CPhill

x + y + 2z  = 1        (1)

-2x + 4y + 2z = 1     (2)

y + z  = 1                 (3)

Multiply (3)  by 2  and rearrange  → 2z = 2 - 2y     (4)

Sub (4)  into  (1)    and simplify

x + y + [2 - 2 y]  = 1

x - y  = -1     (5)

Multiply (2) by -1   and add to (1)

3x - 3y   = 0     divide both sides by 3

x - y  = 0      (6)

But.....we have a contradiction  between  (5)  and (6)......x - y  cannot take on two values.....thus.....this system has no solution

Let  √[4x - 3]  =  q^2

So we have

q^2  +  10/q^2   = 7      multiply both sides by q^2

q^4 + 10  = 7q^2

q^4 - 7q^2 + 10   = 0    factor

(q^2 - 5) ( q^2 - 2)  = 0

So either

q^2  = 5     or    q^2   =  2    .....so....

√[4x - 3]  = 5        square both sides

4x - 3   = 25     add 3 to both sides

4x = 28

x = 7

Or

√[4x - 3]  = 2     square both sides

4x - 3  = 4      add 3 to both sides

4x = 7

x = 7/4

Check  x = 7

√[4x - 3] +  10 / √[4x - 3] = 7  ??

√[4*7 - 3] + 10 / √[4*7 - 3]  = 7  ??

5 + 10/5  = 7      is true

Check  x  = 7/4

√[4(7/4) - 3] +  10 / √[4(7/4) - 3] = 7   ??

√[7-3] +  10 / √[7 - 3]] = 7  ??

2 + 10/2   = 7   is also true

So  x   =  7    or   x  = 7/4

5x + 3z  = 1       (1)

-x + y + z  = 0     (2)

3y + 2z= 5        (3)

Uising (1)  z = [ 1 - 5x] /3      (4)

Using (3)   z =  [5 - 3y] / 2     (5)

Equating (4)   and (5)  we have

[1 - 5x]/ 3  = [ 5-3y ] / 2     cross-multiply

2[ 1 - 5x] = 3 [ 5 - 3y]    simplify

2 - 10x  = 15 - 9y

9y  = 13 + 10x

y = [13 + 10x]/ 9     (6)

Subbing (4) and (6)  into (2), we have

-x +  [13 + 10x]/9  + [1 -5x]/3  = 0

-9x + 13 + 10x + 3  - 15x  = 0

-14x = -16

x = 8/7

And

y = [13 + 10(8/7)]/ 9  = 19/7

And

z = [ 1 - 5(8/7)] /3    =  - 11/7

So   {x, y, z}    =  { 8/7, 19/7, -11/7}

Hey, Mellie......it's tough to see the whole problem, but I guess you want the solutions

[ and their product ] for :

6x^2 + 10x  = 4 - 10x - 6x^2   simplify

12x^2 + 20x - 4   = 0     divide through by 4

3x^2 + 5x - 1   = 0 

The solutions are  :

[ -5 + √37] / 6     and   [ -5 - √37] / 6

The product of the solutions   =   ( [ -5 + √37] / 6] ) *  ( [ -5 - √37] / 6 )  =  [25 - 37] / 36  =  -12/36  =   -1/3

If this is not what you asked for......sorry  !!!!

Centillion  is either 1 followed by 303 zeroes [US definition]  or 1 followed by 600 zeroes  [ Great Britain]

There are many "large" numbers that are greater.......check out "googleplex", or  "Graham's Number"

Do you want to know where  f(x)  = g(x) ???

Here's a graph that might help : https://www.desmos.com/calculator/chlxcf4kpw

There are two intersection points :   at about (-2.46673, 0)  and at (2, 0)

x^2 + y^2 = - 8 – 9x    move the -9x term to the left side  [it changes signs]

x^2 + 9x + y^2   = -8        complete the square on x

x^2 + 9x + 81/4 + y^2  = -8 + 81/4      simplify

(x + 9/2) + y^2   =  -32/4 + 81/4

(x + 9/2)^2  + y^2  = 49/4

The right side is the radius squared......so......take the square root to find the radius  = sqrt(49/4)  = 7/2  = 3.5  units

Here's a graph of the original function :  https://www.desmos.com/calculator/2s9iegula0

83-__=11     subtract 83 from both sides

-___  = - 72

So ___  =  72

19.   Domain  =  (-4, 6]

        Range  =  [-4, 4]

20. Domain  =  [-3, 4 ]   I'm assuming that the function becomes a vertical line when x = 3, but it's hard to tell from the picture.....

Range  =  ( -infinity,   4  ] 

Thanks, qazwsxedc123,  for those good answers  !!!!!