Question

40 PTS + Brainliest

Answer 1

1.

rational numbers are numbers that can be written in form a/b where a and b are integers (-4,-3,-2,-1,0,1,2,3, etc) and b isn't 0

so for case $\frac{a}{b}+\frac{c}{d}$

we need to make the denomenators the same

so for $\frac{a}{b}+\frac{c}{d}$, multiply left one by d/d and multiply right one by b/b to get

$\frac{ad}{bd}+\frac{cb}{bd}=\frac{ad+cb}{bd}$

therefore $\frac{a}{b}+\frac{c}{d}=\frac{ad+cb}{bd}$[/tex] is the rule

2. do the same as previous but do minus

so for case $\frac{a}{b}-\frac{c}{d}$

we need to make the denomenators the same

so for $\frac{a}{b}-\frac{c}{d}$, multiply left one by d/d and multiply right one by b/b to get

$\frac{ad}{bd}-\frac{cb}{bd}=\frac{ad-cb}{bd}$

therefore $\frac{a}{b}-\frac{c}{d}=\frac{ad-cb}{bd}$[/tex] is the rule

3. not sure, I hope this is right

so for case $\frac{a}{b}+\frac{c}{d}=0$

we need to make the denomenators the same

so for $\frac{a}{b}+\frac{c}{d}$, multiply left one by d/d and multiply right one by b/b to get

$\frac{ad}{bd}+\frac{cb}{bd}=\frac{ad+cb}{bd}=0$

if we multiply both sides by bd we get

ad+cb=0

therefore, ad=-cb for the sum to always equal 0

4.  

$(\frac{a}{b})(\frac{c}{d})=1$

$\frac{ac}{bd}=1$

multiply both sides by bd

ac=bd for the product of the 2 fractions to equal 1