p (2/5) -5 (1/5) +6=0

p^(2/5)-5^(1/5)+6=0

p= 32 and 243

(1/2)x-1=(2/3)y

(20/9)=(1/3)x+(1/3)y

x=(14/3)

y=2

Please help me solve these before tomorrow? My teacher gave me the answers but I can't figure out how to get to that point.

Simplify the following:
32^(2/5)-5^(1/5)+6

32^(2/5)  =  (32^2)^(1/5)  =  (2^10)^(1/5)  =  2^2:
2^2-5^(1/5)+6

2^2 = 4:
4-5^(1/5)+6

4-5^(1/5)+6 = 10-5^(1/5):
Answer: |  10 - 5^(1/5)

Simplify the following:
243^(2/5)-5^(1/5)+6

243^(2/5)  =  (243^2)^(1/5)  =  (3^10)^(1/5)  =  3^2:
3^2-5^(1/5)+6

3^2 = 9:
9-5^(1/5)+6

9-5^(1/5)+6 = 15-5^(1/5):
Answer: |  15 - 5^(1/5)

(1/2)*(14/3)-1=(2/3)*2=4/3=4/3

(1/3)*(14/3)+(1/3)*2=20/9

Simplify the following:
(14/3)/3+2/3

(14/3)/3 = 14/(3×3):
14/(3×3)+2/3

3×3  =  9:
14/9+2/3

Put 14/9+2/3 over the common denominator 9. 14/9+2/3  =  14/9+(3×2)/9:
14/9+(3×2)/9

3×2  =  6:
14/9+6/9

14/9+6/9  =  (14+6)/9:
(14+6)/9
Answer: |  20/9

p^(2/5) - 5p^(1/5) + 6  = 0     factor as

[p^(1/5]  - 3)  [ p^(1/5) - 2]  = 0

Set each factor to 0

p^(1/5)  - 3  = 0      add 3 to both sides

p^(1/5)  = 3       take each side to the 5th power

p = 3^5   = 243

p^(1/5) - 2  =   0      add 2 to both sides

p^(1/5)  = 2    take each side to the 5th power

p  = 2^5   = 32

I'm assuming that you want to solve this system :

(1/2)x-1=(2/3)y

(20/9)=(1/3)x+(1/3)y

Multiply the top equation through by 6   and the bottom one through by 9

3x - 6   =  4y     →  3x - 4y   = 6      (1)

20 = 3x +3y    →   3x  +3y  = 20    (2)

Multiply (2)   by -1 all the way through

-3x - 3y  = -20      add this to (1)

-7y = -14      divide both sides by - 7

y = 2

And using (1) to   find x, we have

3x - 4(2)  = 6

3x - 8 =  6     add 8 to both sides

3x = 14    divide both sides by 3

x = 14/3