1.  3/4y = -27
2. -12 = 7-y/3
3. -3(x+5)=8x +18
4. 2-1/4(b-12)=9
5. r/5-3 = 2r/5 +16
6. Negative two thirds of a number is 8/5. What is the number?
7. Solve h=at-0.25vt squared for A
8. Solve A=1/2(b+B)h, for h

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Request for Question Clarification by livioflores-ga on 16 Jun 2003 18:15 PDT

I am working in this question, and need some little clarifications:

7. Solve h=at-0.25vt squared for A

h = at-0.25vt squared = (a.t - 0.25 v.t)^2  and you need to know how
to calculate a if you know h, t and v.


8. Solve A=1/2(b+B)h, for h

A=1/2(b+B)h = 1/2.(b+B).h  and you need to know how to calculate h if
you know A, b and B.

Am I right?

REgards.
livioflores-ga

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Clarification of Question by skywolf-ga on 16 Jun 2003 18:52 PDT

I don't really know... It's been 30 years since I looked at this
stuff!
All I know is that I'm supposed to solve each equation or formula for
the variable specified. Your worst guess is better than my best guess!

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Request for Question Clarification by livioflores-ga on 16 Jun 2003 19:12 PDT

Ok, the common sense said that I am right, so I will work on this
using this interpretation.
Wait for an hour and I will give you the exercises solved.
livioflores-ga

Hi skywolf!!

I will try to answer you showing the work, the idea is that you
understand how to do this task alone in the future.

First of all you must know the notions of operations in both sides of
an equality and inverse operators.
When you are working on an equality, the equality remains true if you
do a valid operation in both sides of it. For example, if you have the
following equality:
2+7 = 5+4 , this equality remains true if I multiply both sides by 2;
(2+7).2 = (5+4).2  because multiplying by 2 is a valid operation
(dividing by zero is not a valid operation).

The inverse operation of a given operation is an operation that do the
opposite of that, for example the opposite operation for the sum is
the subtraction; this relationship is reciprocal: the opposite
operation for the subtraction is the sum. I will give you alist of the
operations used here and their opposites, due the reciprocal
relationship between them I will list them in pairs:
sum <--> subtraction
multiplication <--> division
square <--> square root
 
What is the utility of this concept here?, it is simple, if you have
an equation and you need to simplify it (to solve it), you can use
inverse operations to do that, only if they are valid operations.
For example:
2.k + 18 = 28   (the dot means multiplication)
we want to know k, so we need to simplify the equation in order to
obtain a "clean" k in one side of the equality. First we can eliminate
the term "+18", to do that we must use the inverse operation of the
sum, that is the subtraction. How? subtracting at both sides of the
equality 18:
2.k + 18 - 18 = 28 - 18
2.k + 18 - 18 = 2.k + 0 = 2.k  and 28- 18 = 10, now we have a new
VALID equality:
2.k = 10
Now we have 2 multiplying k, the inverse operation of the
multiplication is the division:
2.k /2 = 10/2  
2.k /2 = k  and 10/2 = 5 then we have another VALID equality:
k = 5  and the problem is solved.

Some tips to do this exercices:
1. Start simplifying the operations that involves all the "unknown
quantity side" of the equation (for example the square on the exercise
7 must be eliminated first using the square root).
2. Remember to "separate" in terms the unknown quantity side of the
equation, you can identify each term easily: they are the quantities
added or subtracted (any of the individual addends of an expression).
For example in the expression 2.x + 3, the terms are 2.x and 3.
3. Eliminate one by one all the terms of the unknown quantity side
except the term wich includes the unknown quantity.

I give you this part first in order to give you time to read and
understand the method while I finish the problems. So when you have
the problems solved you will have some back up to understand how I did
it.

The complete answer is coming, no more than an hour.

Regards.
livioflores-ga

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Clarification of Answer by livioflores-ga on 16 Jun 2003 21:01 PDT

THE EXERCICES:

(Remember that the dots mean multiplication)

1.  3/4y = -27 

I rewrite the equation in a more clear way:
(3/4).y = -27  
We can divide by (3/4) in both sides:
(3/4).y / (3/4) = -27 / (3/4);

(3/4).y / (3/4) = y    and  -27 / (3/4) = -36  ; 
then:
y = -36

----------------------------------------------

2. -12 = 7 - y/3  

First we must eliminate the term 7 by substracting 7 at both sides of
the equation:
-12 - 7 = 7 - y/3 -7 ;

