Tina's change of $1.35 consisted of quarters and dimes only and
she received twice as many dimes as quarters. How many quarters were there?

Let x = the number of quarters

>>...she received twice as many dimes as quarters...<<

That tells us that the number of dimes is twice the number or
quarters, or 2x.

since each quarter is worth 25 cents,

x quarter are worth 25x cents

Since each dime is worth 10 cents,

2x dimes are worth 10 times 2x cents or 20x cents.

>>...Tina's change of $1.35...<<

This says 

the worth of the quarter + the worth of the dimes = 
$1.35 = 135 cents.

                             25x + 20x = 135

Can you solve that for x?

Answer:  x=3, so there are 3 quarters.

Check: There are 3 quarters and twice as many or 6 dimes.  The 
3 quarters are worth 75 cents and the 6 dimes are worth 60 cents, 
and the total is worth $1.35. 

Edwin
AnlytcPhil@aol.com

4/8 < 2/8t - 6/8 < 1/8 

This can have no solution because 4/8 is GREATER than, not LESS than 1/8.
I think you copied the problem wrong.

Edwin
AnlytcPhil@aol.com

What is the slope and y-intercept of 5y+4x=5

Rule: 

1. Solve for y.

2. The slope equals the coefficient of x on the right side.

3. The y-intercept is the constant term on the right side

                      5y + 4x = 5

                           5y = 5 - 4x

Divide every term by 5

                           5y     5     4x
                          ———— = ——— - ————
                            5     5      5

                           1      1
                           5y     5     4x
                          ———— = ——— - ————
                            5     5      5         
                            1     1
                                      4x
                             y = 1 - ————
                                       5

                                      4
                             y = 1 - ———·x
                                      5

Now that we have solved for y, the coefficient of x on 
the right side is -4/5 so that's the slope. The constant
term on the right side in 1, so that's the y-intercept.
Actually the y-intercept is the point (0,1) on the y-axis.

Edwin
AnlytcPhil@aol.com

        (a2-4a-1)2 - 16

        (a2-4a-1)2 - 42

        (a2-4a-1)2 - 42    that's the difference of two squares

  [(a2-4a-1) - 4][(a2-4a-1) + 4]

[a2 - 4a - 1 - 4][a2 - 4a - 1 + 4]

   [a2 - 4a - 5][a2 - 4a + 3]

 [(a - 5)(a + 1)][(a - 3)(a - 1)]

   (a - 5)(a + 1)(a - 3)(a - 1)

Edwin
AnlytcPhil@aol.com

  64x = 1/2

  64x = 1/21

(26)x = 2-1

  26x = 2-1

   6x = -1

    x = -1/6

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

     x4/3 = 625

     x4/3 = 54

(x4/3)3/4 = (54)3/4

       x1 = 53

        x = 125

Edwin
AnlytcPhil@aol.com

  5
————— 
 Ö8

The idea is to rationalize the denominator; that is get the
denominator to be free of roots (radicals).  First we break 
8 into primes like this 2·2·2

   5
————————
 Ö2·2·2
  
Two of those 2's will pair up as 22 which will come out of the 
square root radical since it is a perfect square.

   5
————————
 Ö22·2
  
This will leave one 2 under the radical.  So we need another 2
under that radical so it will pair with the other 2, so we will be able
to take it out too.  To accomplish this we multiply numerator and 
denominator by Ö2
      _
   5·Ö2
——————————
 Ö22·2·Ö2
  
Now we multiply under the radical on the bottom
      _
   5·Ö2
—————————
 Ö22·2·2
 
or 

     _
  5·Ö2
————————
 Ö22·22

  
Now we can take individual square roots on the bottom
     _
  5·Ö2
—————————
 Ö22·Ö22

and get 2·2
      _
   5·Ö2
—————————
   2·2
  
or

   _
 5Ö2
—————
  4

Edwin
AnlytcPhil@aol.com

How do i find the solution set? 
6(r-9)+3 > 7(r+3)-22


Solve it just like an equation with one exception:


6(r - 9) + 3 > 7(r + 3) - 22

 6r - 54 + 3 > 7r + 21 - 22

     6r - 51 > 7r - 1

     6r - 7r > 51 - 1

         -1r > 50

Up to here the proble is exactly the same as if there
had been an " = " instead of a " > "

Her is where it is different.  When you divide (or multiply)
an inequality by a negative number, you must change the 
direction of the inequality. 

We need to divide both sides by -1, the coefficient of r, so we must
change the " > " to " < "
 
-1r    50
——— < ————
-1     -1

  r < -50

You can either express this in set builder notation as

{r|r is real and r < -50}

or in interval notation as  (-oo, -50)

Edwin
AnlytcPhil@aol.com

How do I solve for b? 
16b-squared - 9 = 0

16b2 - 9 = 0

Two ways.  Factoring method:

16b2 - 9 = 0

can be written as

(4b)2 - 32 = 0

This is the difference of two perfect squares. The difference of two perfect
squares factors into the product of the sum and differencve of the two square
roots.

