Inverse addition is repeated addition?

Consider the general case: $\frac {1}{n} - \frac {1}{n+m} = \frac { (n+m) - (n)}{n(n+m)} = \frac {m}{n(n+m)}$

Then

$\frac {m}{n} - \frac {m}{n+m}= \frac {m^2}{n(n+m)} = \frac {m}{n} \times \frac {m}{n+m}$

, in this case $m=3,n=13$ , so it's true! Hi Kevin!

we can simply solve both sides of equation the answer is 9/208=9/208

Why is this lvl. 3. lol, nice problem, but just compute each side. (3/13=48/208)-(3/16=39/208)=9/208 and then (3^2/13*16)=9/208 and 9/208=9/208 therefore the statement is true.

$\frac{3}{13} - \frac{3}{16} = \frac{48}{208} - \frac{39}{208} = \frac{9}{208}$ , and $\frac{3}{13} \times \frac{3}{16} = \frac{9}{208}$ . So they are same number.

Find LCD on both sides of the equation. What I usually do to find this is just multiply the two denominators together, and then check to see if half the product of the two numbers is divisible by both factors to ensure that the result is irreducible.

13*16 = 208, which isn't divisible by BOTH 13 and 16, so it is the LCD.

(3 * 16)/(13 * 16) - (3 * 13)/(13 * 16) = 9/(13 * 16)

Since the denominators are all equal, set the numerators equal on each side once you have the LHS under one denominator.

48/208 - 39/208 = 9/208 9/208 = 9/208, so it's true.

(3/13) - (3/16) = [3(16) - 3(13)]/(13)(16) = (3)(3)/(13)(16) = (3/13) X (3/16)

So , the equation is true

3/13 - 3/16 = 3(16)- 3(13)/ 13 16 = 3( 16-13)/13 16 = 9/13*16.

3X3/13 16 = 9/13 16