How do you solve (2h-3k)square / (10h-15k/6hk) ?
(2h-3k)square / (10h-15k/6hk)
= (2h-3k) * (2h-3k) / [(10h-15k) / 6hk]
You will need to use the rules of multiplying fractions. Switch the divide sign to multiply, and then
invert the denominator of the larger ratio:
(10h-15k)/6hk => you would invert this to become: (6hk) / (10h-15k)
Now rewrite your original expression:
[ (2h-3k) * (2h-3k)] * [ 6hk /(10h-15k) ]
Now factor out 5 from the last term:
[ (2h-3k) * (2h-3k)] * [ 6hk / 5(2h-3k) ]
Now multiply your fractions:
(2h-3k) * (2h-3k) * 6hk ]
-------------------------------
5(2h-3k)
You can now cancel the (2h-3k) from both top and bottom of the fraction, and finish up your work
(2h-3k)^2 / (10h-15k/6hk) =
(2h-3k)^2 / 5(2h-3k/6hk) =
(2h-3k)^2 (6hk) / 5(2h-3k) =
(2h-3k) (6hk) / 5
Comments
(2h-3k)square / (10h-15k/6hk)
= (2h-3k) * (2h-3k) / [(10h-15k) / 6hk]
You will need to use the rules of multiplying fractions. Switch the divide sign to multiply, and then
invert the denominator of the larger ratio:
(10h-15k)/6hk => you would invert this to become: (6hk) / (10h-15k)
Now rewrite your original expression:
[ (2h-3k) * (2h-3k)] * [ 6hk /(10h-15k) ]
Now factor out 5 from the last term:
[ (2h-3k) * (2h-3k)] * [ 6hk / 5(2h-3k) ]
Now multiply your fractions:
(2h-3k) * (2h-3k) * 6hk ]
-------------------------------
5(2h-3k)
You can now cancel the (2h-3k) from both top and bottom of the fraction, and finish up your work
(2h-3k)^2 / (10h-15k/6hk) =
(2h-3k)^2 / 5(2h-3k/6hk) =
(2h-3k)^2 (6hk) / 5(2h-3k) =
(2h-3k) (6hk) / 5