(4b - 3)(4b + 3) = 0

Now use the zero-factor principle, which is to put " = 0 " after each of those
two parenthetical factors and solve for the two answers:

4b - 3 = 0                4b + 3 = 0
    4b = 3                    4b = -3 
     b = 3/4                   b = -3/4

=======================================================

Second way:  Taking square roots:

16b2 - 9 = 0

Add 9 to both sides

16b2 = 9

Take the square roots of both sides, remembering that there are
two square roots.
                         _
square roots of 16b2 = ±Ö9
      
                  4b = ±3

                   b = ±3/4

Edwin
AnlytcPhil@aol.com

Solve for the number of apples 
The number of apples is to 12 oranges as 45 is to 60

Replace the words "is to" by "divided by", and the word "as" by "equals":

So

The number of apples is to 12 oranges as 45 is to 60

becomes

The number of apples divided by 12 oranges equals 45 divided by 60.

Now replace "The number of apples" by "N apples"

N apples divided by 12 oranges equals 45 divided by 60.

                  N/12 = 45/60

Can you solve that?

Answer: N = 9

Edwin
AnlytcPhil@aol.com

3x3(x - 2)(x + 2)

First just work on the 3x3(x-2) only, but don't forget to bring the (x + 2)
each time.

(3x3·x - 3x3·2)(x + 2)

(3x4 - 6x3)(x + 2)

Now you can use FOIL

3x4·x + 3x4·2 - 6x3·x - 12x3

3x5 + 6x4 - 6x4 - 12x3

The middle two terms cancel, and we get

3x5 - 12x3 

Edwin
AnlytcPhil@aol.com

7-(7+3/4-2)+(-4)-8+4-(24)

The order of operations are
1. Do whatever can be done within in the first parentheses, using the order of
operations 2, 3, and 4 below, replacing it with what you got.
2, Do the first exponentiation you come to going left to right, replacing it
with what you got.
3. Do the first multiplication or division you come to going left to right,
replacing it with what you got.
4. Do the first addition or subtraction you com to going left to right,
replacing it with what you got.

7-(7+3/4-2)+(-4)-8+4-(24)

We look for the first parentheses (7+3/4-2).  Now we use the order of
operations above within the parentheses.  There are no exponentiations for step
2, so we go on to step 3.  The first multiplication or division we come to in
(7+3/4-2) is the division 3/4, so we do that division and get 0.75 and replace
the 3/4 by 0.75

7-(7+0.75-2)+(-4)-8+4-(24)

We look for the first parentheses (7+0.75-2).  Now we use the order of
operations above within the parentheses.  There are no exponentiations for step
2, so we go on to step 3.  There are no multiplications or divisions for step
3, so we go on to step 4.  The first addition or subtraction we come to in
(7+0.75-2) is the addition 7+0.75, so we do that addition and get 7.75 and
replace the 7+0.75 by 7.75 

7-(7.75-2)+(-4)-8+4-(24) 

We look for the first parentheses (7.75-2).  Now we use the order of operations
above within the parentheses.  There are no exponentiations for step 2, so we
go on to step 3.  There are no multiplications or divisions for step 3, so we
go on to step 4.  The first addition or subtraction we come to in (7.75-2) is
the subtraction 7+0.75, so we do that subtraction and get 5.75 and replace the
7.75-2 by 5.75 

7-(5.75)+(-4)-8+4-(24)  

Now there are only single numbers inside parentheses, so there is nothing to do
inside any of them. So we co to step 2.  There are no exponensts, so we go to
step 3.  There are no multiplications or divisions, so we go to step 4. The
first addition of subtraction we come to is the sub\traction 7-(5.75).  So we
do that subtraction getting 1.25 and replace the 7-(5.75) by 1.25

1.25+(-4)-8+4-(24) 

Now there are only single numbers inside parentheses, so there is nothing to do
inside any of them. So we co to step 2.  There are no exponensts, so we go to
step 3.  There are no multiplications or divisions, so we go to step 4. The
first addition of subtraction we come to is the addition 1.25+(-4).  So we do
that addition getting -2.75 and replace the 1.25+(-4) by -2.75

-2.75-8+4-(24)

Now there are only single numbers inside parentheses, so there is nothing to do
inside any of them. So we co to step 2.  There are no exponensts, so we go to
step 3.  There are no multiplications or divisions, so we go to step 4. The
first addition of subtraction we come to is the subtraction -2.75-8.  So we do
that subtraction getting -10.75 and replace the -2.75-8 by -10.75.

-10.75+4-(24)

Now there are only single numbers inside parentheses, so there is nothing to do
inside any of them. So we co to step 2.  There are no exponensts, so we go to
step 3.  There are no multiplications or divisions, so we go to step 4. The
first addition of subtraction we come to is the addition -10.75+4.  So we do
that addition getting -6.75 and replace the -10.75+4 by -6.75.

