The numbers are: "11" and "21" .
________________________________________
Explanation:
________________________________________
x + y = 32 ; Note: y is greater than "x";
y = 2x − 1 ;
________________________________________
We have: " x + y = 32" ;
Plug in "(2x − 1)" for "y" ;
________________________________________
→ x + (2x − 1) = 32 ;
→ x + 1(2x − 1) = 32 ;
→ x + 2x − 1 = 32 ;
→ 1x + 2x − 1 = 32 ;
→ 3x − 1 = 32 ;
Add "1" to each side of the equation ;
→ 3x − 1 + 1 = 32 + 1 ;
→ 3x = 33 ;
Divide each side of the equation by "3" ;
to isolate "x" on one side of the equation; & to solve for "x" ;
3x / 3 = 33 / 3 ;
x = 11 ;
_______________________________________
Now, let us solve for "y" ;
_______________________________________
Method 1)
_______________________________________
We have: "x + y = 32" ;
We have "x = 11 " ;
Plug in "11" for "x" into the equation; to solve for "y" ;
_________________________
"x + y = 11 + y = 32 " ;
→ y + 11 = 32 ;
Subtract "11" from each side of the equation;
→ y + 11 − 11 = 32 − 11 ;
→ y = 21 ;
____________________________________
So; the numbers are: "11" and "21" .
____________________________________
Method 2) (and we can do this anyways to confirm our answer):
___________________________________________________
When we have: "x = 11 " ; Let us solve for "y" ;
___________________________________________________
We have the equation: "y = 2x − 1 " ;
Plug in the value, "11", for "x", in this equation; to solve for "y" ;
___________________________________________________
→ y = 2(11) − 1 ;
→ y = 22 − 1 ;
→ y = 21
____________________________________________
So, our answers are: "11" and "21" .
____________________________________________
Note: x + y = ? 32 ? ; → "11 + 21 = ? 32 ? " ; Yes!
____________________________________________
Note: "y" is the larger number; y > x ? ; → "21 > 11 " ? Yes!
____________________________________________
