Algebra II Equation and inequalities?

1) 2(2x-4) + 5 - 10x = 8 - 3x + 2

by simplifying and rearranging terms, we get,

4x-10x+3x = 8+2-5+4

thus, -3x = 9

therefore, x= -3

2) 2/5x - 1/10 = -x/2 + 2

multiplying throughout by 10x, we get,

4 - x = -5 x^2 + 20x

by rearranging,

5x^2 - 21x + 4 = 0

solving this quadratic equation, we get,

x = 4 or x = 1/5

3) |2x-25| = 52

to remove mod, square both sides. so we get,

(2x - 25)^2 = 52^2

4x^2-100x+625 = 2704

thus, 4x^2 -100x - 2079= 0

solving this quadratic equation,

x= 38.5 or x= -13.5

4) -|1-x| - 5 = -20

rearranging,

-|1-x| = -20 + 5

to remove mod, square both sides. so we get,

(1-x)^2 = -15^2

1- 2x + x^2 = 225

rearranging,

x^2 - 2x - 224=0

solving this quadratic equation, we get,

x=16 or x=-14

5) |2x-6|+3 ≤ 21

|2x-6| ≤ 21-3

|2x-6| ≤ 18

to remove mod, square both sides. so we get,

(2x-6)^2 ≤ 324

4x^2 - 24x + 36 ≤ 18 ...................... ................. ............ (eqn 1)

now to solve the inequality, lets first solve the quadratic equation as

4x^2 - 24x + 36= 18

therefore, 4x^2 - 24x + 18=0

we get solution as,

x= 5.12 or x= 0.878

Now between 0.878and 5.12, the function will either be

always greater than 18, or

always less than 18.

to check for our condition of less than 18, let's pick a value in-between and substitute in "eqn 1" above.

say x=1

so,

4x^2 - 24x + 36 = 4 -24 + 36 = 16 < 18

so solution of x lies in between x= 0.878 and x= 5.12

1) [ 3/5 (x-12) > x - 24 ] => 3x-36>5x-a hundred and twenty => eighty four>2x =>forty two>x 2) 6[ 5y - (3y - 1)] > 4 (3y - 7) => 6[2y+1]>12y-28 =>12y+6>12y-28 for all values of y this inequality keep real.

I don't know the exact answer, but use order of operations