-12 - 7 = -19  and   7 - y/3 -7 = y/3
then we have:
y/3 = -19

Now we will obtain y multiplying by 3 both sides of the equation:
3.y/3 = 3.(-19)  
then: 
y = -57

---------------------------------------

3.  (-3).(x+5) = 8.x + 18 

Here we have a more complicated problem. We need to stablish a
strategy to solve it. We can use the distributive property of the
multiplication in the left side, and then start to work:

(-3).(x+5) = (-3).x + (-3).5 = -3.x - 15 ;
Now we can back to the original equation and do the replace:
(-3).(x+5) = -3.x - 15 = 8.x + 18 and we will work with:

-3.x - 15 = 8.x + 18 ; 
in this situation we have that the unknown quantity is in the both
sides of the equation, we must find the way to have only one "unknown
quantity side", but this task now is easy for us because the term "+
8.x" can be eliminated from the right side by substracting "8.x" in
both sides:
-3.x - 15 - 8.x = 8.x + 18 - 8.x

-3.x - 15 - 8.x = -11.x - 15  and  8.x + 18 - 8.x = 18
then:
-11.x - 15 = 18 ;
then:
-11.x - 15 +15 = 18 + 15 
then:
-11.x = 33
then:
-11.x /(-11) = 33/(-11) = -3
then:
x = -3

---------------------------------------------

4. 2 - (1/4).(b-12) = 9

then:
2 - (1/4).(b-12) - 2 = 9 - 2 = 7
then:
-(1/4).(b-12) = (-1/4).(b-12) = 7
then:
(b-12) = 7/(-1/4) = -28
then:
b - 12 + 12 = -28 + 12 = -16
then:
b = -16

---------------------------------------

5. r/5 - 3 = 2.r/5 + 16 

r/5 - 3 - 2.r/5 = 2.r/5 + 16 - 2.r/5 = 16 ;

r/5 - 3 - 2.r/5 = r/5 - 2.r/5 -3 = -r/5 - 3 = (-1).r/5 - 3
then:
(-1).r/5 - 3 = 16
then:
(-1).r/5 - 3 + 3 = 16 + 3 = 19
then:
(-1).r/5 = 19
then:
r/5 = 19/(-1) = -19
then.
5.r/5 = 5.(-19) = -95
then:
r = -95

------------------------------------------

6. Negative two thirds of a number is 8/5. What is the number?

this is a different kind of problem, it is like the puzzles that
appears in some magazines.
What "two thirds of a number" means?
We know two thirds, it is 2/3, and "of a number" means "of x" where x
is the unknown number.
So "two thirds of a number is 8/5" means:
(2/3).x = 8/5 ;
but we have that the "Negative two thirds of a number is 8/5", then we
must solve:
-(2/3).x = 8/5

-(2/3).x /(2/3) = (8/5)/(2/3) = 12/5
then:
-x = 12/5
then:
x = -12/5

------------------------------------------

7. Solve h = (a.t - 0.25 v.t)^2  for a

In this exercise you must think that all the values except a are
known, then you will work with them like common numbers.

If we apply the square root to the both sides of the equation we have:
sqr(h) = a.t - 0.25 v.t  where sqr(h) means the square root of h.
Then:
sqr(h) + 0.25 v.t = a.t - 0.25 v.t + 0.25 v.t = a.t

More clearly:
a.t = sqr(h) + 0.25 v.t

Now follows the division at both sides of the equation by t, but this
operation is valid only if t is different of zero. You must keep this
in mind.
Then:
a = (sqr(h) + 0.25 v.t)/t   WITH t DIFFERENT TO ZERO.

Note: may be you want to know what happen when t is zero; well in this
case is not relevant, because if you look at the initial equation, if
t=0 then h=0 whatever be the value of a.

---------------------------------------

8. Solve A = 1/2.(b+B).h  for h

A = 1/2.(b+B).h
then:
A /(1/2) = 1/2.(b+B).h /(1/2) = (b+B).h
then:
2.A = (b+B).h
then, if (b+B) is different to zero:
2.A /(b+B) = (b+B).h /(b+B) = h
then:
h = 2.A /(b+B)

----------------------------------

I hope this helps. If you need an answer clarification please post a
request for it, the same if you find a possible error in the solutions
or if one of the exercises was  misinterpreted. I will redo the
exercises if necessary.