-6.75-(24)

Now the only thing left is that subtraction, so we get

-30.75

You can shorten the process by doing more than one thing at a time, but you
have to be careful that you follow the rule of priorities of operations.

Edwin
AnlytcPhil#aol.com

                          -7 = 9 + 2g

To get used to algebra, at first make sure every term has a sign either a + or
a - on its left side.  The -7 already has a sign on its left.  The 9 does not
have a sign left of it so we put a + left of it

                          -7 = +9 + 2g 

There is one term on the left, -7, and two terms on the right, +9 and +2g. 

There are no terms that will combine on the left or right.

1. We want to get rid of all terms on the left which DON'T contain the unknown
letter g.

2. 1. We want to get rid of all terms on the right which DO contain the unknown
letter g.

         The -7 does not contain a g and it is on the left side, so we need to
get rid of it by adding its opposite, +7 to both sides

                          -7 = +9 + 2g
                          +7   +7
                        ------------------

Add vertically:

                          -7 =  +9 + 2g
                          +7    +7
                        ------------------
                           0 = +16 + 2g

Now e have a new equation:

                           0 = +16 + 2g

The first thing we come to is the +16. It does not contain h and it is on the
right side, so we leave it alone.

The next thing we come to is the +2g on the right side. It contains a g so we
must get rid of the +2g on the right by adding its opposite -2g to both sides

                           0 = +16 + 2g
                      -2g          - 2g 
                     --------------------

Add vertically

                           0 = +16 + 2g
                     -2g           - 2g 
                     --------------------
                     -2g + 0 = +16 +  0

We can eliminate the zeros

                         -2g = +16

When we get it down to just one term on the left with the letter and just one
term on the right with just a number, that's when we DIVIDE both sides by
the coefficient of the unknown letter on the left:

                          -2g    +16
                         ————— = ————
                           -2     -2

Now you can cancel the -2's on the left and divide +16 by -2, getting -8 on the
right                             
                           1      -8
                          -2g    +16
                         ————— = ————
                           -2     -2                        
                            1      1
and get

                             g = -8 
____________________________________________________________________________
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

solve. check your solution.

                    5h + 2 - h = 22

The terms on the left are: +5h, +2, and -h.  There is only one term on the
right, namely +22.  Write the 5h as +5h, -h as -1h, and the 22 as +22

                  +5h + 2 - 1h = +22

The terms +5h and the term -1h combine as +4h since +5 and -1 give +4.

                       +4h + 2 = +22

The first term we come to is the +4h. It contains h and it is on the left, so
we leave it as is.

The next term is the +2.  It does not contain h and is on the left side so we
must get rid of it by adding its opposite, -2, to both sides:

                       +4h + 2 = +22
                            -2 =  -2
                     —————————————————
                       +4h + 0 = +20

Get rid of the 0
                           +4h = +20

When we get it down to just one term on the left with the letter and just one
term on the right with just a number, that's when we DIVIDE both sides by
the coefficient of the unknown letter on the left:

                          +4h    +20
                         ————— = ————
                           +4     +4

Now you can cancel the +4's on the left and divide +20 by +4, getting +5 on the
right                             
                           1      +5
                          +4g    +20
                         ————— = ————
                           +4     +4                        
                            1      1
and get

                             g = +5 

Edwin
AnlytcPhil@aol.com

Find an equation of the hyperbola such taht for any point on
the hyperbola. the difference between its distance from the
points (2,2) and (10,2) is 6.

This hyperbola has a horizontal transverse axis, i'e', it
looks sort of like this: ")(", so it has equation of the form

(x - h)²   (y - k)²
———————— - ———————— = 1
   a²         b²

where (h-a,k) and (h+a.k) are the left and right vertices,
respectively

where (h-c,k) and (h+c,k) are the left and right foci,
respectively

where c = absolute value distance between the foci and the
center.  

where a² + b² = c²

where a = absolute value distance between the center and the
vertex = semi-transverse axis.

where 2a = transverse axis

where b = semi-conjugate axis

and where 2b = conjugate axis

The given points are the foci.  The center (h,k) of the
hyberbola is the midpoint between these foci or 
(h,k) = ((2+10)/2, (2+2)/2)) = (6,2).
So far we have

(x - 6)²   (y - 2)²
———————— - ———————— = 1
   a²         b²

The vertices are " a " units from the center, so their
coordinates are

(h-a, 2) and (h+a, 2)

and since h = 6, the vertices are

(6-a, 2) and (6+a, 2)

Also c = absolute value distance between center and focus,
and the distance between (6,2) and (2,2) is 4, so c=4

The vertices are themselves points on the hyperbola.
Therefore the difference (in absolute value) between each one
of these vertices' distance from the points (2,2) and (10,2)
is 6.

