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egoroff_w [7]
3 years ago
7

Can someone help me with this problem plz

Mathematics
2 answers:
myrzilka [38]3 years ago
8 0
5/3+35
5+105 over 3
x= 36 2/3
Nutka1998 [239]3 years ago
6 0
The answer is: 110/3 = 36 ⅔ ; or, write as: 110/3 ; or, write as 36.66666667;
__________________________
Explanation: (Two methods):
_____________________________________________
Method 1):
____________
Since: "(1/3)x" and "7x" are "like terms" , add them together, to get:
____________________________________________________
"(7 + 1/3)x "
________________
→ 7 = 7/1 = ?/3 ; → (7*3)/ (1*3) = 21/3;  
→ 7 = 21/3 ; 

→ 7  1/3 = 7 + 1/3 = 21/3 + 1/3 = 22/3; 

→ (7 1/3)x  = (22/3) x

→ We have: "(22/3)x" ;  x = 5 ;

→  So substitute the "5" for "x"; and solve the expression:
→  (22/3)x = (22/3)*5 = (22*5) / (3*1) = 110/3 = 36 ⅔
____________
Method 2):
________________________________________
Either: 1):
__________________________________
 Given:  (1/3)x + 7x ;  x = 5; Solve for "x" .
_____________________________________________
    → Substitute the given value, "5", for "x"; to solve the expression:
____________________________
    →  (1/3)*5  +  7* 5  = (5/3)  + 35 = 1 ⅔ + 35 = 36 ⅔ ; or write as "110/3"; or, write as: 36.66667.
___________________
Or:  2):  
_____________________________________
 Given:  (1/3)x + 7x ;  x = 5; Solve for "x" .
_____________________________________________
    → Substitute the given value, "5", for "x"; to solve the expression:
Given:  (1/3)x + 7x ;  x = 5; Solve for "x" .
    →{Note: Similar to, a bit different from, the above method):
_____________________________________________
  →Here is what is done differently:
    → (1/3)x + 7x → = (x/3) + 7x) ;
_______________________________
Now, we have: 
_______________________________
→ (x/3) + 7x ; x = 5; → Plug in the value, "5" for the "x" values:
_______________________________
→ (5/3) + (7*5) = (5/3) + 35 = 1 ⅔ + 35 = 36 ⅔ ; or write as "110/3"; or, write as: 36.66667.
_________________________________
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\qquad\qquad\huge\underline{{\sf Answer}}♨

Let's solve for x ~

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