Solving Equations

Date: 3/23/96 at 22:15:3
From: Anonymous
Subject: Solving equations

Dear Dr. Math,

I have a problem.  Can you help?

    4        3         1
  ----- -  ------ = -------
    b       b+2      b(b+2)

Date: 3/24/96 at 13:15:29
From: Doctor Jodi
Subject: Re: Solving equations

Hi there! 

Since each of these terms has b and/or b+2 in the denominator, 
first I would multiply by b* (b+2).  That gives us:

4b(b+2)  - 3b(b+2)   = 1b (b+2)
------     ------      -------
  b          b+2        b(b+2)

Now we can rewrite these fractions as

4(b+2) * b   - 3b * b+2  = 1 * b  * b+2
         -          ---        --   ---
         b          b+2        b    b+2

Since b/b and (b+2)/(b+2) are both equal to 1 (do you see why?) we 
can rewrite this again:

(4(b+2) * 1) - 3b*1 = 1*1*1

which also equals

4(b+2) - 3b = 1

Now we can expand 4(b+2) by multiplying both b and 2 times 4 to get 
4b+8.

Does that make sense?

Now let's put that back into our equation above.

4b+8 - 3b = 1

So, subtracting 3b from 4b, we get

b + 8 = 1

Now, let's subtract 8 from each side:

b + 8 = 1
  - 8  -8
__________

b  = -7

Write back if you have any questions!

-Doctor Jodi,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 3/25/96 at 1:30:20
From: Anonymous
Subject: Solving equations

Dear Dr. Math,

I have another problem:

During one week in a particular city 5 percent more boys than girls 
were born. if the total number of babies born was 410, find the number 
of each.

Can you help?

Also:

A batch of articles is sold for $810. if each article cost $43.20 and 
the profit made was 25 percent, how many articles were there in the 
batch?

Thanks.

Date: 3/25/96 at 11:14:15
From: Doctor Ethan
Subject: Re: Solving equations

Hello,

These problems are very similar, so I will work one for you 
and explain each step, and then I will let you work the other one.

Let's try to make a chart, okay?

Boys  *  Girls * Total
***********************
      *        *  410


We want to find out how many of each were born, so choose a letter to 
represent one of them.  Let's let y be the number of girls born.

Then how many boys were born? Well, the problem says that 5 percent 
more were born - this means that the number of boys =   

y + 5%y  =   y + .05y  =  1.05 y 

So now we can fill in the chart.

Boys   *  Girls * Total
***********************
 1.05y *   y    *  410


So 1.05y + y = 410

So 2.05 y = 410.

Now we just divide both sides by 2.05 and get  y = 200 . 

Hope that method helps you with the next one.

-Doctor Ethan,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 4/24/96 at 7:23:31
From: Anonymous
Subject: Solving equations

Dear Dr. Math,

I have been given these equations to solve using the easiest method. 
I have tried all different ways but it never seems to work. Please 
explain.

2(x2+2x) = 4

(The two next to the x in the bracket means to the power of two.)

3b2-2b+4 = 0

(The two next to the b is also to the power of two.)

Thank you for your help

Date: 7/16/96 at 17:52:32
From: Doctor Alain
Subject: Re: Solving equations

Hello!

Let's look at the first equation,

2(x^2+2x) = 4         if we divide by 2 we get
x^2+2x = 2            if we substract 2 we get
x^2+2x-2 = 0.

To find the solution to this equation we will use the quadratic 
formula, which you have probably learned at school. 

This formula says that if

ax^2 + bx + c = 0

then 

(i)   x=(-b + Sqrt(b^2-4ac))/(2a)   or
(ii)  x=(-b - Sqrt(b^2-4ac))/(2a).

In the case in question we are trying to solve a=1, b=2 and c=-2.
So we have 2 solutions,

(i)  x = (-2 + Sqrt(2^2-4 (1) (-2)))/(2 (1))
       = (-2 + Sqrt(4 + 8))/2
       = -2/2 + Sqrt(12)/2
       = -1 + Sqrt(12/4)
       = -1 + Sqrt(3)

or

(ii) x = (-2 - Sqrt(2^2-4 (1) (-2)))/(2 (1))
       = (-2 - Sqrt(4 + 8))/2
       = -2/2 - Sqrt(12)/2
       = -1 - Sqrt(12/4)
       = -1 - Sqrt(3).

The other problem is similar and uses the same quadratic formula,
but be careful, the b in the formula must not be confused with the
b in the problem.

-Doctor Alain,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/