The vertices are between the foci, so

The distance between (6-a, 2) and (2,2) is (6-a)-2 or 4-a

The distance between (6-a, 2) and (10,2) is 10-(6-a) or a+4

The difference between these in absolute value = (a+4)-(4-a)
or 2a

So 2a = 6 or a = 3 

Therfore we have

(x - 6)²   (y - 2)²
———————— - ———————— = 1
   3²         b²

or

(x - 6)²   (y - 2)²
———————— - ———————— = 1
   9          b²

All that's left is b², and we get that from

a² + b² = c²
3² + b² = 4²
 9 + b² = 16
     b! = 7

So we replace b² by 7 and we are done:

 (x - 6)²   (y - 2)²
———————— - ———————— = 1
   9           7

Edwin
AnlytcPhil@aol.com

      5       3
     ——— x + ——— = x
      8       5

Multiply thru by LCD = 40 

    5         3
40·———x + 40·——— = 40·x
    8         5

Cancel
          
5   5     8   3
40·———x + 40·——— = 40·x
    8         5
     1         1 

        25x + 24 = 40x

            -15x = -24

               x = -24/(-15)

               x = 8/5

Edwin
AnlytcPhil@aol.com

(x/x+y)+(x/x-y)

What you typed means this

 x         x
--- + y + --- - y
 x         x

I'm sure that's not what you meant as that is too easy:

1 + y + 1 - y 

which equals 2.

You didn't put in enough parentheses.  To type an expression all on one line,
you must begin every numerator and every denominator with a "(" and end it with
a "}"; otherwise you cannot tell where it starts and ends.  The only times you
don't need a pair of parentheses around a numerator or denominator is when it
contains only one number or one letter.

What I think you meant was this

  x       x
----- + -----
 x+y     x-y

which you should have typed on one line as:

x/(x+y) + x/(x-y)

==========================================

  x       x
----- + -----
 x+y     x-y

When the denominators have no common factor, as here, you may use
this rule:

     A     C     AD ± CB
    --- ± --- = ---------
     B     D        BD


  x       x      x(x-y) + x(x+y)
----- + ----- = ----------------- = (CAREFUL! NO CANCELLING BETWEEN
 x+y     x-y       (x+y)(x-y)        NUMERATOR AND DENOMINATOR ALLOWED HERE!)


      x² - xy + x² + xy
      -----------------
         (x+y)(x-y)


             2x²
        ------------
         (x+y)(x-y)


Edwin
AnlytcPhil@aol.com

Solve each inequality. State the solution set using interval notation and 
graph it. 

 x^2 – x – 20 < 0


1. Make sure only 0 is on the right.

   This is already the case

2. Factor left side

   (x - 5)(x + 4) < 0

1. Find the critical points.  These are found by setting the 
left hand side = 0 and solving for x

    (x-5)(x+4) = 0
  x-5=0 gives critical value x=5
  x+4=0 gives critical value x = -4

2. Draw a number line and circle the critical values

——————————————o———————————————————————————————————o———————————
 -7  -6  -5  -4  -3  -2  -1   0   1   2   3   4   5   6   7
 
3.  Choose any number left of the leftmost critical point, substitute
it in the factored form of the inequality. If the result is true, shade
that part of the number line, otherwise do not shade it. 

Say we choose -5. substitute in:

    (x - 5)(x + 4) < 0
  (-5 - 5)(-5 + 4) < 0
         (-10)(-1) < 0
                10 < 0

This is false, so we do not shade the region to the left of -4.

4.  Choose any number between the first and second critical points, substitute
it in the factored form of the inequality. If the result is true, shade that
part of the number line, otherwise do not shade it. 

Say we choose 0. substitute in:

    (x - 5)(x + 4) < 0
    (0 - 5)(0 + 4) < 0
           (-5)(4) < 0
               -20 < 0

This is true, so we shade the region between -4 and +5.

——————————————o===================================o———————————
 -7  -6  -5  -4  -3  -2  -1   0   1   2   3   4   5   6   7

5.  Choose any number left of the rightmost critical point, substitute
it in the factored form of the inequality. If the result is true, shade
that part of the number line, otherwise do not shade it. 

Say we choose 6. substitute in:

    (x - 5)(x + 4) < 0
    (6 - 5)(6 + 4) < 0
           (1)(10) < 0
                10 < 0

This is false, so we do not shade the region to the right of 5.

