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 <span>− 1 - x ; ____________________________________________________ </span>→ 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) . _______________________________________________