CPhill

Simplify these as

(1/2)x - 2y =  -12      (1)

12x + 7y =  97         (2) 

Multiply (1) through by 2

x  - 4y  = -24   →  x =  4y - 24    (3)

Sub (3)  into (2) for x

12 (4y -24) + 7y  = 97

48y - 288 +7y   = 97

55y  = 385

y = 385/ 55  =   7

x = 4(7) - 24  =  4

(x, y)  = (4, 7)

Simplify as

(2/5)x - 1 = -3x + 7

2x - 5  = -15x + 35

17x  = 40

x = 40 / 17  = r

Since r >  1, this  series does not converge to any value

\(\frac{a + 2}{a + 1} = \frac{b}{8} \)

Rewrite  as

8(a + 2)  / (a + 1)  =  b

As a approaches  -infinity and  + infinity,  b  approaches 8

a         b

0       16

1       12

3       10

7        9

-2       0

-3       4

-5       6

-9       7

\(\frac{3-z}{z+1} \ge 2 + \frac{z}{z-7} \)

Rewrite  using x for z

[3 - x ] / [ x + 1 ] - 2 -[ x ]  / [ x - 7]   ≥  0     

Multiply through by ( x + 1) ( x -7)     and  make this  an equality

(3 - x) (x - 7)   - 2 (x + 1)(x -7) - x ( x + 1)  =  0

-x^2+ 10x  - 21  - 2 (x^2 - 6x - 7)  - x^2 - x   =  0      

-x^2 + 10x - 21  - 2x^2 + 12x + 14 - x^2 - x =  0

-4x^2 + 21x  - 7  =   0

Using the Q Formula

x =  [ -21 ± sqrt [ 21^2  - 4(-4)(-7) ] ] / -8

x = [-21 + sqrt [ 329 ]  ] / -8   = [21 - sqrt [329] ]/ 8     ≈  .358     

or   

x =  [ -21 - sqrt [329] [ / -8  =   [ 21 + sqrt [ 329] ] /  8 ≈   4.89

We  have these intervals to consider  ( x not defined at -1, 7 )

Interval                  Test Point        Test point makes the original inequality

(-inf , -1)                    x =  -2                       False

( - 1 , .358]                x = 0                          True

(.358  , 4.89)             x = 1                         False

[ 4.89 , 7 )                 x = 5                          True

(7, inf)                       x = 8                          False

z =  (-1 , [21 - sqrt [329] ] / 8 ]   U  [  [21 + sqrt [ 329] ] / 8 , 7 )

4t^2  ≤ -9t + 12 - 14t + 28      simplify as

4t^2 + 23t - 40   ≤  0          (1)            set  as an equality

4t^2 + 23t  -40  =  0

Using the Q Formula

t =    [  -23   ± sqrt ( 23^2  - 4 * 4 * -40 )  / 8

t = [ - 23 ± sqrt [ 1169 ]  / 8

t  ≈  - 7.14      or    t  ≈  13.99

0  lies in the  interval between these two  values......if t = 0 , then (1)  is  true 

Then  t  = (  [-23 - sqrt (1169) ] / 8   ,   [ -23 + sqrt (1169) ] / 8  )

x^2 + 4x -12 le  3x + 7

x^2 + x -19 le 0      set up as an equality  , complete the square on x

x^2 + x  + 1/4  =   19 + 1/4

(x + 1/2)^2  =  77/4     take both roots

x =  [ -1 - sqrt 77 ]  /  2   ≈   -4.89    or        x =  [ -1+ sqrt 77 ] / 2    ≈ 3.89

Test 0 in  the original inequaity  and it makes it true

Then  the  answer  comes from  the interval   ( [ -1 - sqrt(77) ] / 2 , [ -1 + sqrt (77) ]  /2 )

n^2 < 144 + 40n

n^2 - 40n  - 144   < 0

Make this  an equality   and  complete the square on n

n^2  - 40n   = 144

n^2  -40n + 400  = 144 + 400

(n - 20)^2 =  544       take  both roots

n - 20  = -sqrt (544)

n = 20  - sqrt (544)           and      n=   20 +  sqrt (544)

n ≈ -3.32                                       n ≈  43.32

The   answer  either comes from  the interval between these two  numbers or from  the two intervals on  either side of these two numbers

n = 0   is in the first interval and   it solves the inequality  because 0 < 144 + 40(0) is  true

So

The answer is   20 -sqrt 544 < n < 20 + sqrt 544

I don't find an easier way, either, EP.

I did it with just some  Geometry

Here's my solution  (but it  might not  be  any  easier)

Like EP, I found BT to be 30

We can find the coordinates of T  by finding the intersection of two circles  (this is the key !!!)

One with a center of (0,30) and a  radius of 30

One with a center of (15,0) and a  radius of 15

The equations are

x^2  + ( y - 30)^2  = 900    →  x^2 + y^2 - 60y   =  0   (1)

(x - 15)^2 + y^2  = 225     →  x^2 - 30x + y^2  = 0     (2)

Subtract these

-60y + 30x  =  0

x = 2y

Sub  this into (2)

4y^2 - 60y + y^2  = 0

5y^2 -60y  =  0

y - 12  = 0

y =12

x = 2(12)   = 24

T = (24,12)

Slope   of   line through BP = ( 30 - 12) / ( 0 - 24) = -3/4

Equation of this  line  is  y =(-3/4)x + 30

Now, as EP  said, put this into  the  equation of the circle  x^2 + y^2  =900  and find the x coordinateof P  as  28.8

And  -(3/4)(28.8) + 30  =   8.4 = the y coordinate of P

TP = sqrt [ (28.8 -24)^2  + (12 - 8.4)^2 ]  = sqrt [ 4.8^2  + 3.6^2 ]  =  sqrt (36)   = 6

Nice job, EP!!!

\($x \mathbin{\spadesuit} y = \frac{x^2}{y}$ for all $x$ and $y$ such that $y\neq 0$ \)

\(Find all values of $a$ such that $a \mathbin{\spadesuit} (a + 1) = 9$. \)

a (spade) (a + 1)  =       a^2 / (a + 1)   =  9

a^2 = 9(a + 1)

a^2  = 9a + 9

a^2  - 9a   - 9    =  0

a^2 - 9a =   9             complete the square on  a

a^2  - 9a  + 81/4  =  9 + 81/4

(a -  9/2)^2  =  117  / 4             take  both roots

a - 9/2  = sqrt (117) / 2

a = [ 9 + sqrt (117) ] / 2         or        a =   [ 9 - sqrt (117) ]  / 2

Not possible

If the axis  of symmetry is x  =1 , then the parabola   turns  either upward or  downward

If the x intercept is at 0, then y at that point  also = 0  

Then  we can't also have a y intercept at -1 because this parabola would fail the vertical line test  because it would have two y intercepts  at (0, -1)  and (0, 0)