The interval notation is found by putting the endpoints left to right in
parentheses, with a comma between

   Answer:        (-4, 5)
 
Edwin
AnlytcPhil@aol.com

 Simplify each complex fraction
(1/3-1/4)/(1/3+1/6) 

  1     1
 ——— - ———
  3     4
———————————
  1     1
 ——— + ———
  3     6  

Look at all four denominators 3,4,3,and 6.  The LCD of all
four of these is 12.  Therefore we multiply every fraction
in the top and in the bottom by 12/1

  12     1     12     1
 ———— · ——— - ———— · ———
   1     3      1     4
—————————————————————————
  12     1     12     1
 ———— · ——— + ———— · ———
   1     3      1     6

Now we cancel:

  4            3
  12     1     12     1
 ———— · ——— - ———— · ———
   1     31     1     4 1
—————————————————————————
 412     1    212     1
 ———— · ——— + ———— · ———
   1     31     1     61

All that's left is:

 4 - 3
———————
 4 + 2

or
   1
  ———
   6

===================================================
===================================================

  1     3
 ——— - ———
  x     2
———————————
  3     1
 ——— + ———
  4     x  

Look at all four denominators x,2,4,and x.  The LCD of all
four of these is 4x.  Therefore we multiply every fraction
in the top and in the bottom by 4x/1

  4x     1     4x     3
 ———— · ——— - ———— · ———
   1     x      1     2
—————————————————————————
  4x     3     4x     1
 ———— · ——— + ———— · ———
   1     4      1     x

Now we cancel:

                          2
  4x     1     4x     3
 ———— · ——— - ———— · ———
   1     x1     1     21
—————————————————————————
 14x     3    24x     1
 ———— · ——— + ———— · ———
   1     41     1     x1

All that's left is:

 4 - 2x·3
——————————
 x·3 + 4

or

   4 - 6x
  ————————
   3x + 4

Edwin
AnlytcPhil@aol.com

how to round off 8,008,278 to the nearest ten thousand.

First let's name all the digits, starting at the far right and moving
to the left.

8 (on the far right) is the one's digit
7 is the ten's digit
2 is the hundred's digit
8 (the 8 in the middle)is the thousand's digit
0 (the 0 farthest to the right) is the ten-thousand's digit
0 (the 0 farthest to the left) is the hundred-thousand's digit
8 (the 8 on the far left) is the million's digit

Since you are to round off to the nearest ten thousand, I will color
the ten-thousand's digit, the right-most 0, red,
and the digits to the right of it blue:

                  8,008,278
 
We look at the digit just right of that 0.  It is an 8.  Since 8 is 5 or more
we add one to the 0 in the ten-thousand's place, getting 1, so we replace that
0 in the ten-thousand's place by 1.  Then we replace ALL digits to the
right of that ten-thousand's digit with 0's

Final result:
                  8,010,000

Edwin
AnlytcPhil@aol.com

Why does the degree of x2 + abcx equal 4?

The degree of a polynomial is the largest number of letters that are
multiplied together in a term.

First write out the polynomial without any exponents showing

x2 + abcx

becomes

xx + abcx

The first term has 2 letters multiplied together and the second term 
has 4 letters multiplied together. 4 is the larger of the two so the 
degree of the polynomial is 4.

Edwin
AnlytcPhil@aol.com

Hi, can you help me solve

3x-2      2
————— - —————
 x+2     x-2

We want to cause both those denominators to become the LCD, which is
(x+2)(x-2).  We use the principle:

If you multiply the numerator and denominator of a fraction by the
same thing you don't change the value of the fraction. You only change
its looks. 

The first fraction has denominator x+2.  It needs to become the LCD, 
which is (x+2)(x-2), so it needs to be multiplied by (x-2). 

I can do this without changing the fraction's value if I'll also 
multiply the numerator by (x-2).

So the first fraction:

3x-2     
————— 
 x+2 

becomes 

(3x-2)(x-2)    
——————————— 
 (x+2)(x-2)

FOIL the top out but DON'T FOIL the bottom!

 3x²-8x+4
——————————
(x+2)(x-2) 

Now the second fraction has denominator x-2.  It needs to become 
the LCD, which is (x+2)(x-2), so it needs to be multiplied by (x+2). 

I can do this without changing the fraction's value if I'll also 
multiply the numerator by (x+2) also.

Now the second fraction:

  2     
————— 
 x-2 

becomes 

  2(x+2)    
——————————— 
 (x-2)(x+2)

Multiply the top out but not the bottom!

   2x+4
——————————
(x-2)(x+2)

Now back to the original problem:

3x-2      2
————— - —————
 x+2     x-2

becomes

 3x²-8x+4       2x+4
—————————— - ——————————
(x+2)(x-2)   (x-2)(x+2)
 
Now the denominators are equal.

So we subtract the numerators, placed first in parentheses,
then write that over the common denominator.


(3x²-8x+4) - (2x+4)
———————————————————
    (x+2)(x-2)

Now remove the parentheses on top and collect terms. Leave
the bottom as it is:

  3x²-8x+4-2x-4
———————————————————
    (x+2)(x-2)

     3x²-10x
———————————————————
    (x+2)(x-2)

That's good enough.  You can leave it like that.  But if you
like you can factor out x on the top. 

    x(3x-10)
—————————————————
    (x+2)(x-2) 

In some problems, doing that produces something that will cancel,
but that didn't happen this time. For instance, if that (3x-10)
had been (x+2) or (x-2) instead it would have canceled. But it
wasn't so it didn't. 