Best regards.
livioflores-ga

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Clarification of Answer by livioflores-ga on 16 Jun 2003 21:12 PDT

Please excuse me, a typo led me to a mistake in the exercise 2:

I wrote:

-12 = 7 - y/3   
 
First we must eliminate the term 7 by substracting 7 at both sides of
the equation:
-12 - 7 = 7 - y/3 -7 ; 
 
-12 - 7 = -19  and   7 - y/3 -7 = y/3 
then we have: 
y/3 = -19   HERE IS THE MISTAKE!!! it must be -y/3 = -19

It follows: 
Now we will obtain y multiplying by (-3) both sides of the equation: 
-3.(-y/3) = (-3).(-19)   
then:  
y = 57 

I hope that this does not give you a headache.
Regards.
livioflores-ga

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Clarification of Answer by livioflores-ga on 16 Jun 2003 21:59 PDT

Please let me know if the answer works.
This is in order to know if our work satisfy you, and help us to bring
a better service.

livioflores-ga

Excellent Thanks for the help!

I don't mean to preempt livioflores-ga, but I decided to work these
out to brush up on my algebra. Here are my answers (checked on a
TI-89, so they should be right):

Question 1 

(3/4)y = -27
Multiply both sides by 4
3y = -108
Divide both by 3
y = -36


Question 2

-12 = 7 - y/3
Subtract 7 from both sides
-19 = -y/3
Multiply both sides by 3
-57 = -y
Multiply both sides by -1, to flip the signs
y = 57


Question 3

-3(x + 5) = 8x + 18
Distribute the -3 to the x and 5
-3x - 15 = 8x + 18
Subtract 18 from both sides
-3x - 33 = 8x
Add 3x to from both sides
-33 = 11x
Divide both sides by 11
x = -3


Question 4

2 - 1/4(b - 12) = 9
Distribute the 1/4 to the b and -12
2 - (1/4)b + 3 = 9
Subtract 5 from both sides (canceling out the 3 and 2)
-(1/4)b = 4
Multiply both sides by -4 (since 1/4 times 4 is 1)
b = -16


Question 5

r/5 - 3 = 2r/5 + 16
Add 3 to both sides
r/5 = 2r/5 + 19
Subtract 2r/5 form both sides
r/5 - 2r/5 = 19
Multiply both sides by 5
r - 2r = 95
Simplify the r - 2r
-r = 95
Multiply both sides by -1
r = -95


Question 6

Put into equation form, what you're asking is (with x as the number)
(-2/3)x = 8/5
To cancel out the -2/3, multiply both sides by -3/2
x = 8/5 * -3/2
Multiply the fractions (top * top = top, bottom * bottom = bottom)
x = -24/10
Simplify the fraction by dividing by 2
x = -12/5 (or -2.4, same thing)


Question 7

First, take the square root of both sides. Remember, there are two
different answers to any square root - the negative and the positive.
For example, -2 squared and 2 squared both equal positive 4. This
means we'll end up with two equations, both producing valid solutions
for A. I'll use sqrt to mean square root.

sqrt(h) = at - 0.25vt   OR   sqrt(h) = -at + 0.25vt

Divide both sides by t

sqrt(h)/t = a - 0.25v   OR   sqrt(h)/t = -a + 0.25v

Now move the 0.25v away from the A, by adding it to both sides in the
first equation and subtracing it in the second one.

A = 0.25v + sqrt(h)/t   OR   A = 0.25v - sqrt(h)/t

or, in a prettier version:

              ___                      ___
             / h                      / h
A = 0.25v + ----- +   OR  A = 0.25v — -----
              t                         t


Question 8

A = 1/2(b+B)h
Divide both sides by 1/2(b+B). I'll turn 1/2 into 0.5 to make it
easier to read.
A/0.5(b+B) = h
That's a valid solution right there, but it can be simplified.
Multiply both parts of the fraction by 2.
h = 2A/(b+B)

or, in a prettier version:

     2A
h = -----
     b+B

livioflores's solution for #2 is wrong because he/she forgets that
we're dealing with a negative y. -y equals -57, so flipping the signs
dives y = 57.

Also, the solution for #7 is partially correct, but only one of the
two solutions is given. Read my other comment to see why this is
important.

Thank you, but as you can see I noted the mistake before you post your
second comment, and also you are right about the square root, I forgot
that into all the explanataions, in the intention of give to skywolf a
detailed work.
Damian always take care to take the square root to positive numbers,
you didnīt clarify this point; also remember to add to your answers 7
and 8 the conditions that leave you divide both sides of the
equations.In the question 7 the condition is t different to zero and
in the question 8 the condition is (b+B) different to zero.

livioflores-ga

You're right, I should have remembered to state the domain of the
functions. Another condition for #7 is that h must be >= 0 for #7 to
get a real result.