Question 264508
Just break down what you are given, like this:


One number (make it "x") 
is (=) 
four greater (4 +) 
than another number (make it "y"):
EQUATION:   x = 4 + y



when three times the smaller number (the smaller number is y, so 3 times y: 3y) is added (3y +)to 
twice the larger number (2 times the larger:  2x) 
the result is (=) 43.
EQUATION:  3y + 2x = 43



Now you have TWO equations:  
x = 4 + y
3y + 2x = 43


Solve this system.  Do you see that x = 4 + y?   SO instead of "X" in the 2nd equation, put in 4 + y, like  this:


3y + 2x = 43 (original equation)
3y + 2(4 + y) = 43 (Plugged in 4 + y for the "X" variable)
3y + 8 + 2y = 43 (distributed 2 to the 4 and 2 to the y)
5y + 8 = 43 (combined like terms)
5y = 35 (subtracted 8 from both sides to isolate the y)
y = 7 (divided both sides by 5 to further isolate the y)


Now we know y = 7.  


Plug in 7 for the "Y" variable in the 1st equation

x = 4 + Y
x = 4 + 7
x = 11.  


Soooooooo one number is 7 and one number is 11.


11 is 4 greater than 7 (the smaller number) and......


3 times the smaller number (3)(7) is 21 added to 2 times the larger (2)(11) is 22, is 43.  

21 + 22 = 43.  Yay. 


I hope this helps you.