminate the same variable from a different "pair" of
equations. Using eq. 1 and eq. 3...
1. a + b - 3c = 8
3. 2a - 3b + c = -6
Multiply the 1st by -2 and add them...
-2a - 2b + 6c = -16
2a - 3b + c = -6
---------------------
5. -5b + 7c = -22
Now, eq. 4 and eq. 5 is a system of two equations with two variables:
4. b + 7c = -4
5. -5b + 7c = -22
Assuming she can handle this with no problem.........b=3 and c=-1.
Since she now knows the value of two of the three variables, she can
simply substitute the values into any equation containing a and solve
for a. It doesn't matter which she chooses, but I would choose the
one that looks like it will be the easiest to solve for a (minimizing
the risk of making a silly error.) Using equation 1...
1. a + b - 3c = 8
...and substituting 3 for b and -1 for c...
a + 3 - 3(-1) = 8
a + 3 + 3 = 8
a + 6 = 8
a = 2
So (a,b,c) = (2,3,-1).
Darrell
Jon Miller
> My daughter is trying to figure out the value of a, b, and c for the
> following set of problems. What I actually need is the method to help show
> her how to do it. I know it's fairly simple but I guess I'm getting old.
> Thanks for the help.
>
> (1) a + b -3c = 8
> (2) 3a + 4b - 2c = 20
> (3) 2a - 3b + c = -6
This should be in her book.
I have taken the liberty of adding labels to your equations.
The idea is, you want to manipulate the equations to get them into the form,
a=something, b=something, c=something. Well, if you multiply the first
equation by 3, you get 3a+3b-9c=24. But 3a+4b-2c=20. Equals subtracted from
equals yields equals, so b+7c=-4. Let's call this equation (2a). Just for
the sake of describing what we did, we will say that (2a)=(2)-3(1), equation
2a is equation 2 - 3 times equation 1.
Now, let's rewrite the system as its equivalent system
(1) a + b -3c = 8
(2a) b + 7c = -4
(3) 2a - 3b + c = -6
Notice that you've gotten rid of the a in equation 2. Now do the same for
equation 3. Now you can (temporarily) ignore equation 1, and continuing on
the same way, get rid of b in equation 3. So that tells you c=whatever. Now
go back and substitute in equation 2z (or whatevery you've gotten to) to solve
for b, and then back to equation 1 to solve for a. Then check your answers in
the original equations (it's very easy to make a mistake in this chain of
calculations).
There are faster ways (it's the same method, but you can combine steps and
avoid writing "unnecessary" details like a,b,c,=...) These also should be in
your daughter's book.
Jon Miller
Dale
13-year-old, she "left it at school", which I believe as the assignment was
on paper instead of from the book. We've talked about the necessity of the
book at home. I'll talk to her about the solving of the equation. Again,
thanks for the help.
Dale
"Jon Miller" <jonatha...@home.com> wrote in message
news:3BE772A0...@home.com...
Dale
was all miscalculations on our part. We tried your method on another set of
problems and it worked flawlessly. Thanks so much. Most importantly, my
daughter understands it again.
Dale
"Darrell" <nospam...@hotmail.com> wrote in message
news:ehKF7.67499$tb2.5...@bin2.nnrp.aus1.giganews.com...
Jon Miller
> Thanks. You're right, it's probably in her book, but being a good 13-year-old,
> she "left it at school",
Accidentally, of course.
> which I believe as the assignment was on paper instead of from the book. We've
> talked about the necessity of the book at home.
At one extreme, there are schools that provide the students with two sets of
books, one for home and one for school. I think the major concern here is to
lighten the load in the backpack (and there are injuries occurring, so this is
not a stupid thing to worry about) rather than to lighten the load on the
student's head.
At the other extreme, there are schools that do not allow the children to take
their books home. This is to keep the books from getting "lost". It also makes
doing their homework difficult.
Actually, just about any book will do. But, if you're going to spend money on
it, you probably ought to get one you like. Some people like Schaum's outlines,
some don't. They tend to be, IMO, good for calculation practice, but no good at
all for learning the concepts.
I started to use _Algebra the Easy Way_ (I forget the author) with my daughter,
but she got bored (it's got stories, which slow you down, but make the concepts
more interesting [for some people]), so we switched to something that moved
along faster. Still, I recommend looking at it.
Math books tend to be either good at explanations or good references (complete,
but probably fear of making the book too long prohibits good explainers from
including a lot and good compilers from explaining well) or neither. There are
a few exceptions, but I don't know any at this level.
Jon Miller
nutsascherz...@gmail.com
Frederick Williams
>
> 3a-8=a+6
I wouldn't call that multi variable, but never mind, the rule "whatever
is done to one side of the equation must be done to the other" applies:
Add 8 to get 3a = a + 14,
subtract a to get 2a = 14,
divide by 2 to get a = 7.
The adding and subtracting is also summarized in "change side, change
sign" which I seem to recall from my school days:
Move -8 from left to right 3a = a + 6 + 8 = a + 14,
move a from right to left 3a - a = 14
i.e. 2a = 14.
I recall no rule cognate to "change side, change sign" for dealing with
multiplication/division.
However you arrive at a = 7, it is worth checking by substituting it
back in the original equation:
On the left 3a - 8 = 3(7) - 8 = 13.
On the right 7 + 6 = 13.
--
When a true genius appears in the world, you may know him by
this sign, that the dunces are all in confederacy against him.
Jonathan Swift: Thoughts on Various Subjects, Moral and Diverting
Pfs...@aol.com
nutsascherz...@gmail.com wrote:
>3a-8=a+6
First of all it is NOT a multi-variable equation. There is one
variable "a".
This is a trivial 7th grade problem!
What happens if you add 8 to both sides and subtract "a" from
both sides?