pre-algerba hw help?

Question

solve 3/4r-1/2= 1 1/4r+8s for r - it's one n one-fourth
solve 3/2k+L^2 =6L2 for k - it's L to the second power
solve 4m-2n^2 = 2m-72 for m - it's 2 n to the second power
solve z^2 =y = 5y for z - z to the second power

thank you sooooooo much

Answer 1

3/4r-1/2=1
adding 12/
3/4r=3/2
multiplying by 4/3
r=2

2.3/2k+L^2=6L^2
adding-L^2
3/2k=5L^2
dividing by 5
3/10 k=L^2
multiplyingby10/3
k=(10/3)l^2

3.4m-3n^2=2m-72
adding -2m+3n^2
2m=3n^2-72
dividing by2
m=(1/2)3n^2-72

4.z^2=y+5y
z^2=6y
z=sq.rt.(6y)

Answer 2

solve 3/4r-1/2= 1 1/4r+8s for r - it's one n one-fourth
3/4(r) -(5/4)r = 8s +1/2
-(1/2)r = 8s +1/2
r = -4s -1

solve 3/2k+L^2 =6L2 for k - it's L to the second power
(3/2)k +L^2 = 6L^2
(3/2)k = 5L^2
k = (2/3)(5L^2) = (10 L^2)/3

solve 4m-2n^2 = 2m-72 for m - it's 2 n to the second power
4m -2m =2n^2 -72
2m-m =n^2 -36
m= n^2 - 36

solve z^2 =y = 5y for z - z to the second power
z^2 + y = 5y
z^2 = 4y
z = +/- 2*sqrt(y)

Answer 3

3/4r - 1/2 = 5/4r + 8s
-r = 8s +.5; r = -(8s +1/2)

3/2k+ L^2 = 6L^2
3/2 k = 5L^2; k = 10/3L^2

4m - 2n^2 = 2m-72
2m = 2n^2 - 72; m = n^2 - 36

I assume this is supposed to have been z^2 + y = 5y
z^2 = 4y; z = 2 * Sqrt(y)