Edwin
AnlytcPhil@aol.com

A class of 14 students is made up of 6 girls and 8 boys
Determine the number of different groups of 5 that can be formed
if there must be at most 1 boy in each group (there could be
0 or 1 boy in each group). 

A. 23
B. 30
C. 120
D. 126

If there are no boys, then we only need to eliminate one of the 6 girls.
We can choose the girl to eliminate any of 6 ways

If there is 1 boy, then we may choose the boy any of 8 ways.  For each of
these 8 ways to choose the boy we can choose 4 girls to go with him C(6,4)
ways or 6!/(4!2!) = 720/(24·2) = 15. So that 8·15 
or 120 ways.

The grand total is 6+120 or 126, choice D

Edwin
AnlytcPhil@aol.com

x=1+the square root of x squared + x + 2x times the 
square root of x - the square root of x + 1 + 2 times 
the square root of x
         ____      _    ___     _ 
x = 1 + Öx2+x + 2xÖx - Öx+1 + 2Öx

Check to see if 0 is a solution

         ____        _    ___     _ 
0 = 1 + Ö02+0 + 2(0)Ö0 - Ö0+1 + 2Ö0

0 = 1 + 0 + 0 - 1 + 0

0 = 0

Yup, sure is. That's the solution. So x=0

Edwin
AnlytcPhil@aol.com

2x^-2y^0(x^2y^0-4x^-6y^4

Is this what you mean?

2x-2y0(x2y0 - 4x-6y4)

If so, replace both y0's by 1.

2x-2·1(x2·1 - 4x-6y4)

2x-2(x2 - 4x-6y4)

Now use the distributive principle:

2x-2·x2 - 8x-2·4x-6y4

Add the exponents of x in the first term: -2 + 2 = 0,
so replace x-2·x2 by x0

2x0 - 8x-2·x-6y4

But x0 is just 1, so replace x0 by 1

2·1 - 8x-2x-6y4

2 - 8x-2x-6y4

Now add the exponents of x in the second term:
-2 + -6 = -8. so replace x-2x-6 by x-8

2 - 8x-8y4

To get rid of the negative exponent in x-8, replace
x-8 by 1/x8

2 - 8(1/x8)y4

Simplify the second term:

2 - 8y4/x8
 
Edwin
AnlytcPhil@aol.com

Please help me to solve this equation:
1+2(x^2-4)=5-2x 

1 + 2(x2 - 4) = 5 - 2x

  1 + 2x2 - 8 = 5 - 2x

      2x2 - 7 =  5 - 2x
          + 7   +7
     -------------------
      2x2     = 12 - 2x
          -12  -12
     --------------------
     2x2 - 12 =     -2x
        +2x         +2x
     --------------------
  2x2 + 2x - 12 = 0

  2(x2 + x - 6) = 0

2(x + 3)(x - 2) = 0         

x + 3 = 0 gives solution x = -3
x - 2 = 0 gives solution x = 2

Edwin
AnlytcPhil@aol.com

What is the equation of a line, that passes thru (0,4) and is parallel to the
line 4y-20=3x. And the equation must be in standard form.

4y - 20 = 3x

Place this in standard form

-3x + 4y = 20

Multiply by -1 to make first term positive:

 3x - 4y = -20

Parallel lines in standard form have the same left side. So the
equation you want also has left side

 3x - 4y

Substitute the point (0,4) into that expression

 3(0) - 4(4) 

   0 - 16

     -16

Therefore the right side must be -16.

Answer:

  3x - 4y = -16

Edwin
AnlytcPhil@aol.com

Write an equation for the line containing
1.(3,17) and (-5,-23)
2.(3,-10)(7,-22)

For these you need 
1. the slope formula:

     y2 - y1
m = —————————
     x2 - x1

2.  The point-slope form of the equation of a line:

y - y1 = m(x - x1)

1.(3,17) and (-5,-23)

Here

 (x1,y1) = (3,17) so x1 = 3 and y1 = 17
 (x2,y2) = (-5,-23) so x2 = -5 and y2 = -23

     y2 - y1
m = —————————
     x2 - x1

     (-23) - (17)
m = ——————————————
      (-5) - (3)

     -40
m = ————— = 5
     -8

Now use

y - y1 = m(x - x1)

Substitute only for x1, y1 and m.  Do not substitute anything
for x or y.

y - (17) = 5(x - 3)

  y - 17 = 5x - 15

       y = 5x + 2

2.(3,-10)(7,-22)

This is exactly like the first one.

The answer is 

       y = -3x - 1

Edwin
AnlytcPhil@aol.com

can u help me solvethese 3 problems?
1.  x times 1 over y?

