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what class is this for 

= k2ho76

Solve for y:
2 sqrt(2) y-3 = 5 1/sqrt(2) y+1

Rationalize the denominator. (5 y)/sqrt(2) = (5 y)/sqrt(2)×(sqrt(2))/(sqrt(2)) = (5 y sqrt(2))/(2):
2 sqrt(2) y-3 = (5 sqrt(2) y)/(2)+1

Put each term in (5 y sqrt(2))/2+1 over the common denominator 2: (5 y sqrt(2))/2+1 = (5 sqrt(2) y)/2+2/2:
2 sqrt(2) y-3 = (5 sqrt(2) y)/2+2/2

(5 sqrt(2) y)/2+2/2 = (5 sqrt(2) y+2)/(2):
2 sqrt(2) y-3 = (5 sqrt(2) y+2)/(2)

Multiply both sides by 2:
2 (2 sqrt(2) y-3) = (2 (5 sqrt(2) y+2))/(2)

(2 (5 sqrt(2) y+2))/(2) = 2/2×(5 sqrt(2) y+2) = 5 sqrt(2) y+2:
2 (2 sqrt(2) y-3) = 5 sqrt(2) y+2

Expand out terms of the left hand side:
4 sqrt(2) y-6 = 5 sqrt(2) y+2

Subtract 5 sqrt(2) y from both sides:
(4 sqrt(2) y-5 sqrt(2) y)-6 = (5 sqrt(2) y-5 sqrt(2) y)+2

4 (sqrt(2) y)-5 (sqrt(2) y) = -(sqrt(2) y):
-sqrt(2) y-6 = (5 sqrt(2) y-5 sqrt(2) y)+2

5 sqrt(2) y-5 sqrt(2) y = 0:
-(sqrt(2) y)-6 = 2

Add 6 to both sides:
(6-6)-sqrt(2) y = 2+6

6-6 = 0:
-sqrt(2) y = 2+6

2+6 = 8:
-sqrt(2) y = 8

Divide both sides of -sqrt(2) y = 8 by -sqrt(2):
-(sqrt(2) y)/(-sqrt(2)) = 8/(-sqrt(2))

(-sqrt(2))/(-sqrt(2)) = 1:
y = 8/(-sqrt(2))

Multiply numerator and denominator of 8/(-sqrt(2)) by -1:
y = (-8)/sqrt(2)

Rationalize the denominator. (-8)/sqrt(2) = (-8)/sqrt(2)×(sqrt(2))/(sqrt(2)) = (-8 sqrt(2))/(2):
y = (-8 sqrt(2))/(2)

(-8)/2 = (2 (-4))/2 = -4:
Answer: |y = -4 sqrt(2)