h/3 = 17 i dont understand how the left side cancels out by multiplying leaving only h

Question

h/3 = 17 , i know to just cancel out the 3 on the left and multiply by 3 on the right to get h=51 but how does the 3 on the left of the equation cancel out? Im multiplying on the left so shouldn't i end up with more then just h?

Answer 1

A number may be multiplied by the "multiplicative identity" (the number 1) without changing the original number.  That's why both sides of an equation by be multiplied by the same number without changing the equation.

 

         if a = b   then   (3/3)*a = b       and so   3*a = 3*b

 

Your problem starts with   h/3 = 17   then (3/3)*(h/3) = 17    and also    3*(h/3) = 3*17

 

Rearranging that last equation,     (3/3)*h = 3*17        but 3/3  = 1     that's what 'cancels out' means (it goes away)

   So, we are left with   h = 3*17 = 51

 

Hope this helps!

Answer 2

h/3=17

the left side is the same as h(1/3)

h(1/3)=17

multiply both sides by 3

h(1/3)(3)=17(3)

h[(1/3)*(3)]=51

h[1]=51

h=51 because (1/3)*(3)=1; they are reciprocals of each other

another way to look at it is that "one operation undoes the other"

h is being divided by 3 and multiplication by 3 undoes division by 3

 

Answer 3

h/3 can be thought of as h divided by 3. The opposite of dividing by 3 is multiplying by 3. Dividing and multiplying are inverse operations and always cancel out to give you the number 1. 1 times h would then be written as just h.

 

You can also think about it as cancellation of fractions. For example:

 

5/7 * 7 is the same as 5/7 * 7/1 and you can cancel out the 7 from both the numerator and denominator and then multiply to get 5/1 or 5.....so

 

h/3 * 3/1 can cancel out the 3 from top and bottom and then multiply and get h/1 or just h.

 

Hope this helped.