     1     x     1     1·x      x
x · ——— = ——— · ——— = —————— = ———
     y     1     y     y·1      y

2.what is a 16oz of bottle of water costs $1.44.what is the cost per ounce? 

1.44 dollars    144 cents     144   cents      cents
———————————— = ———————————  = ——— · —————— = 9 ————— = 9¢/oz
  16 ounces      16 ounces     16   ounce      ounce


3.Evaluate 3(x3-5x)+6for x=3?

     3(x3 - 5x) + 6

Change the parentheses "{ }" to brackets "[ ]"

     3[x3 - 5x] + 6

In place of each "x" write "(3)"

     3[(3)3 - 5(3)] + 6

(3)3 = 3·3·3 = 27 so replace the (3)3 by 27

     3[27 - 5(3)] + 6

5(3) = 15, so replace the 5(3) by 15

     3[27 - 15] + 6

27 - 15 = 12 so replace the 27 - 15 by 12

     3[12] + 6

3[12] = 36 so replace the 3[12] by 36

      36 + 6

36 + 6 = 42, so replace the 36 + 6 by 42

      42 

Edwin
AnlytcPhil@aol.com

<How to show a*b = b*a?

This page has the proof of that as well as to 
other similar theorems:

http://mathforum.org/library/drmath/view/51563.html

Edwin
AnlytcPhil@aol.com

Find the quotient using long division. (I have to show work.)

(8m^3 + 38m^2 - 6m + 20)/(m + 5)

Answers being: (A)8m^2+2m+4 (B)m^2+2m+8 (C)8m^2-2m+4 (D) m^2+3m+4 
Second question:


Start with this:

       ----------------------
  m + 5) 8m3 + 38m2 - 6m + 20

Divide m into 8m3, which amounts to 8m3/m or 8m2. Place that above
the line above 38m2:

                8m2
       ----------------------
  m + 5) 8m3 + 38m2 - 6m + 20

Multiply 8m2 by the 5, getting 40m2 and place it under the 38m2

                8m2
       ----------------------
  m + 5) 8m3 + 38m2 - 6m + 20 
             + 40m2

Multiply 8m2 by the m, getting 8m3 and place this under the other 8m3

                8m2
       ----------------------
  m + 5) 8m3 + 38m2 - 6m + 20 
         8m3 + 40m2

Draw a line underneath

                8m2
       ----------------------
  m + 5) 8m3 + 38m2 - 6m + 20 
         8m3 + 40m2
         ----------

Subtract vertically by mentally changing the sign of the 
8m3 and the 40m2 and adding, getting 0 and -2m2.  Write only
the -2m2

                8m2
       ----------------------
  m + 5) 8m3 + 38m2 - 6m + 20 
         8m3 + 40m2
         ----------
               -2m2

Bring down the -6m

                8m2
       ----------------------
  m + 5) 8m3 + 38m2 - 6m + 20 
         8m3 + 40m2
         ----------
               -2m2 - 6m 

Divide m into -2m2, which amounts to -2m2/m or -2m. Place that above
the line above " - 6m ":

                8m2 - 2m
       ----------------------
  m + 5) 8m3 + 38m2 - 6m + 20 
         8m3 + 40m2
         ----------
               -2m2 - 6m 

Multiply -2m by the 5, getting -10m and place it under the "- 6m"

                8m2 -  2m
       -----------------------
  m + 5) 8m3 + 38m2 -  6m + 20 
         8m3 + 40m2
         ----------
               -2m2 -  6m 
                    - 10m

Multiply -2m by the m, getting -2m2 and place this under the other -2m2  

                8m2 -  2m
       -----------------------
  m + 5) 8m3 + 38m2 -  6m + 20 
         8m3 + 40m2
         ----------
               -2m2 -  6m 
               -2m2 - 10m

Draw a line underneath:

                8m2 -  2m
       -----------------------
  m + 5) 8m3 + 38m2 -  6m + 20 
         8m3 + 40m2
         ----------
               -2m2 -  6m 
               -2m2 - 10m
               ----------

Subtract vertically by mentally changing the sign of the -2m2 and the -10m
and adding, getting 0 and 4m.  Write only the 4m.

