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write and find 2 mixed numbers so the rules apply,

1:one number is larger than the other by 4 2/3
2:The difference of the numbers is equal to the smaller number.
3:The sum of the numbers is natural number.

 Nov 12, 2017
edited by Guest  Nov 12, 2017

Best Answer 

 #1
avatar+7352 
+2

Let's call the larger number  L , and the smaller number  S .

 

1.     L  =  S + 4\(\frac23\)

 

2.     L - S  =  S

                                               Substitute  S + 4\(\frac23\)  in for  L .

S + 4\(\frac23\)  -  S  =  S

                                               Cancel the  S  and  -S  on the left side.

\(4\frac23\)  =  S

 

 

L  =  S + 4\(\frac23\)

                                               Substitute  4\(\frac23\)  in for  S .

L  =  4\(\frac23\)  +  4\(\frac23\)

 

L  =  8 +  \(\frac43\)

 

L  =  8 +  \(\frac33\) +  \(\frac13\)

 

L  =  9\(\frac13\)

 

And...let's check that their sum is a natural number.

 

L + S   =   9\(\frac13\)  +  4\(\frac23\)   =   13 \(\frac33\)   =   14   Yep!  smiley

 Nov 12, 2017
 #1
avatar+7352 
+2
Best Answer

Let's call the larger number  L , and the smaller number  S .

 

1.     L  =  S + 4\(\frac23\)

 

2.     L - S  =  S

                                               Substitute  S + 4\(\frac23\)  in for  L .

S + 4\(\frac23\)  -  S  =  S

                                               Cancel the  S  and  -S  on the left side.

\(4\frac23\)  =  S

 

 

L  =  S + 4\(\frac23\)

                                               Substitute  4\(\frac23\)  in for  S .

L  =  4\(\frac23\)  +  4\(\frac23\)

 

L  =  8 +  \(\frac43\)

 

L  =  8 +  \(\frac33\) +  \(\frac13\)

 

L  =  9\(\frac13\)

 

And...let's check that their sum is a natural number.

 

L + S   =   9\(\frac13\)  +  4\(\frac23\)   =   13 \(\frac33\)   =   14   Yep!  smiley

hectictar Nov 12, 2017

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