The answer is: (1, 2) .
________________________________________
Explanation:
______________________________________________
Given:
y = 2x ;
y = x + 1; Solve for "x and y" ; Write the solution as an "ordered pair".
___________________________________________________
Start with: y = x + 1 ;
Since: y = 2x ; Substitute "2x" for "y" ; in the equation: y = x + 1 ;
_________________________________________________________
→ 2x = x + 1 ; Subtract "1" ; and subtract "x" ; from each side ;
________________________________________________________
→ 2x − 1 − x = x + 1 − 1 - x ;
____________________________________________________
→ 1x − 1 = 0 ; ↔ x − 1 = 0 ; Add "1" to each side of the equation ;
______________________________________________________
→ x − 1 + 1 = 0 + 1 ;
______________________________________________________
to get: → x = 1 .
______________________________________________________
Now that we know that " x = 1 " ; we can solve for "y" ; using either (or both!) of 2 (TWO) methods. Let's us both methods — for demonstration purposes—and to confirm that "y" is the same value when "x = 1" (as extra assurance that "x = 0" makes sense!) ;
___________________________________________________
1) y = 2x ; so, when "x = 1"; what does "y" equal?
→ Plug in "1" for "x" ; and solve for "y" ;
______________________________________________
→ y = 2*(1) = 2 . So, x = 1; y = 2 .
______________________________________________
2) y = x + 1; so, when "x = 1"; what does "y" equal?
→ Plug in "1" for "x" ; and solve for "y" ;
______________________________________________
→ y = 1 + 1 = 2 . So, x = 1 ; y = 2 .
______________________________________________
→ We write this answer as an "ordered pair"; that is:
______________________________________________
The answer is: (1, 2) .
_______________________________________________