                8m2 -  2m
       -----------------------
  m + 5) 8m3 + 38m2 -  6m + 20 
         8m3 + 40m2
         ----------
               -2m2 -  6m 
               -2m2 - 10m
               ----------
                       4m

Bring down the +20

                8m2 -  2m
       -----------------------
  m + 5) 8m3 + 38m2 -  6m + 20 
         8m3 + 40m2
         ----------
               -2m2 -  6m 
               -2m2 - 10m
               ----------
                       4m + 20

Divide m into 4m, which amounts to 4m/m or "+ 4". Place that above
the line above "+ 20":

                8m2 -  2m +  4
       -----------------------
  m + 5) 8m3 + 38m2 -  6m + 20 
         8m3 + 40m2
         ----------
               -2m2 -  6m 
               -2m2 - 10m
               ----------
                       4m + 20

Multiply 4 by the 5, getting +20, and place it under the other + 20

                8m2 -  2m +  4
       -----------------------
  m + 5) 8m3 + 38m2 -  6m + 20 
         8m3 + 40m2
         ----------
               -2m2 -  6m 
               -2m2 - 10m
               ----------
                       4m + 20
                          + 20

Multiply 4 by the m, getting 4m and place this under the other 4m

                8m2 -  2m +  4
       -----------------------
  m + 5) 8m3 + 38m2 -  6m + 20 
         8m3 + 40m2
         ----------
               -2m2 -  6m 
               -2m2 - 10m
               ----------
                       4m + 20
                       4m + 20

Draw a line underneath

                8m2 -  2m +  4
       -----------------------
  m + 5) 8m3 + 38m2 -  6m + 20 
         8m3 + 40m2
         ----------
               -2m2 -  6m 
               -2m2 - 10m
               ----------
                       4m + 20
                       4m + 20
                       -------

Subtract vertically by mentally changing the sign of the 4m and the +20
and adding, getting 0 and 0.  Write only the second 0.

                8m2 -  2m +  4
       -----------------------
  m + 5) 8m3 + 38m2 -  6m + 20 
         8m3 + 40m2
         ----------
               -2m2 -  6m 
               -2m2 - 10m
               ----------
                       4m + 20
                       4m + 20
                       -------
                             0

Since the remainder is 0, the quotient is 8m2 - 2m + 4

So the correct choice is (C)

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

Multiply: (3y+11)(8y²-2y-9) (I have to show work.) 

  (3y+11)(8y²-2y-9)

Answers being:
(A) 24y^3-6y^2-27y+11
(B) 112y^2-28y-126
(C) 24y^3+28y^2-49y-99
(D) 24y^3+94y^2+49y+99 
Thank you.

Write the second parenthetical expression (8y²-2y-9)
as [(8y²-2y)-9]

(3y+11)[(8y²-2y)-9]

I colored the (8y²-2y) red so you can see that you are using "FOIL"
Consider the parenthetical red factor as just ONE single term.

 FIRSTS    +  OUTERS  +   INNERS     +  LASTS
    |            |           |            |              
3y(8y²-2y) + (3y)(-9) + (11)(8y²-2y) + (11)(-9)

(24y³-6y²) +  (-27y)  +  (88y²-22y)  +  (-99)

   24y³ - 6y² - 27y + 88y² - 22y - 99

Combining like terms and arranging terms in descending order:

    24y³ + 82y² - 49y - 99 

That answer is correct even though you don't have it listed.  My guess
is that it is (C) and you inadvertently reversed the digits of 82 as 28.

Edwin
AnlytcPhil@aol.com

 6(2 + y) - 4 = -10 

6·2 + 6·y - 4 = -10     *remove parentheses using distributive principle

 12  + 6y - 4 = -10

       8 + 6y = -10          *combine 12 and -4 to get 8 on left

   8 + 6y - 8 = -10 - 8      *subtract 8 from both sides

           6y = -18          *combine -10 and -8 on right

           6y    -18
           -- = -----        *divide both sides by -6
            6     6

            y = -3

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

 7(y-2) + 8 = 3(y-4) - 2    
 
7y - 14 + 8 = 3y - 12 - 2   *distributive principle

     7y - 6 = 3y - 14       *combining terms 
 
 7y - 6 + 6 = 3y - 14 + 6   *adding 6 to both sides

         7y = 3y - 8 

    7y - 3y = 3y - 8 - 3y   *subtracting 3y from both sides

         4y = -8            *combining terms

         4y     -8
        ---- = ----         *divide both sides by 4 
         4       4

           y = -2

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

   6(3+5y) + 5(y-8) = 13 

 18 + 30y + 5y - 40 = 13         *distributive principle

          -22 + 35y = 13         *combining terms
   
     -22 + 35y + 22 = 13 + 22    *adding 22 to both sides

                35y = 35         *combining terms

               35y     35
              ----- = ----       *dividing both sides by 35
                35     35

                  y = 1

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


    2(f+3) - 2.7 = 3(f+1.1)

    2f + 6 - 2.7 = 3f + 3.3        *distributive principle

        2f + 3.3 = 3f + 3.3        *combining terms  

  2f + 3.3 - 3.3 = 3f + 3.3 - 3.3  *subtracting 3.3 from both sides

              2f = 3f              *combining terms

         2f - 3f = 3f - 3f         *subtracting 3f from both sides

             -1f = 0               *combining terms

            -1f     0
           ----- = ---             *dividing both sides by -1
            -1     -1

              f = 0 

Edwin
AnlytcPhil@aol.com