Pre-Algebra 8th Grade

\(\frac{1}{3h}-(\frac{2}{3h}-3)=\frac{2}{3h}-6\)

expand brackets

\(\frac{1}{3h}-\frac{2}{3h}+3=\frac{2}{3h}-6\)

multiply by 3h

\(1-2+3(3h)=2-6(3h)\)

take \(h\)s to one side (ie -3(3h))

\(1-2=2-6(3h)-3(3h)\)

-2

\(1-2-2=-6(3h)-3(3h)\)

times out brackets

\(1-2-2=-18h-9h\)

combine

\(-3=-27h\)

\(\times\)-1

\(3=27h\)

multiply by \(\frac{1}{27}\)

\(\frac{3}{27}=\frac{27h}{27}\)

simplify

\(h=\frac{3}{27}\)

common factors

\(h=\frac{3(1)}{3(9)}\)

cancel out

\(h=\frac{1}{9}\)

ta dah!

Solve for h:
h/3-4 ((2 h)/(3)-3) = (2 h)/(3)-6

Put each term in (2 h)/(3)-3 over the common denominator 3: (2 h)/(3)-3 = (2 h)/3-(9)/3:
h/3-4(2 h)/3-(9)/3 = (2 h)/(3)-6

(2 h)/3-(9)/3 = (2 h-9)/3:
h/3-4(2 h-9)/3 = (2 h)/(3)-6

h/3-(4 (2 h-9))/3 = (h-4 (2 h-9))/3:
(h-4 (2 h-9))/3 = (2 h)/(3)-6

-4 (2 h-9) = 36-8 h:
(h+36-8 h)/3 = (2 h)/(3)-6

h-8 h = -7 h:
(-7 h+36)/3 = (2 h)/(3)-6

Put each term in (2 h)/(3)-6 over the common denominator 3: (2 h)/(3)-6 = (2 h)/3-(18)/3:
(36-7 h)/3 = (2 h)/3-(18)/3

(2 h)/3-(18)/3 = (2 h-18)/3:
(36-7 h)/3 = (2 h-18)/3

Multiply both sides by 3:
36-7 h = 2 h-18

Subtract 2 h from both sides:
36+(-7 h-2 h) = (2 h-2 h)-18

-7 h-2 h = -9 h:
-9 h+36 = (2 h-2 h)-18

2 h-2 h = 0:
36-9 h = -18

Subtract 36 from both sides:
(36-36)-9 h = -18-36

36-36 = 0:
-9 h = -18-36

-18-36 = -54:
-9 h = -54

Divide both sides of -9 h = -54 by -9:
(-9 h)/(-9) = (-54)/(-9)

(-9)/(-9) = 1:
h = (-54)/(-9)

The gcd of -54 and -9 is -9, so (-54)/(-9) = (-9×6)/(-9×1) = (-9)/(-9)×6 = 6:
Answer: |h = 6