can someone help me with these?

Question

1) Solve the following equations algebraically. Show all work.

a) cuberoot(x) -5=3
b) x 3/2=27
c) 3/4=1 -3x-2/x+1

Answer 1

a) cbrt(x) -5=3
cbrt x = 8
x = 8^3 = 512

b) x 3/2=27
is that x^(3/2) or 3x/2?
x^(3/2) = 27
x^(1/2) = cube root of 27
x^(1/2) = 3
x = 3^2 = 9

c) 3/4=1 -3x-2/x+1
Do you mean
3/4 = 1 - [(3x-2)/(x+2)]
3/4 - 1 = -(3x-2)/(x+2)
-1/4 = -(3x-2)/(x+2)
cross multiply
-1(x + 2) = (-4)(3x - 2)
-x - 2 = -12x + 8
11x = 10
x = 10/11

If I didn't correctly guess at what you were asking, you'll need to repost the questions with more notations so we know what you want.

Answer 2

x^(1/3) - 5 = 3

x^(1/3) = 8

cube both sides

x = 8^3 = 512

x^(3/2) = 27 take each side to the 2/3 power, leaving x alone on the left:

x = 27^(2/3) = 9 (cube root of 27 = 3, then squared = 9)


Maybe you mean 3/4 = 1 - 3*x - 2/(x + 1)

Multiply through by (x+1):

(3/4)*(x+1) = (x+1) - 3*x*(x+1) - 2

expand product terms:

(3/4)*x + (3/4) = x + 1 - 3*x^2 - 3*x - 2

Collect terms:

3*x^2 + 0.75*x + 1.75 = 0

Solve using quadratic equation

Answer 3

a) cuberoot(x) - 5 = 3
cuberoot x = 8
x = 8^3 = 512

b) x 3/2 = 27
x = 27*2/3 = 18

c) 3/4 = 1 - 3x - 2/x + 1 (multiply both terms by x)
3x/4 = x - 3x^2 - 2 + x
3x^2 + 3x/4 - 2x + 2 = 0 (multiply both terms by 4)
12x^2 + 3x - 8x + 8 = 0
12x^2 - 5x + 8 = 0
x = [5 +- Sq Rt (25 - 384)]/24
As there are no Sq Rt of negative numers, x will never = 0

Answer 4

a)
(x)^1/3 - 5 = 3
x^1/3 = 8
(x^1/3)^3 = 8^3
x^(1/3*3) = 8^3
x^(1) = 512
x = 512

b)
x * 3/2=27
x = 27 * 2/3
x = 54/3
x = 18

c)
I am not sure where the parentheses are supposed to go in this one, can you be more specific?